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%I #15 Jan 27 2023 19:01:16
%S 1,2,3,5,7,9,11,13,15,19,21,25,27,31,41,45,49,81,85,129,133,135,139,
%T 357,361,429,431,433,435,447,451,507,511,567,569,571,573
%N Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).
%e 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).
%t Select[Range[140],Divisible[DivisorSigma[2,Fibonacci[#]^2+1],Fibonacci[#]]&]
%o (PARI) isok(k) = my(f=fibonacci(k)); sigma(f^2 + 1, 2) % f == 0; \\ _Michel Marcus_, Jan 26 2023
%Y Cf. A000045, A001157, A067719, A245236, A245306, A338762, A360105.
%K nonn,more
%O 1,2
%A _Michel Lagneau_, Jan 26 2023
%E a(24)-a(37) from _Daniel Suteu_, Jan 27 2023