A360105 Numbers k such that sigma_2(k^2 + 1) == 0 (mod k).

1, 2, 5, 7, 13, 25, 34, 52, 89, 93, 100, 200, 233, 338, 610, 850, 915, 1028, 1352, 1508, 1918, 2105, 3918, 4181, 5540, 6396, 6728, 7250, 9282, 10100, 10132, 10946, 15507, 16609, 17125, 32708, 32776, 37107, 42568, 47770, 58218, 61230, 72125, 74948, 75025, 78608

Offset: 1

Author

Michel Lagneau, Jan 26 2023

Keywords

• nonn

Comments

Conjecture: the sequence contains infinitely many Fibonacci numbers (see A360107).

For k < 10^7, we observe only 6 prime numbers in the sequence: {2, 5, 7, 13, 89, 233} including the Fibonacci numbers {2, 5, 13, 89, 233} and the Lucas number {7}.

Examples

			7 is in the sequence because the divisors of 7^2+1 = 50 are {1, 2, 5, 10, 25, 50}, and 1^2 + 2^2 + 5^2 + 10^2 + 25^2 + 50^2 = 3255 = 7*465 == 0 (mod 7).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..222

Crossrefs

Cf. A000032, A000045, A001157, A067719, A360107.

Programs

Maple

filter:= k -> NumberTheory:-SumOfDivisors(k^2+1,2) mod k = 0:
select(filter, [$1..10^5]); # Robert Israel, Feb 19 2024

Mathematica

Select[Range[50000], Divisible[DivisorSigma[2, #^2+1], #]&]

PARI

isok(k) = sigma(k^2 + 1, 2) % k == 0; \\ Michel Marcus, Jan 26 2023