A360107 Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).

1, 2, 3, 5, 7, 9, 11, 13, 15, 19, 21, 25, 27, 31, 41, 45, 49, 81, 85, 129, 133, 135, 139, 357, 361, 429, 431, 433, 435, 447, 451, 507, 511, 567, 569, 571, 573

Offset: 1

Author

Michel Lagneau, Jan 26 2023

Keywords

• nonn
• more

Examples

			7 is in the sequence because the divisors of Fibonacci(7)^2 + 1 = 13^2 + 1 = 170 are {1, 2, 5, 10, 17, 34, 85, 170}, and 1^2 + 2^2 + 5^2 + 10^2 + 17^2 + 34^2 + 85^2 + 170^2 = 37700 = 13*2900 == 0 (mod 13).
		

Crossrefs

Cf. A000045, A001157, A067719, A245236, A245306, A338762, A360105.

Programs

Mathematica

Select[Range[140],Divisible[DivisorSigma[2,Fibonacci[#]^2+1],Fibonacci[#]]&]

PARI

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

Extensions

a(24)-a(37) from Daniel Suteu, Jan 27 2023