## 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 11

^{k}.

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 13

^{k}.

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).