A289629 Smallest positive k such that (k+1)^n + (-k)^n is divisible by a square greater than 1.

3, 7, 113, 14, 3, 23, 19, 7, 1, 2, 113, 75, 3, 7, 765, 36, 3, 2476, 87, 1, 3, 165, 19, 14, 2, 7, 28, 149, 1, 2972, 151, 2, 3, 14, 113, 977, 3, 5, 19, 34, 3, 135, 113, 7, 3, 136, 335, 23, 1, 7, 113, 11, 3, 2, 19, 2, 3

Offset: 2

Author

Juri-Stepan Gerasimov, Sep 02 2017

Keywords

• nonn
• more

Comments

From Kevin P. Thompson, Dec 03 2021: (Start)

Most of the factorizations to prove known terms of this sequence have been uploaded to factordb.com.

For known terms up to n = 100, the square that divides (k+1)^n + (-k)^n is very small, less than 1500^2, with only one example greater than 1000^2.

a(20m+10) = 1 for m >= 0. With k = 1 and starting at n = 20*0 + 10 = 10, (k+1)^n + (-k)^n = 2^10 + 1 = 1024 + 1 = 1025 which is divisible by 5^2. Since the last two digits of 2^n repeat in a cycle of length 20, (k+1)^n + (-k)^n will always be divisible by 5^2 for n = 20m + 10.

Conjecture: (k+1)^n + (-k)^n is not squarefree for the following (n, k) patterns, with m >= 1: (22m-11, 2), (20m-6, 3), (20m-2, 3), (3^m, 7), (15m, 7), (20m-15, 14), (16m-8, 19), (42m-35, 23), and (8m-4, 113). In each case, the value of a(n) in this sequence is usually equal to the value specified for k, but sometimes this value is not the smallest such k. For example, in the (n, k) = (20m-2, 3) case, a(20m-2) = 3 for m = 1..3, but at m = 4, a(20*4-2) = a(78) = 1.

(End)

Examples

			a(2) = 3 because (1+1)^2 + (-1)^2 = 5 is squarefree, (2+1)^2 + (-2)^2 = 13 is squarefree, and (3+1)^2 + (-3)^2 = 25 is divisible by 5^2.
		

Links

• Kevin P. Thompson, Table of n, a(n) for n = 2..150 with unconfirmed terms (term 31 confirmed by _Hugo Pfoertner_ and term 48 confirmed by _Jon E. Schoenfield_ and _Hugo Pfoertner_)

Crossrefs

Cf. A280547, A285929.

Cf. A280302, A289985.

Programs

Mathematica

Table[k=1;While[SquareFreeQ[(k+1)^n+(-k)^n],k++];k,{n,2,15}] (* Giorgos Kalogeropoulos, Dec 03 2021 *)

PARI

a(n) = my(k=1); while (issquarefree((k+1)^n + (-k)^n), k++); k; \\ Michel Marcus, Dec 04 2021

Formula

a(2n+1) = A280547(2n+1).

Extensions

Offset corrected; a(16), a(32), a(36), a(44), and a(48) corrected; and a(50)-a(58) added by Kevin P. Thompson, Dec 05 2021.