Theorems
This section contains major proved results that cover several conjectures.
Theorem (1)
Let $p$ be an odd prime number and $b$ be a base such that, among the distinct prime factors $q_i^{e_i}$ of $b$, $p$ does not equal $q_i$ or divide $q_i - 1$ (for any $i$). If $p$ appears in the factorization of index 1 of the sequence starting with $b^m$, for a given $m$, then $p$ will appear as a factor in index 1 for all sequences starting with $b^{m+(p-1)k}$, for any positive $k$.
Final version by Alexander Jones. Original idea by MDaniello in post #2660 of the MersenneForum thread.Let $n = m + k(p - 1)$ be the exponent. We prove that $s(b^n) \equiv 0 \pmod{p}$.
By the hypothesis, we know that $s(b^m) = \sigma(b^m) - b^m = \frac{\displaystyle\prod_i (q_i^{e_i*m+1)} - 1)}{\displaystyle\prod_i (q_i - 1)} - b^m \equiv 0 \pmod{p}$. Let $x \equiv b^m \pmod{p}$.
By the definition of the aliquot sum, $\sigma(b^n) = \frac{\displaystyle\prod_i (q_i^{e_i*n+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$. From this, we get $\sigma(b^n) = \frac{\displaystyle\prod_i (q_i^{e_i*n+1} - 1)}{\displaystyle\prod_i (q_i - 1)} = \frac{\displaystyle\prod_i (q_i^{(e_i*k(p-1))+e_i*m+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$. By Fermat's little theorem, the $e_i*k(p-1)$ term cancels, so we get $\frac{\displaystyle\prod_i (q_i^{e_i*m+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$, which by the hypothesis is equivalent to $b^m \bmod p$.
Also applying Fermat's little theorem, $b^n = b^{m + k(p - 1)} \equiv b^m \pmod{p}$. As such, $\sigma(b^n) - b^n \equiv \sigma(b^m) - b^m \equiv 0 \pmod{p}$. This completes the proof.
Lemma (1)
The odd prime number $p$ appears in the factorization of the terms of index 1 of sequences for exponent $p - 1$ and base $b$ described in Theorem (1).
By Alexander Jones.Let $n = p - 1$ be the exponent. We prove that $s(b^n) \equiv 0 \pmod{p}$.
By definition of the aliquot sum, $s(b^n) = \sigma(b^n) - b^n$. By the definition of the sum of divisors function and the formula for sums of geometric series, we have $\sigma(b^n) = \displaystyle\prod_i \frac{q_i^{e_i*n+1} - 1}{q_i - 1}$. Multiplying the numerator and denominator through, we get $\frac{\displaystyle\prod_i (q_i^{e_i*n+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$.
Taking the numerator first, we factor a $q_i$ out of $q_i^{e_i*n+1}$ to get $\displaystyle\prod_i (q_i \cdot q_i^{e_i*n} - 1)$. Since we defined $n = p - 1$, by Fermat's little theorem, $q_i^{e_i*n} \equiv 1 \pmod{p}$, so this reduces to $\displaystyle\prod_i (q_i - 1)$. Note that this is the same as the denominator. Therefore, they cancel out, and $\sigma(b^n) \equiv 1 \pmod{p}$.
To determine $b^n \bmod{p}$, we again note that $n$ is divisible by ($p - 1$). Again applying Fermat's little theorem, we end up with $b^n \equiv 1 \pmod{p}$. As a result, $s(b^n) = \sigma(b^n) - b^n \equiv 1 - 1 = 0 \pmod{p}$. This completes the proof.
Corollary (1)
The odd prime number $p$ appears in the factorization of the terms of index 1 of sequences for exponent $k(p - 1)$, for all positive integers $k$, and base $b$ described in Theorem (1)
By Alexander Jones.Proof by induction. The base case is Lemma (1), and the inductive case is Theorem (1), with $m = p - 1$.
These theorems are not generally true if $p$ is equal to or a factor of $q_i$ or $q_i - 1$ (for any $i$), as the form of Fermat's little theorem used in the proof does not hold in that case. Furthermore, if $p$ is a factor of $q_i - 1$, the formula $\frac{\displaystyle\prod_i (q_i^{e_i*n+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$ leads to a division by 0 in $\mathbb{Z}_p$, which is also illegal.
Theorem (2)
A prime $p$ never appears in the first term of an aliquot sequence starting at $p^n$, for any exponent $n$.
By Alexander Jones.$s(p^n) = \frac{p^n - 1}{p - 1} \equiv \frac{-1}{-1} = 1 \pmod{p}$. Therefore, $s(p^n)$ is never divisible by $p$.
Conjectures
Conjectures (1) to (133) published on August 19, 2020, on the Mersenne forum, see post #447
In all the statements below, $k$ is an integer and $M_n$ is the $n\text{-th}$ Mersenne number.
Note: Several of these conjectures motivated my request to Edwin Hall to push the calculations further for some exponents $i = 36k$, $i = 60k$, $i = 70k$, $i = 72k$, $i = 90k$.
Conjecture (1)*
The prime number 3 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{2k}$.
$s(2^{2k}) = 2^{2k} - 1 = M_{2k}$, which is divisible by $M_{2} = 3$.
Alternatively, this is also a result of Corollary (1).
Conjecture (2)**
The prime number 3 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{4k}$.
$s(2^{4k}) = 2^{4k} - 1 = (2^{2k} - 1)(2^{2k} + 1)$.
$2^{2k} + 1 \equiv 2 \pmod{3}$, so there exist a prime $p \equiv 2 \pmod{3}$ and some $m$ such that $p^{2m-1} \mid 2^{2k} + 1$, but $p^{2m} \nmid 2^{2k} + 1$.
$\gcd(2^{2k} + 1, 2^{2k} - 1) = gcd(2, 2^{2k} - 1) = 1$, so $p \nmid 2^{2k} - 1$.
Hence $p^{2m-1}$ preserved 3 for $s(2^{4k})$, so $3 \mid s(s(2^{4k}))$.See successively on the Mersenne forum, posts #476, #480 and #481
Conjecture (3)
The prime number 3 appears in the factorization of the terms of indexes 1 through 7 of all sequences that begin with the integers $2^{36k}$.
The case of index 1 follows directly from Conjecture (1).
Conjecture (4)
The prime number 3 appears in the factorization of the terms of indexes 1 through 18 of all sequences that begin with the integers $2^{126k}$.
The case of index 1 follows directly from Conjecture (1).
Conjecture (5)*
The prime number 5 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{4k}$.
$s(2^{4k}) = 2^{4k} - 1 = M_{4k}$, which is divisible by $M_{4} = 15 = 3 \cdot 5$.
Alternatively, this is also a result of Corollary (1).
Conjecture (6)**
The prime number 5 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{28k}$, $2^{44k}$, $2^{76k}$, $2^{92k}$, $2^{116k}$.
The case of index 1 follows directly from Conjecture (5).
Conjecture (7)
The prime number 5 appears in the factorization of the terms of indexes 1, 2, 3, 4 of all sequences that begin with the integers $2^{36k}$.
The case of index 1 follows directly from Conjecture (5).
Conjecture (8)
The prime number 5 appears in the factorization of the terms of indexes 1, 2, 3, 4 of all sequences that begin with the integers $2^{132k}$.
The case of index 1 follows directly from Conjecture (5).
Conjecture (9)*
The prime number 7 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{3k}$.
$s(2^{3k}) = 2^{3k} - 1 = M_{3k}$, which is divisible by $M_{3} = 7$.
Conjecture (10)**
The prime number 7 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{12k}$.
The case of index 1 follows directly from Conjecture (9).
Conjecture (11)
The prime number 7 appears in the factorization of the terms of indexes 1, 2, 3, 4 of all sequences that begin with the integers $2^{60k}$.
The case of index 1 follows directly from Conjecture (9).
Conjecture (12)*
The prime number 11 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{10k}$.
$s(2^{10k}) = 2^{10k} - 1 = M_{10k}$, which is divisible by $M_{10} = 3 \cdot 11 \cdot 31$.
Alternatively, this is also a result of Corollary (1).
Conjecture (13)**
The prime number 11 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{120k}$, $2^{130k}$.
The case of index 1 follows directly from Conjecture (12).
Conjecture (14)
The prime number 11 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $2^{70k}$.
The case of index 1 follows directly from Conjecture (12).
Conjecture (15)
The prime number 13 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{12k}$.
$s(2^{12k}) = 2^{12k} - 1 = M_{12k}$, which is divisible by $M_{12} = 3^2 \cdot 5 \cdot 7 \cdot 13$.
Alternatively, this is also a result of Corollary (1).
Conjecture (16)**
The prime number 13 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{60k}$.
The case of index 1 follows directly from Conjecture (15).
Conjecture (17)*
The prime number 17 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $2^{8k}$.
$s(2^{8k}) = 2^{8k} - 1 = M_{8k}$, which is divisible by $M_{8} = 3 \cdot 5 \cdot 17$.
Conjecture (18)**
The prime number 17 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $2^{144k}$.
The case of index 1 follows directly from Conjecture (17).
Conjecture (19)
The prime number 19 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $2^{72k}$.
Conjecture (20)
The prime number 31 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $2^{90k}$.
Conjecture invalidated by Edwin Hall's calculations.
Conjecture (21)**
The prime number 79 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $2^{156k}$.
The case of index 1 is an immediate result of Corollary (1) ($79 \mid 2^{78k} \mid 2^{156k}$ for all positive $k$).
Conjecture (22)**
The prime number 2089 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $2^{87k}$.
Conjecture (23)**
The prime number 4051 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $2^{100k}$.
Conjecture (24)
The prime number 15121 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $2^{540k}$.
Conjecture (25)*
The prime number 5 appears in the factorization of the terms of index 1 of all sequences starting with the integers $3^{4k}$.
$s(3^{4k}) = \frac{3^{4k} - 1}{2}$, which is divisible by $\frac{3^4 - 1}{2} = 2^3 \cdot 5$.
Alternatively, this is also a result of Corollary (1).
Conjecture (26)**
The prime number 5 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $3^{4+8k}$.
The case of index 1 follows directly from Conjecture (25).
Conjecture (27)*
The prime number 7 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $3^{6k}$.
$s(3^{6k}) = \frac{3^{6k} - 1}{2}$, which is divisible by $\frac{3^6 - 1}{2} = 2^2 \cdot 7 \cdot 13$.
Alternatively, this is also a result of Corollary (1).
Conjecture (28)*
The prime number 7 appears in the factorization of the terms of many consecutive indexes of all sequences that begin with the integers $3^{6+12k}$.
This conjecture is poorly stated, and it is thus marked as "disproven" on those grounds. Below is an (unreviewed) proof for indices 1 and 2.
The case of index 1 follows directly from Conjecture (27), so we proceed to index 2.
An integer has the driver $2^{2} \cdot 7$ as a factor when it is divisible by $2^{2} \cdot 7 = 28$, but not by $2^{3} \cdot 7 = 56$.
However, $s(3^{6+12k}) = \frac{(3^{6+12k} - 1)}{2}$ is divisible by $28$ because $3^{6+12k} - 1 = (3^{6})^{1+2k} - 1^{1+2k}$ is divisible by $3^{6}-1$, which is divisible by $56$ because $3^{6} - 1 = 728 = 56 \cdot 13$.
One $2$ is lost when dividing by $2$, so we can say with certainty that our number is divisible by $\frac{56}{2} = 28$.
Note that $3^{6+12k} = 3^{6} \cdot 3^{12k} = 729 \cdot (3^{12})^{k} \equiv 57 \cdot 1^{k} = 57 \pmod{112}$, implying that $3^{6+12k} - 1$ is never divisble by 112. Hence, $\frac{3^{6+12k} - 1}{2}$ is not divisible by $\frac{112}{2} = 56$, which completes the proof.
Here is the observation that led to this conjecture:
Code:prime 7 in sequence 3^6 at index i for i from 1 to 5 prime 7 in sequence 3^18 at index i for i from 1 to 50 prime 7 in sequence 3^30 at index i for i from 1 to 25 prime 7 in sequence 3^42 at index i for i from 1 to 86 prime 7 in sequence 3^54 at index i for i from 1 to 179 prime 7 in sequence 3^66 at index i for i from 1 to 39 prime 7 in sequence 3^78 at index i for i from 1 to 124 prime 7 in sequence 3^90 at index i for i from 1 to 171 prime 7 in sequence 3^102 at index i for i from 1 to 72 prime 7 in sequence 3^114 at index i for i from 1 to 45 prime 7 in sequence 3^126 at index i for i from 1 to 60 prime 7 in sequence 3^138 at index i for i from 1 to 230 prime 7 in sequence 3^150 at index i for i from 1 to 148 prime 7 in sequence 3^162 at index i for i from 1 to 228 prime 7 in sequence 3^174 at index i for i from 1 to 219 prime 7 in sequence 3^186 at index i for i from 1 to 9 prime 7 in sequence 3^198 at index i for i from 1 to 105 prime 7 in sequence 3^210 at index i for i from 1 to 194 prime 7 in sequence 3^222 at index i for i from 1 to 98 prime 7 in sequence 3^234 at index i for i from 1 to 87 prime 7 in sequence 3^246 at index i for i from 1 to 38We have just demonstrated that sequences beginning with $s(3^{6+12k})$ has the driver $2^2 \cdot 7$ as a factor.
It is therefore normal that this driver persists in so many consecutive terms.
Conjecture (29)*
The prime number 11 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $3^{5k}$.
$s(3^{5k}) = \frac{3^{5k} - 1}{2}$, which is divisible by $\frac{3^5 - 1}{2} = 11^2$.
Conjecture (30)*
The prime number 13 appears in the factorization of index 1 terms in all sequences that begin with the integers $3^{3k}$.
$s(3^{3k}) = \frac{3^{3k} - 1}{2}$, which is divisible by $\frac{3^3 - 1}{2} = 13$.
Conjecture (31)**
The prime number 13 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $3^{51k}$.
The case of index 1 follows directly from Conjecture (30).
Conjecture (32)*
The prime number 17 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $3^{16k}$.
$s(3^{16k}) = \frac{3^{16k} - 1}{2}$, which is divisible by $\frac{3^{16} - 1}{2} = 2^5 \cdot 5 \cdot 17 \cdot 41 \cdot 193$.
Alternatively, this is also a result of Corollary (1).
Conjecture (33)**
The prime number 17 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $3^{48k}$.
The case of index 1 follows directly from Conjecture (32).
Conjecture (34) (* if only index 1)
The prime number 19 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $3^{18k}$.
Conjecture invalidated by warachwe on August 9, 2021. For the sequence $3^{18*37}$, the factor 19 is not maintained at the second iteration.
Conjecture (35) (* if only index 1 and 2)
The prime number 19 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $3^{36k}$.
Conjecture invalidated by warachwe on August 9, 2021. For the sequence $3^{36*37}$, the factor 19 is not maintained at the second iteration.
Conjecture (36)*
The prime number 23 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $3^{11k}$.
$s(3^{11k}) = \frac{3^{11k} - 1}{2}$, which is divisible by $\frac{3^{11} - 1}{2} = 23 \cdot 3851$.
Conjecture (37)**
The prime number 31 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $3^{30k}$.
The case of index 1 follows from Corollary (1).
Conjecture (38)*
The prime number 37 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $3^{18k}$.
$s(3^{18k}) = \frac{3^{18k} - 1}{2}$, which is divisible by $\frac{3^{18} - 1}{2} = 2^2 \cdot 7 \cdot 13 \cdot 19 \cdot 37 \cdot 757$.
Conjecture (39)**
The prime number 37 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $3^{36k}$.
The case of index 1 follows from both Corollary (1) and Conjecture (38).
Conjecture (40)**
The prime number 79 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $3^{78k}$.
The case of index 1 follows from Corollary (1).
Conjecture (41)**
The prime number 547 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $3^{14k}$.
Conjecture (42)**
The prime number 398581 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $3^{26k}$.
Conjecture (43)
The prime number 3 appears in the factorization of the terms of index 1 of all sequences starting with the integers $5^{2k}$.
$s(5^{2k}) = \frac{5^{2k} - 1}{4}$, which is divisible by $\frac{5^2 - 1}{4} = 2 \cdot 3$.
Alternatively, this is also a result of Corollary (1).
Conjecture (44)
The prime number 3 appears in the factorization of many consecutive indexes of all sequences that begin with the integers $5^{2+4k}$.
For example, 3 appears in the factorization of the terms in indexes 1 through 786 of the sequence that begins with $5^{58}$.
Conjecture (45)*
The prime number 5 never appears in the factorization of the terms at index 1 of all sequences beginning with the integers $5^k$.
This is an immediate result of Theorem (2).
Conjecture (46)*
The prime number 7 appears in the factorization of terms at index 1 of all sequences that begin with the integers $5^{6k}$.
$s(5^{6k}) = \frac{5^{6k} - 1}{4}$, which is divisible by $\frac{5^6 - 1}{4} = 2 \cdot 3^2 \cdot 7 \cdot 31$.
Alternatively, this is also a result of Corollary (1).
Conjecture (47)**
The prime number 7 appears in the factorization of index 1 and index 2 terms of all sequences that begin with the integers $5^{12k}$.
The case of index 1 follows from Conjecture (46).
Conjecture (48)*
The prime number 11 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $5^{5k}$.
$s(5^{5k}) = \frac{5^{5k} - 1}{4}$, which is divisible by $\frac{5^5 - 1}{4} = 11 \cdot 71$.
Conjecture (49)**
The prime number 11 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $5^{35k}$.
The case of index 1 follows from Conjecture (48).
Conjecture (50)**
The prime number 11 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $5^{40k}$.
The case of index 1 follows from Conjecture (48).
Conjecture (51)**
The prime number 11 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $5^{65k}$.
The case of index 1 follows from Conjecture (48).
Conjecture (52)*
The prime number 13 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $5^{4k}$.
$s(5^{4k}) = \frac{5^{4k} - 1}{4}$, which is divisible by $\frac{5^4 - 1}{4} = 2^2 \cdot 3 \cdot 13$.
Conjecture (53)*
The prime number 17 appears in the factorization of index 1 terms in all sequences that begin with the integers $5^{16k}$.
$s(5^{16k}) = \frac{5^{16k} - 1}{4}$, which is divisible by $\frac{5^{16} - 1}{4} = 2^4 \cdot 3 \cdot 13 \cdot 17 \cdot 313 \cdot 11489$.
Alternatively, this is also a result of Corollary (1).
Conjecture (54)*
The prime number 19 appears in the factorization of index 1 terms in all sequences that begin with the integers $5^{9k}$.
$s(5^{9k}) = \frac{5^{9k} - 1}{4}$, which is divisible by $\frac{5^9 - 1}{4} = 19 \cdot 31 \cdot 829$.
Conjecture (55)**
The prime number 19 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $5^{18k}$.
The case of index 1 follows from Corollary (1) and Conjecture (54).
Conjecture (56)*
The prime number 31 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $5^{3k}$.
$s(5^{3k}) = \frac{5^{3k} - 1}{4}$, which is divisible by $\frac{5^3 - 1}{4} = 31$.
Conjecture (57)**
The prime number 31 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $5^{30k}$.
The case of index 1 follows from Corollary (1) and Conjecture (56).
Conjecture (58)
The prime number 31 appears in the factorization of the terms of many consecutive indexes of all sequences that begin with the integers $5^{48+96k}$.
Here is the observation that led to this conjecture:
Code:prime 31 in sequence 5^48 at index i for i from 1 to 447 prime 31 in sequence 5^144 at index i for i from 1 to 32The same remark can be made here as for the Conjecture (28).
And it should be shown here that $s(5^{48+96k})$ has the driver $2^4 \cdot 31$ as a factor.
Conjecture (59)*
The prime number 71 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $5^{5k}$.
$s(5^{5k}) = \frac{5^{5k} - 1}{4}$, which is divisible by $\frac{5^5 - 1}{4} = 11 \cdot 71$.
Conjecture (60)**
The prime number 71 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $5^{45k}$.
The case of index 1 follows from Conjecture (59).
Conjecture (61)*
The prime number 521 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $5^{10k}$.
$s(5^{10k}) = \frac{5^{10k} - 1}{4}$, which is divisible by $\frac{5^{10} - 1}{4} = 2 \cdot 3 \cdot 11 \cdot 71 \cdot 521$.
Conjecture (62)**
The prime number 521 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $5^{50k}$.
The case of index 1 follows from Conjecture (61).
Conjecture (63)
The prime number 3 appears in the factorization of the terms of index 1 of all sequences that start with the integers $6^{1+2k}$.
$\sigma(6^{2k+1}) = \frac{(2^{2k+2} - 1)(3^{2k+2} - 1)}{2} = \frac{(4^{k+1} - 1)(9^{k+1} - 1)}{2}$. $(4^{k+1} - 1) \equiv 0 \pmod{3}$ and $(9^{k+1} - 1) \equiv 0 \pmod{8}$, so $(4^{k+1} - 1)(9^{k+1} - 1)$ is divisible by 24, and thus $\sigma(6^{2k+1})$ is divisible by 12. Therefore, $6 = 2*3 \mid \sigma(6^{2k+1}) - 6^{2k+1}$.
Conjecture (64)*
The prime number 5 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $6^{2k}$.
$\sigma(6^{2k}) = \frac{2^{2k+1} - 1}{2 - 1}\frac{3^{2k+1} - 1}{3 - 1} = \frac{(2^{2k+1} - 1)(3^{2k+1} - 1)}{2}$. Modulo 5, we have $\frac{(2^{2k+1} - 1)(3^{2k+1} - 1)}{2} \equiv \frac{(2^{2k+1} - 1)((-2)^{2k+1} - 1)}{2} \equiv \frac{(2^{2k+1} - 1)(-2^{2k+1} - 1)}{2} \equiv \frac{(2^{2k+1})(-2^{2k+1}) + 1}{2} = \frac{2}{2} = 1 \pmod{5}$. Since $6^{2k} \equiv 1 \pmod{5}$, we have $\sigma(6^{2k}) - 6^{2k} \equiv 1 - 1 = 0 \pmod{5}$.
Conjecture (65)*
The prime number 7 appears in the factorization of index 1 terms of all sequences that begin with the integers $6^{6k}$.
$\sigma(6^{6k}) = \frac{2^{6k+1} - 1}{2 - 1}\frac{3^{6k+1} - 1}{3 - 1} = \frac{(2^{6k+1} - 1)(3^{6k+1} - 1)}{2}$. Modulo 7, we have $\frac{(2^{6k+1} - 1)(3^{6k+1} - 1)}{2} \equiv \frac{(2 \cdot 3)^{6k+1} - 2^{6k+1} - 3^{6k+1} + 1}{2} \equiv \frac{6 - 2 - 3 + 1}{2} \equiv \frac{2}{2} \equiv 1 \pmod{7}$. Since $6^{6k} \equiv 1 \pmod{7}$, we have $\sigma(6^{6k}) - 6^{6k} \equiv 1 - 1 = 0 \pmod{7}$.
Alternatively, this is also a result of Corollary (1).
Conjecture (66)*
The prime number 11 appears in the factorization of index 1 terms of all sequences that begin with the integers $6^{10k}$ and $6^{2+10k}$.
This follows from Theorem (1) ($11 \mid s(6^2)$) and Corollary (1).
Conjecture (67)*
The prime number 13 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $6^{12k}$.
This is an immediate result of Corollary (1).
Conjecture (68)*
The prime number 19 appears in the factorization of index 1 terms of all sequences that begin with the integers $6^{18k}$ and $6^{10+18k}$.
This follows from Theorem (1) ($19 \mid s(6^{10})$) and Corollary (1).
Conjecture (69)*
The prime number 23 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $6^{11k}$.
The statement is true, and can be proven using the properties of the Legendre's symbol $(\frac{a}{p})$.
$(\frac{a}{p})\equiv a^{\frac{p-1}{2}} \pmod {p}$, and the Legendre symbol is multiplicative so $(\frac{6}{p}) = (\frac{2}{p})(\frac{3}{p})$.
Knowing that $(\frac{2}{p})$ is $1$ if $p\equiv\pm 1\pmod{8}$ and $(\frac{3}{p})$ is $1$ if $p \equiv \pm 1 \pmod{12}$, we get that $2^{\frac{p-1}{2}} \equiv 3^{\frac{p-1}{2}} \equiv 6^{\frac{p-1}{2}} \equiv 1 \pmod{p}$ for all $p \equiv \pm 1 \pmod{24}$.So $2^{\frac{p-1}{2}} \equiv 3^{\frac{p-1}{2}} \equiv 6^{\frac{p-1}{2}} \equiv 1 \pmod{p}$ for all $p \equiv \pm 1 \pmod{24}$. So $2^{11} \equiv 3^{11} \equiv 6^{11} \equiv 1 \pmod{23}$, and $s(6^{11k}) \equiv (2^{11k+1}-1)(\frac{3^{11k+1}-1}{2}) - 6^{11k} \equiv (2\cdot 1^k-1)(\frac{3\cdot 1^k-1}{2}) - 1^k \equiv 0 \pmod{23}$
Conjecture (70)*
The prime number 29 appears in the factorization of index 1 terms in all sequences that begin with the integers $6^{28k}$.
This is an immediate result of Corollary (1).
Conjecture (71)*
The prime number 31 appears in the factorization of index 1 terms of all sequences that begin with the integers $6^{30k}$, $6^{11+30k}$ and $6^{17+30k}$.
This follows from Theorem (1) ($31 \mid s(6^{11})$, $31 \mid s(6^{17})$) and Corollary (1).
Conjecture (72)**
The prime number 37 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $6^{36k}$
The case of index 1 follows from Corollary (1).
Conjecture (73)*
The prime number 37 appears in the factorization of the terms of index 1 of all sequences starting with the integers $6^{14+36k}$.
This follows from Theorem (1) ($37 \mid s(6^{14})$) and Corollary (1).
Conjecture (74)*
The prime number 59 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $6^{58k}$, $6^{8+58k}$, $6^{35+58k}$ and $6^{53+58k}$.
This follows from Theorem (1) ($59 \mid s(6^8)$, $59 \mid s(6^{35})$, and $59 \mid s(6^{53})$) and Corollary (1).
Conjecture (75)*
The prime number 61 appears in the factorization of the terms of index 1 of all sequences beginning with the integers $6^{60k}$, $6^{44+60k}$ and $6^{55+60k}$.
This follows from Theorem (1) ($61 \mid s(6^{44})$ and $61 \mid s(6^{55})$) and Corollary (1).
Conjecture (76)*
The prime number 71 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $6^{70k}$, $6^{11+70k}$, $6^{32+70k}$, $6^{35+70k}$, $6^{46+70k}$ and $6^{67+70k}$.
This follows from Theorem (1) ($71 \mid s(6^{11})$, $71 \mid s(6^{32})$, $71 \mid s(6^{35})$, $71 \mid s(6^{46})$, and $71 \mid s(6^{67})$) and Corollary (1).
Conjecture (77)*
The prime number 601 appears in the factorization of the terms of index 1 of all sequences beginning with the integers $6^{75k}$.
The statement is true, and it's a consequence of the fact that $2^{75} \equiv 3^{75} \equiv 1 \pmod{601}$.
This is because $s(6^{75k}) \equiv (2^{75k+1}-1)(\frac{3^{75k+1}-1}{2}) - 6^{75k} \equiv (2\cdot (2^{75})^k-1)(\frac{3\cdot (3^{75})^k-1}{2}) - (2^{75}\cdot 3^{75})^k \equiv ( 2\cdot 1^k-1)(\frac{3\cdot 1^k-1}{2}) - 1^k \equiv 0 \pmod{601}$.
Conjecture (78)
The prime number 3 appears in the factorization of the terms of index 1 of all sequences starting with the integers $7^{3k}$.
$s(7^{3k}) = \frac{7^{3k} - 1}{6}$, which is divisible by $\frac{7^3 - 1}{6} = 3 \cdot 19$.
Conjecture (79)
The prime number 3 appears in the factorization of the terms from index 1 to 10 for all sequences that begin with the integers $7^{6+12k}$ and $7^{21k}$.
The case of index 1 for both forms follows from Conjecture (78).
Conjecture (80)*
The prime number 5 appears in the factorization of the terms of index 1 for all sequences that begin with the integers $7^{4k}$.
$s(7^{4k}) = \frac{7^{4k} - 1}{6}$, which is divisible by $\frac{7^4 - 1}{6} = 2^4 \cdot 5^2$.
Alternatively, this is also a result of Corollary (1).
Conjecture (81)*
The prime number 7 appears in the factorization of index 1 terms in all sequences that begin with the integers $7^{3k}$.
Conjecture invalidated by Garambois on August 8, 2021. Conjecture in contradiction with Conjecture (82). It was probably an error of inattention!
Conjecture (82)*
The prime number 7 never appears in index 1 of all sequences that begin with the integers $7^k$.
This is an immediate result of Theorem (2).
Conjecture (83)*
The prime number 11 appears in the factorization of the terms in index 1 of all sequences that begin with the integers $7^{10k}$.
$s(7^{10k}) = \frac{7^{10k} - 1}{6}$, which is divisible by $\frac{7^{10} - 1}{6} = 2^3 \cdot 11 \cdot 191 \cdot 2801$.
Alternatively, this is also a result of Corollary (1).
Conjecture (84)**
The prime number 13 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $7^{12k}$.
The case of index 1 follows from Corollary (1).
Conjecture (85)
The prime number 13 appears in the factorization of the terms of many indexes of all sequences that begin with the integers $7^{72k}$.
27 consecutive indexes for $7^{72}$ and 9 consecutive indexes for $7^{144}$.
The case of index 1 follows from Conjecture (84).
Conjecture (86)*
The prime number 17 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $7^{16k}$.
$s(7^{16k}) = \frac{7^{16k} - 1}{6}$, which is divisible by $\frac{7^{16} - 1}{6}$ = $2^6 \cdot 5^2 \cdot 17 \cdot 1201 \cdot 169553$.
Alternatively, this is also a result of Corollary (1).
Conjecture (87)**
The prime number 17 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $7^{32k}$.
The case of index 1 follows from Conjecture (86).
Conjecture (88)*
The prime number 19 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $7^{3k}$.
$s(7^{3k}) = \frac{7^{3k} - 1}{6}$, which is divisible by $\frac{7^3 - 1}{6} = 3 \cdot 19$.
Conjecture (89)**
The prime number 19 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $7^{9k}$.
The case of index 1 follows from Conjecture (88).
Conjecture (90)*
The prime number 31 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $7^{15k}$.
$s(7^{15k}) = \frac{7^{15k} - 1}{6}$, which is divisible by $\frac{7^{15} - 1}{6} = 3 \cdot 19 \cdot 31 \cdot 2801 \cdot 159871$.
Conjecture (91)**
The prime number 67 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $7^{66k}$.
The case of index 1 follows from Corollary (1).
Conjecture (92)*
The prime number 419 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $7^{19k}$.
$s(7^{19k}) = \frac{7^{19k} - 1}{6}$, which is divisible by $\frac{7^{19} - 1}{6} = 419 \cdot 4534166740403$.
Conjecture (93)**
The prime number 419 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $7^{38k}$.
Final version by Alexander Jones. Original proof by fish224 in post #2674 of the MersenneForum thread.The case of index 1 follows from Conjecture (92), and it is true for all positive $k$.
To analyze the case of index 2, we start by noting that $s(7^{38k}) = \frac{7^{38k} - 1}{6}$, and that $419 \mid (7^{38} - 1) \mid (7^{38k} - 1)$, so $419 \mid s(7^{38k})$. We note also that $351121 \mid (7^{38} - 1)$ and $419 \mid (351121 + 1)$.
So 419 always appears in the factorization of the second term of aliquot sequence starting with $7^{38k}$, except when $351121^{2} \mid (7^{38k} - 1)$. The smallest exponent $i$ at which $7^{i} \equiv 1 \pmod{351121^2}$ is $i = \varphi(351121^2)/8 = 15410700690$. Therefore, this conjecture does not hold for index 2 when $i \mid 38k$, unless $s(7^{38k})$ is also factored by another, odd, power of 351121. (This is beyond the scope of this proof.)
Conjecture (94)
The prime number 3 appears in the factorization of the terms of index 1 of all sequences starting with the integers $10^{2k}$.
This is an immediate result of Corollary (1).
Conjecture (95)*
The prime number 5 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $10^{3+4k}$.
This follows from Theorem (1) ($5 \mid s(10^3)$).
Conjecture (96)*
The prime number 13 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $10^{12k}$.
This is an immediate result of Corollary (1).
Conjecture (97)**
The prime number 13 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $10^{2+12k}$.
Conjecture (98)**
The prime number 19 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $10^{18k}$.
The case of index 1 follows from Corollary (1).
Conjecture (99)*
The prime number 61 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $10^{52+60k}$ and $10^{54+60k}$.
This follows from Theorem (1) ($61 \mid s(10^{52})$ and $61 \mid s(10^{54})$).
Conjecture (100)**
The prime number 61 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $10^{60k}$.
The case of index 1 follows from Theorem (1).
Conjecture (101)*
The prime number 7 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $11^{3k}$.
$s(11^{3k}) = \frac{11^{3k} - 1}{10}$, which is divisible by $\frac{11^3 - 1}{10} = 7 \cdot 19$.
Conjecture (102)**
The prime number 7 appears in the factorization of the terms of indexes 1, 2, 3 of all sequences that begin with the integers $11^{6k}$.
The case of index 1 follows from Theorem (1) and Conjecture (101).
Conjecture (103)*
The prime number 11 never appears in the factorization of the terms of index 1 of all sequences that begin with the integers 11k.
This is an immediate result of Theorem (2).
Conjecture (104)**
The prime number 13 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $11^{12k}$.
The case of index 1 follows from Corollary (1).
Conjecture (105)*
The prime number 19 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $11^{3k}$.
$s(11^{3k}) = \frac{11^{3k} - 1}{10}$, which is divisible by $\frac{11^3 - 1}{10} = 7 \cdot 19$.
Conjecture (106)**
The prime number 19 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $11^{6k}$.
Conjecture invalidated by warachwe on August 9, 2021. For the sequence $11^{6*37}$, the factor 19 is not maintained at the second iteration.
Conjecture (107)**
The product of prime $19 \cdot 79 \cdot 547$ appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $11^{39k}$.
REMARKABLE, checked up to $k = 12$. See prime 79 bases 2 and 3.
Conjecture (108)
The prime number 3 never appears in the factorization of the terms of index 1 of all sequences that start with the integers $12^k$.
The statement is equivalent to "$s(12^{k})$ is never congruent to $0 \pmod{3}$", so we'll prove that instead.
$s(12^{k}) = ( 2^{2k+1}-1)(\frac{3^{k+1}-1}{2})-12^{k} = ( 2(2^{2})^k-1)(\frac{3^{k+1}-1}{2})-12^{k}$, so by reducing it modulo $3$ we get $s(12^{k}) \equiv ( 2-1)(\frac{-1}{2}) \equiv 1 \pmod{3}.$
Conjecture (109)*
The prime number 17 appears in the factorization of index 1 terms in all sequences that begin with the integers $12^{16k}$ and $12^{6+16k}$.
This follows from Theorem (1) ($17 \mid s(12^6)$) and Corollary (1).
It is difficult to notice for base 12, other behaviors different from the bases already presented so far.
Conjecture (110)*
The prime number 3 appears in the factorization of the terms of index 1 of all sequences that start with the integers $13^{3k}$.
$s(13^{3k}) = \frac{13^{3k} - 1}{12}$, which is divisible by $\frac{13^3 - 1}{12} = 3 \cdot 61$.
Conjecture (111)
The prime number 3 appears in the factorization of the terms of indexes 1 through 6 of all sequences that begin with the integers $13^{6k}$.
The case of index 1 follows directly from Conjecture (110).
Conjecture (112)*
The prime number 5 appears in the factorization of the terms in index 1 of all sequences that begin with the integers $13^{4k}$.
$s(13^{4k}) = \frac{13^{4k} - 1}{12}$, which is divisible by $\frac{13^4 - 1}{12} = 2^2 \cdot 5 \cdot 7 \cdot 17$.
Alternatively, this is also a result of Corollary (1).
Conjecture (113)
The prime number 5 appears in the factorization of the terms of many consecutive indexes of all sequences that begin with the integers $13^{8+16k}$.
Conjecture (114)*
The prime number 7 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $13^{2k}$.
$s(13^{2k}) = \frac{13^{2k} - 1}{12}$, which is divisible by $\frac{13^2 - 1}{12} = 2 \cdot 7$.
Conjecture (115)
The prime number 7 appears in the factorization of the terms of many consecutive indexes of all sequences that begin with the integers $13^{4+8k}$.
Conjecture (116)*
The prime number 13 never appears in the factorization of the terms of index 1 of all sequences that begin with the integers 13k.
This is an immediate result of Theorem (2).
Conjecture (117)*
The prime number 19 appears in the factorization of index 1 terms in all sequences that begin with the integers $13^{18k}$.
$s(13^{18k}) = \frac{13^{18k} - 1}{12}$, which is divisible by $\frac{13^{18} - 1}{12} = 2 \cdot 3^2 \cdot 7 \cdot 19 \cdot 61 \cdot 157 \cdot 271 \cdot 937 \cdot 1609669$.
Alternatively, this is also a result of Corollary (1).
Conjecture (118)**
The prime number 19 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $13^{36k}$.
The case of index 1 follows from Conjecture (117).
Conjecture (119)*
The prime number 29 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $13^{14k}$.
$s(13^{14k}) = \frac{13^{14k} - 1}{12}$, which is divisible by $\frac{13^{14} - 1}{12} = 2 \cdot 7^2 \cdot 29 \cdot 22079 \cdot 5229043$.
Conjecture (120)**
The prime number 29 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $13^{42k}$.
The case of index 1 follows from Conjecture (119).
Conjecture (121)*
The prime number 61 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $13^{3k}$.
$s(13^{3k}) = \frac{13^{3k} - 1}{12}$, which is divisible by $\frac{13^3 - 1}{12} = 3 \cdot 61$.
Conjecture (122)**
The prime number 61 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $13^{21k}$.
The case of index 1 follows from Conjecture (121).
Conjecture (123)**
The prime number 3 appears in the factorization of the terms of indexes 1 to 4 of all sequences starting with the integers $14^{6k}$.
The case of index 1 follows from Corollary (1) ($3 \mid 14^{2k} \mid 14^{6k}$, for all positive $k$).
Conjecture (124)*
The prime number 5 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $14^{4k}$.
This is an immediate result of Corollary (1).
Conjecture (125)**
The prime number 5 appears in the factorization of the terms of indexes 1 to 4 of all sequences that begin with the integers $14^{1+4k}$.
It is difficult to notice for base 14, other behaviors different from the bases already presented so far...
Conjecture (126)*
The prime number 7 appears in the factorization of the terms of index 1 of all sequences which start with the integers $15^{2k}$, except for the $15^{8+12k}$.
7 does not appear in the factorization of the terms of index 1 of sequences starting with integers $15^{2+12k}$. See Conjecture (126a) for a corrected conjecture (which has been proven).
Conjecture (126a)
The prime number 7 appears in the factorization of the terms of index 1 of all sequences which start with the integers $15^{2k}$, except when $k \equiv 1 \pmod{3}$.
This is a correction of Conjecture (126) formulated by MDaniello in post #2652 of the MersenneForum thread.All of the following calculations are modulo 7.
$s(15^{2k}) \equiv \left(\frac{3^{2k+1}-1}{2}\right)\left(\frac{5^{2k+1}-1}{4}\right) - 15^{2k} \equiv \left(\frac{3^{2k+1}-1}{2}\right)\left(\frac{5^{2k+1}-1}{4}\right) - 1$.
If we represent $k \bmod 3$ as $k = r + 3q \pmod{7}$ for some $q$, $r$, we have $\left(\frac{3^{2k+1}-1}{2}\right)\left(\frac{5^{2k+1}-1}{4}\right) - 1 \equiv \left(\frac{3^{2r+6q+1}-1}{2}\right)\left(\frac{5^{2r+6q+1}-1}{4}\right) - 1 \equiv \left(\frac{3^{6q} * 3^{2r+1} - 1}{2}\right)\left(\frac{5^{6q} * 5^{2r+1} - 1}{4}\right) - 1$. By Fermat's little theorem, $a^{6q} \equiv 1$ for all $a$ coprime to 7, so this is equivalent to $\left(\frac{3^{2r+1} - 1}{2}\right)\left(\frac{5^{2r+1} - 1}{4}\right) - 1$.
We now have three cases:
- $\left(\frac{3^{2(0)+1} - 1}{2}\right)\left(\frac{5^{2(0)+1} - 1}{4}\right) - 1 \equiv \left(\frac{3^1 - 1}{2}\right)\left(\frac{5^1 - 1}{4}\right) - 1 \equiv 1 * 1 - 1 \equiv 0$.
- $\left(\frac{3^{2(1)+1} - 1}{2}\right)\left(\frac{5^{2(1)+1} - 1}{4}\right) - 1 \equiv \left(\frac{3^3 - 1}{2}\right)\left(\frac{5^3 - 1}{4}\right) - 1 \equiv 6 * 3 - 1 \equiv 3$.
- $\left(\frac{3^{2(2)+1} - 1}{2}\right)\left(\frac{5^{2(2)+1} - 1}{4}\right) - 1 \equiv \left(\frac{3^5 - 1}{2}\right)\left(\frac{5^5 - 1}{4}\right) - 1 \equiv 2 * 4 - 1 \equiv 0$.
Therefore, $s(15^{2k}) \equiv 0 \pmod{7}$ if and only if $k \not\equiv 1 \pmod{3}$.
Conjecture (127)**
The prime number 3 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $17^{2k}$.
The case of index 1 follows from Corollary (1).
Conjecture (128)**
The prime number 5 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $17^{4k}$.
The case of index 1 follows from Corollary (1).
Conjecture (129)**
The prime number 7 appears in the factorization of the terms of indexes 1, 2 of all sequences that begin with the integers $17^{6k}$.
The case of index 1 follows from Corollary (1).
Conjecture (130)*
The prime number 19 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $17^{9k}$.
$s(17^{9k}) = \frac{17^{9k} - 1}{16}$, which is divisible by $\frac{17^9 - 1}{16} = 19 \cdot 307 \cdot 1270657$.
Conjecture (131)**
The prime number 19 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $17^{36k}$.
The case of index 1 follows from Conjecture (130).
Conjecture (132)*
The prime number 229 appears in the factorization of the terms of index 1 of all sequences that begin with the integers $17^{19k}$.
$s(17^{19k}) = \frac{17^{19k} - 1}{16}$, which is divisible by $\frac{17^{19} - 1}{16} = 229 \cdot 1103 \cdot 202607147 \cdot 291973723$.
Conjecture (133)**
The prime number 229 appears in the factorization of the terms of index 1, 2 of all sequences that begin with the integers $17^{38k}$.
The case of index 1 follows from Conjecture (132).
Conjectures (134) to (137) published on March 2, 2021, on the Mersenne forum, see post #921
Conjecture (134)
If a base $b = p\#$ is primorial ($p\text{ is prime, } p \gt 7$), then the sequence that starts with the integer $b^{14}$ is increasing from index 1 for a few iterations.
Note 1 : In general, this does not seem to be the case with the other exponents, except exponent 8 (see Conjecture (135)).
Note 2 : We believe that if we compute the following larger primorial bases, the growth phenomenon will occur with other exponents. To check.
Note 3 : This conjecture is completed and replaced by Conjecture (140).
Conjecture (135)
If a base $b = p\#$ is primorial ($p\text{ is prime, } p \gt 29$), then the sequence that starts with the integer $b^8$ is increasing from index 1 for a few iterations.
Note 1 : In general, this does not seem to be the case with the other exponents, except exponent 14 (see Conjecture (134)).
Note 2 : The same as for Conjecture (134).
Note 3 : This conjecture is completed and replaced by Conjecture (140).
Conjecture (136)
If a base $b = p\# / 2$ is primorial without the factor 2 ($p\text{ is prime, } p \gt 3$), then some sequences of this base grow from index 1 for a few iterations.
Note 1 : Until March 2021, we have only found three other odd bases for which this is also the case :
231 (3 * 7 * 11), 3003 (3 * 7 * 11 * 13) and 51051 (3 * 7 * 11 * 13 * 17),
to be seen as a primorial numbers without the factors 2 and 5 ?
Because $3003^5$ and $51051^{11}$ also have this property !
Note 2 : It is possible that this is just an illusion, maybe there are many other odd numbers that have the property ?
Note 3 : many of the exponents are prime numbers (especially 11 and 23) and not prime exponents are often equal to 7 * 5.
Conjecture (137)
Base 2 sequences starting with $2^{12k}$, $2^{40k}$, $2^{90k}$, $2^{140k}$, $2^{210k}$, $2^{220k}$, $2^{330k}$, are increasing from index 1 for a few iterations.
Note 1 : This is not the case for the other exponents we have examined.
Note 2 : We believe that there must be other exponents of the form $z \cdot k$ (with $z \gt 330$) which have this property.
Note 3 : We think that this phenomenon is related to the theorem which says that if $p$ is prime, $s(p^i)$ is a factor of $s(p^{im})$ for every positive integer $m$, see post #466.
This theorem ensures, for example, that exponents multiple of 12 have many prime factors in their factorization (like $2^{12}$ itself), which ensures them growth for a few iterations.
Because of this mechanism, this conjecture looks more like those of post #447, see below.
Note 4 : The demonstration of this conjecture is trivial when we go from index 1 to index 2 only. Because $s(p^{i})$ divides $s(p^{im})$, and multiples of abundant numbers are themselves abundant, is enough to show that $s(p^{i})$ is abundant, which is true for the items addressed in the conjecture.
Conjectures (138) to (139) published on March 6, 2021, on the Mersenne forum, see post #952
Conjecture (138)
Base 6 sequences starting with $6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k}$ are increasing at least from index 1 to 2.
If we prove that:
$2^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k + 1} - 1 \equiv 1 \pmod{d}$
$\frac{3^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k + 1} - 1}{2} \equiv 1 \pmod{d}$
$6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k} \equiv 1 \pmod{d}$
Then $s(6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k}) = (2^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k + 1} - 1) \cdot \frac{3^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k + 1} - 1}{2} - 6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k} = 1 \cdot 1 - 1 \pmod{d}$.
For $p$ prime, and a such that $\gcd(a, p) = 1$, we must have $a^{p-1} \equiv 1 \pmod{p}$.
So if $p - 1 \mid 2^3 \cdot 3^2 \cdot 5 \cdot 7$, it means $2^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k} \equiv 1^k \equiv 1 \pmod{p}$, hence $2^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k + 1} - 1 \equiv 1^k \cdot 2 - 1 \equiv 1 \pmod{p}$.
Same thing for 3 and 6.
That takes care of $p$ = 5+, 7+, 11, 13, 19, 29, 31, 37, 41, 43, 61, 71, 73, 127, 181, 211, 281, 421, 631
(+ This only prove that 5, 7 divide $s(6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k})$, not $5^2$, $7^2$)
For $p = 337$, we have $\frac{p - 1}{2} \mid 2^3 \cdot 3^2 \cdot 5 \cdot 7$. Possible values for $a^(\frac{p - 1}{2})$ are $\pm 1 \pmod{p}$. I guess that mean $2^{(337 - 1)/2}$, $3^{(337 - 1)/2}$, $6^{(337 - 1)/2}$ are happened to be $1 \bmod 337$.
To prove that $5^2$, $7^2$ divide $s(6^{(2^3 \cdot 3^2 \cdot 5 \cdot 7)k})$, we use the fact that if $a \equiv 1 \pmod{p}$, then $a^p \equiv 1 \pmod{p^2}$.
$(5 - 1) \mid 2^3 \cdot 3^2 \cdot 7$, so $a^{2^3 \cdot 3^2 \cdot 5 \cdot 7} \equiv 1 \pmod{5^2}$ for $\gcd(a, p) = 1$
$(7 - 1) \mid 2^3 \cdot 3^2 \cdot 5$, so $a^{2^3 \cdot 3^2 \cdot 5 \cdot 7} \equiv 1 \pmod{7^2}$ for $\gcd(a, p) = 1$
See successively on the Mersenne forum, posts #953, #954 and #960
Conjecture (139)
There exists for each base $b$ a starting exponent $i$, such that for any integer $k$, the sequences $b^{ik}$ are increasing from index 1 during at least one iteration.
Conjectures (140) published on April 7, 2021, on the Mersenne forum, see post #1074
Conjecture (140)
If a base $b = p\#$ is primorial ($p\text{ is prime, } p \gt 41$), then $s(b^{2 + 6k})$ is abundant.
This new conjecture completes and replaces Conjecture (134) and Conjecture (135).