%I #41 May 22 2020 09:46:41

%S 0,1,2,1,7,1,20,1,54,1,143,1,376,1,986,1,2583,1,6764,1,17710,1,46367,

%T 1,121392,1,317810,1,832039,1,2178308,1,5702886,1,14930351,1,39088168,

%U 1,102334154,1,267914295,1,701408732,1,1836311902,1,4807526975,1,12586269024

%N a(n) = Fibonacci(n) * Fibonacci(n+1) mod Fibonacci(n+2).

%H Colin Barker, <a href="/A333599/b333599.txt">Table of n, a(n) for n = 0..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/dOcagnesIdentity.html">d'Ocagne's Identity</a>.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-1,3,3,-1,-1).

%F 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.

%F a(n) = F(n) * F(n+1) mod (F(n) + F(n+1)) since F(n+2) := F(n+1) + F(n).

%F From _Colin Barker_, Mar 28 2020: (Start)

%F G.f.: x*(1 + 3*x - x^3) / ((1 + x)*(1 + x - x^2)*(1 - x - x^2)).

%F a(n) = -a(n-1) + 3*a(n-2) + 3*a(n-3) - a(n-4) - a(n-5) for n>4.

%F (End)

%e a(0) = 0*1 mod 1 = 0;

%e a(1) = 1*1 mod 2 = 1;

%e a(2) = 1*2 mod 3 = 2;

%e a(3) = 2*3 mod 5 = 1;

%e a(4) = 3*5 mod 8 = 7.

%t With[{f = Fibonacci}, Table[Mod[f[n] * f[n+1], f[n+2]], {n, 0, 50}]] (* _Amiram Eldar_, Mar 28 2020 *)

%o (Python)

%o def a(n):

%o f1 = 0

%o f2 = 1

%o for i in range(n):

%o f = f1 + f2

%o f1 = f2

%o f2 = f

%o return (f1 * f2) % (f1 + f2)

%o (PARI) a(n) = if (n % 2, 1, fibonacci(n+2) - 1); \\ _Michel Marcus_, Mar 29 2020

%o (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

%Y Equals A035508 interleaved with A000012.

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

%K nonn,easy

%O 0,3

%A _Adnan Baysal_, Mar 28 2020