A349989 a(n) is the smallest k such that k^n + (k+1)^n is divisible by a square > 1.

4, 3, 1, 113, 2, 3, 3, 19, 1, 1, 4, 113, 4, 3, 1, 765, 4, 3, 4, 87, 1, 3, 4, 19, 2, 2, 1, 28, 4, 1, 4, 151, 1, 3, 2, 113, 4, 3, 1, 19, 4, 3, 4, 113, 1, 3, 4, 335, 3, 1, 1, 113, 4, 3, 1, 19, 1, 3, 4, 87, 4, 3, 1, 379, 2, 3, 4, 1, 1, 1, 4, 19, 4, 3, 1, 113, 3, 1, 4

Offset: 1

Author

Jon E. Schoenfield, Dec 07 2021

Keywords

• nonn

Comments

a(64) <= 379; a(76) <= 113. Terms a(65)..a(79): 2, 3, 4, 1, 1, 1, 4, 19, 4, 3, 1, a(76), 3, 1, 4.

At k=4, k^n + (k+1)^n = 4^n + 5^n is a multiple of 9 for all odd n, and at k=3, k^n + (k+1)^n = 3^n + 4^n is a multiple of 25 for all n == 2 (mod 4). Thus, a(n) <= 4 if n is not a multiple of 4.

Links

• Kevin P. Thompson, Factorizations to support a(n) for n = 1..79

Crossrefs

Cf. A280302, A280547, A289629, A289985, A349988.

Programs

PARI

a(n) = my(k=1); while(issquarefree(k^n + (k+1)^n), k++); k; \\ Michel Marcus, Dec 09 2021

Python

from sympy.ntheory.factor_ import core
def squarefree(n): return core(n, 2) == n
def a(n):
    k = 1
    while squarefree(k**n + (k+1)**n): k += 1
    return k
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Dec 09 2021

Formula

a(n) = A289629(n) if n is even.

a(k) = 1 for k in A049096.

a(n) <= 4 if 4 does not divide n; among terms where 4 divides n, certain terms appear repeatedly. E.g.,

a(n) <= 113 for n == 4 (mod 8): for all such n, 17^2 divides 113^n + 114^n;

a(n) <= 19 for n == 8 (mod 16): for all such n, 17^2 divides 19^n + 20^n;

a(n) <= 765 for n == 16 (mod 32): for all such n, 97^2 divides 765^n + 766^n;

a(n) <= 87 for n == 20 (mod 40): for all such n, 41^2 divides 87^n + 88^n;

a(n) <= 28 for n == 68 (mod 136): for all such n, 17^2 divides 28^n + 29^n;

a(n) <= 151 for n == 32 (mod 64): for all such n, 257^2 divides 151^n + 152^n;

a(n) <= 335 for n == 48 (mod 96): for all such n, 769^2 divides 335^n + 336^n.

Extensions

a(64)-a(79) from Kevin P. Thompson, Feb 23 2022