A022522 -id:A022522 - cp's OEIS frontend

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

Samuel Harkness, May 09 2022

nonn

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).

			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.

Samuel Harkness, MATLAB program

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

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

A275615 Decimal expansion of 22/111.

Original entry on oeis.org

1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8, 1, 9, 8

Offset: 0

Peter M. Chema, Aug 03 2016

Period 3: repeat [1,9,8]. Also the digital root of Nexus numbers A022522.

Other variants of this period are decimal expansions of 91/111 (.819 repeating) and 109/111 (.981 repeating).

			0.198198198198198198198198198198198198198198198198198198198...

Cf. A010888, A022522.

First@ RealDigits@ N[22/111, 120] (* or *)
PadRight[{}, 120, {1, 9, 8}] (* or *)
FixedPoint[Total@ IntegerDigits@ # &, #] & /@ Table[(n + 1)^6 - n^6, {n, 0, 120}] (* Michael De Vlieger, Aug 04 2016 *)

a(n) = my(x=[1, 9, 8]); x[n%3+1] \\ Felix Fröhlich, Aug 04 2016

a(n) = A010888(A022522(n)).

a(n) = 6 - 5*cos(2*n*Pi/3) + sin(2*n*Pi/3)/sqrt(3). - Wesley Ivan Hurt, Oct 04 2018

cp's OEIS Frontend

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.

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