A353890 a(n) is the period of the binary sequence {b(m)} defined by b(m) = 1 if (m+1)^n - m^n and (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise.

1, 1, 5, 11, 91, 1247, 3485, 263017, 852841, 1241058127, 74966255, 243641132605417, 181556731572385303, 718802057694183783881, 6582662048285, 943422576750791493013356207217, 487331778345355477261, 607088607861933740557075591887834842297

Offset

2,3

Comments

For any n, consecutive n-th powers will never share a divisor > 1, so now consider the second differences. Specifically, each m > 0, define the binary sequence {b(m)} as follows: b(m) = 1 if the first difference (m+1)^n - m^n and the second difference (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise. I conjecture that {b(m)} is periodic with period a(n).

If m^n mod p == (m+1)^n mod p == (m+2)^n mod p, then p is in the prime factorization of a(n).

All primes p >= 5 belong to a prime factorization for a(n). p will always belong to the prime factorization of n=p-1 due to Fermat's Little Theorem.

I conjecture that the greatest prime factor for any prime n >= 5 is phi(2^n+1)/2 + 1 = Jacobsthal(n). n*A069925 + 1 = A001045(n).

I conjecture that all prime factors "f" are f=n*k+1, unless n is composite, in which case additionally all prime factors for any divisor of n will also be included in the prime factorization for a(n).

Links

Table of n, a(n) for n=2..19.

Samuel Harkness, MATLAB program

Example

For n=2 and n=3, the first and second differences are coprime for all m. Each of their sequences {b(m)} consist only of 1's, which can be described trivially as [1] with a period of 1, so a(2) = a(3) = 1.
For n > 3, the first and second differences are coprime for some m values, but not for all. Each repeating periodic sequence {b(m)} begins at m=1, and can be used to predict what b(m) will be at any higher m value for that power n.
n=4 has the 5-term repeating sequence, beginning at m=1:
  [0 0 1 1 1], so a(4) = 5.
The sequence is repeating, so for example, f(41)..f(45) is also [0 0 1 1 1].
n=5 has the 11-term repeating sequence
  [1 1 0 1 1 0 1 1 1 1 1]
so a(5) = 11.
n=6 has the 91-term repeating sequence
  [0 0 0 0 0 0 1 0 0 0 0 1 1
   1 0 0 0 0 0 1 1 0 0 0 0 1
   1 1 0 0 0 0 1 1 1 0 0 0 0
   1 1 1 0 0 0 0 1 1 1 0 0 0
   0 1 1 1 0 0 0 0 1 1 1 0 0
   0 0 1 1 0 0 0 0 0 1 1 1 0
   0 0 0 1 0 0 0 0 0 0 1 1 1]
so a(6) = 91.
The period for higher n values has yet to be found. If they exist, it seems they would be quite large given the large expansion from 5, 11, to 91.
Example: the 233rd term in the sequence of values for n=6 is calculated by using m=233 and n=6. Define the first difference for the 233rd term as 234^6 - 233^6 = 4164782373647. The second difference for the 233rd term is 235^6 - 2*234^6 + 233^6 = 89948228762. The terms 4164782373647 and 89948228762 share a common factor, so the 233rd term of the sequence for 6th powered terms is denoted 0 (not coprime). Because the 6th powered terms repeat their tendency of being coprime or not every 91 terms, we could instead look at 233 mod 91 = 51, and from the table for n=6 above, the 51st term is 0.

Prog

(MATLAB) See Links section.

Crossrefs

Cf. A005408, A007395, A003215, A008588, A005917, A005914, A022521, A068236, A022522, A069473, A069925, A001045, A002587.

Sequence in context: A368013 A188514 A301923 * A120778 A042761 A372126

Adjacent sequences: A353887 A353888 A353889 * A353891 A353892 A353893

Keyword

nonn

Author

Samuel Harkness, May 09 2022

Extensions

a(7)-a(19) from Jon E. Schoenfield, May 10 2022

Status

approved