A333599 a(n) = Fibonacci(n) * Fibonacci(n+1) mod Fibonacci(n+2).
0, 1, 2, 1, 7, 1, 20, 1, 54, 1, 143, 1, 376, 1, 986, 1, 2583, 1, 6764, 1, 17710, 1, 46367, 1, 121392, 1, 317810, 1, 832039, 1, 2178308, 1, 5702886, 1, 14930351, 1, 39088168, 1, 102334154, 1, 267914295, 1, 701408732, 1, 1836311902, 1, 4807526975, 1, 12586269024
Offset: 0
Examples
a(0) = 0*1 mod 1 = 0; a(1) = 1*1 mod 2 = 1; a(2) = 1*2 mod 3 = 2; a(3) = 2*3 mod 5 = 1; a(4) = 3*5 mod 8 = 7.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Eric Weisstein's World of Mathematics, d'Ocagne's Identity.
- Index entries for linear recurrences with constant coefficients, signature (-1,3,3,-1,-1).
Programs
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Mathematica
With[{f = Fibonacci}, Table[Mod[f[n] * f[n+1], f[n+2]], {n, 0, 50}]] (* Amiram Eldar, Mar 28 2020 *) -
PARI
a(n) = if (n % 2, 1, fibonacci(n+2) - 1); \\ Michel Marcus, Mar 29 2020
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PARI
concat(0, Vec(x*(1 + 3*x - x^3) / ((1 + x)*(1 + x - x^2)*(1 - x - x^2)) + O(x^45))) \\ Colin Barker, Mar 29 2020
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Python
def a(n): f1 = 0 f2 = 1 for i in range(n): f = f1 + f2 f1 = f2 f2 = f return (f1 * f2) % (f1 + f2)
Formula
a(2n+1) = 1, and a(2n) = F(2n+2) - 1, and lim(a(2n+2)/a(2n)) = phi^2 by d'Ocagne's identity.
a(n) = F(n) * F(n+1) mod (F(n) + F(n+1)) since F(n+2) := F(n+1) + F(n).
From Colin Barker, Mar 28 2020: (Start)
G.f.: x*(1 + 3*x - x^3) / ((1 + x)*(1 + x - x^2)*(1 - x - x^2)).
a(n) = -a(n-1) + 3*a(n-2) + 3*a(n-3) - a(n-4) - a(n-5) for n>4.
(End)