Theorems
This section contains major proved results that cover several conjectures.
Theorem (1):
The prime number $p$ appears in the factorization of the terms of index 1 of all sequences with a square-free base, coprime to $p$, and an exponent divisible by $p - 1$.
By Alexander Jones.Let $q_i$ be the (distinct) prime factors of the base $b$, and let $n = k(p - 1)$ 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 $\sigma(b_n) = \displaystyle\prod_i \frac{q_i^{n+1} - 1}{q_i - 1}$. Multiplying the numerator and denominator through, we get $\frac{\displaystyle\prod_i (q_i^{n+1} - 1)}{\displaystyle\prod_i (q_i - 1)}$.
Taking the numerator first, we factor a $q_i$ out of $q_i^{n+1}$ to get $\displaystyle\prod_i (q_i \cdot q_i^n - 1)$. Since we defined $n = k(p - 1)$, by Fermat's little theorem, $q_i^n \equiv 1 \pmod{p}$, 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.
This theorem is not generally true if $p$ is one of the factors of the base, as the form of Fermat's little theorem used in the proof does not hold in that case.
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 Theorem (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 Theorem (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 Theorem (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 Theorem (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 first case is an immediate result of Theorem (1).
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 Theorem (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 Theorem (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 $3^{6+12k}$.
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 38But on reflection, this conjecture is not extraordinary.
7 is a prime number which is in the factorization of the $2^2 \cdot 7$ driver.
It is therefore normal that it persists in so many consecutive terms.
On the other hand, it should be shown here that $s(3^{6+12k})$ has the driver $2^2 \cdot 7$ as a factor.
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 Theorem (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 Theorem (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 Theorem (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 Theorem (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)**, already known conjecture, see previous posts :
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 Theorem (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 Theorem (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 Theorem (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 Theorem (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 Theorem (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 Theorem (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 $6^{2+10k}$.
The first case is an immediate result of Theorem (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 Theorem (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 $6^{10+18k}$.
The first case is an immediate result of Theorem (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 $6^{11k}$.
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 Theorem (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 $6^{17+30k}$.
The first case is an immediate result of Theorem (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 $6^{36k}$
This is an immediate result of Theorem (1).
Conjecture (73)* :
The prime number 37 appears in the factorization of the terms of index 1 of all sequences starting with the integers $6^{14+36k}$.
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 $6^{53+58k}$.
The first case is an immediate result of Theorem (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 $6^{55+60k}$.
The first case is an immediate result of Theorem (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 $6^{67+70k}$.
The first case is an immediate result of Theorem (1).
Conjecture (77)* :
The prime number 601 appears in the factorization of the terms of index 1 of all sequences beginning with the integers $6^{75k}$.
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 Theorem (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 Theorem (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 Theorem (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 Theorem (1).
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 Theorem (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 Theorem (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 $7^{38k}$.
Checked up to $k = 23$.
The case of index 1 follows from Conjecture (92).
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 Theorem (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 $10^{3+4k}$.
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 Theorem (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 Theorem (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 $10^{54+60k}$.
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 Theorem (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 $12^k$.
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 $12^{6+16k}$.
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 Theorem (1).
Conjecture (113) :
The prime number 5 appears in thefactorization 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 Theorem (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 Theorem (1).
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 Theorem (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 $15^{8+12k}$.
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 Theorem (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 Theorem (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 Theorem (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.
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).