A245515 a(n) = n*floor(mod((gcd(n, Fibonacci((-1)^n + n))), 1 + n)/n) for n>=2.
1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, 29, 0, 31, 0, 0, 0, 0, 0, 0, 0, 0, 0, 41, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 59, 0, 61, 0, 0, 0, 0, 0, 0, 0, 0, 0, 71, 0, 0, 0, 0, 0, 0, 0, 79, 0, 0, 0, 0, 0, 0
Offset: 1
Keywords
Examples
For n=1, a(1)=1; for n=2, a(2)=2.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
- José de Jesús Camacho Medina, Table with 140 first nonzero terms of this sequence (Una interesante fórmula generadora de 140 primos).
Programs
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Magma
[n*((Gcd(n, Fibonacci((-1)^n+n)) mod (1+n)) div n): n in [1..100]]; // Vincenzo Librandi, Dec 17 2016
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Maple
f:= n -> n*floor(modp((igcd(n, combinat:-fibonacci((-1)^n + n))), 1 + n)/n): seq(f(n), n=1..100); # Robert Israel, Jul 25 2014
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Mathematica
Table[n*Floor[Mod[(GCD[n, Fibonacci[(-1)^n + n]]), 1 + n]/n], {n, 1, 1890}]
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PARI
a(n) = n*((gcd(n, fibonacci((-1)^n + n)) % (1 + n))\n); \\ Michel Marcus, Jul 25 2014
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PARI
a(n)=gcd(n, lift(((Mod([1,1;1,0],n))^(n+(-1)^n))[1,2]))\n*n \\ Charles R Greathouse IV, Jul 25 2014
Formula
a(n) = n*floor(mod((gcd(n, fibonacci((-1)^n + n))), 1 + n)/n) for n>=1.
Comments