A348591 a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.
0, 1, 0, 3, 5, 3, 18, 3, 52, 3, 141, 3, 374, 3, 984, 3, 2581, 3, 6762, 3, 17708, 3, 46365, 3, 121390, 3, 317808, 3, 832037, 3, 2178306, 3, 5702884, 3, 14930349, 3, 39088166, 3, 102334152, 3, 267914293, 3, 701408730, 3, 1836311900, 3, 4807526973, 3, 12586269022, 3, 32951280096, 3, 86267571269, 3
Offset: 0
Examples
a(5) = L(5)*L(6) mod F(7) = 11*18 mod 13 = 3.
Links
- Robert Israel, Table of n, a(n) for n = 0..4761
- Index entries for linear recurrences with constant coefficients, signature (-1,3,3,-1,-1).
Programs
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Maple
F:= combinat:-fibonacci: L:= n -> F(n-1)+F(n+1): map(n -> L(n)*L(n+1) mod F(n+2), [$0..30]);
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Mathematica
a[n_] := Mod[LucasL[n] * LucasL[n + 1], Fibonacci[n + 2]]; Array[a, 50, 0] (* Amiram Eldar, Jan 26 2022 *)
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Python
from gmpy2 import fib, lucas2 def A348591(n): return (lambda x,y:int(x[0]*x[1] % y))(lucas2(n+1),fib(n+2)) # Chai Wah Wu, Jan 26 2022
Formula
a(n) = 3 if n >= 3 is odd.
a(n) = A000045(n+2)-3 if n >= 2 is even.
a(n) + a(n+1) - 3*a(n+2) - 3*a(n+3) + a(n+4) + a(n+5) = 0 for n >= 2.
G.f.: -x*(2*x^5-5*x^3-x-1)/((x+1)*(x^2+x-1)*(x^2-x-1)). - Alois P. Heinz, Jan 26 2022