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A360107
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Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).
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1
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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
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OFFSET
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1,2
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LINKS
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EXAMPLE
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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).
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MATHEMATICA
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Select[Range[140], Divisible[DivisorSigma[2, Fibonacci[#]^2+1], Fibonacci[#]]&]
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PROG
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(PARI) isok(k) = my(f=fibonacci(k)); sigma(f^2 + 1, 2) % f == 0; \\ Michel Marcus, Jan 26 2023
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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