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

Author

Adnan Baysal, Mar 28 2020

Keywords

• nonn
• easy

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).

Crossrefs

Equals A035508 interleaved with A000012.

Cf. A182126, A000045, A022461, A072565, A022462.

Programs

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

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

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)