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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.
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%I #35 Nov 28 2022 01:50:44

%S 1,1,5,11,91,1247,3485,263017,852841,1241058127,74966255,

%T 243641132605417,181556731572385303,718802057694183783881,

%U 6582662048285,943422576750791493013356207217,487331778345355477261,607088607861933740557075591887834842297

%N 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.

%C 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).

%C 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).

%C 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.

%C 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).

%C 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).

%H Samuel Harkness, <a href="/A353890/a353890.m.txt">MATLAB program</a>

%e 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.

%e 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.

%e n=4 has the 5-term repeating sequence, beginning at m=1:

%e [0 0 1 1 1], so a(4) = 5.

%e The sequence is repeating, so for example, f(41)..f(45) is also [0 0 1 1 1].

%e n=5 has the 11-term repeating sequence

%e [1 1 0 1 1 0 1 1 1 1 1]

%e so a(5) = 11.

%e n=6 has the 91-term repeating sequence

%e [0 0 0 0 0 0 1 0 0 0 0 1 1

%e 1 0 0 0 0 0 1 1 0 0 0 0 1

%e 1 1 0 0 0 0 1 1 1 0 0 0 0

%e 1 1 1 0 0 0 0 1 1 1 0 0 0

%e 0 1 1 1 0 0 0 0 1 1 1 0 0

%e 0 0 1 1 0 0 0 0 0 1 1 1 0

%e 0 0 0 1 0 0 0 0 0 0 1 1 1]

%e so a(6) = 91.

%e 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.

%e 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.

%o (MATLAB) See Links section.

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

%K nonn

%O 2,3

%A _Samuel Harkness_, May 09 2022

%E a(7)-a(19) from _Jon E. Schoenfield_, May 10 2022