cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A348592 a(n) = F(n)*F(n+1) mod L(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

Original entry on oeis.org

0, 1, 2, 6, 15, 11, 10, 45, 99, 79, 65, 312, 675, 545, 442, 2142, 4623, 3739, 3026, 14685, 31683, 25631, 20737, 100656, 217155, 175681, 142130, 689910, 1488399, 1204139, 974170, 4728717, 10201635, 8253295, 6677057, 32411112, 69923043, 56568929, 45765226, 222149070, 479259663, 387729211, 313679522
Offset: 0

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Author

J. M. Bergot and Robert Israel, Jan 25 2022

Keywords

Examples

			a(5) = F(5)*F(6) mod L(7) = 5*8 mod 29 = 11.
		

Crossrefs

Programs

  • Maple
    F:= combinat:-fibonacci:
    L:= n -> F(n-1)+F(n+1):
    seq(F(n)*F(n+1) mod L(n+2), n=0..20);
  • Mathematica
    a[n_] := Mod[Fibonacci[n] * Fibonacci[n + 1], LucasL[n + 2]]; Array[a, 50, 0] (* Amiram Eldar, Jan 26 2022 *)

Formula

For n >= 1, a(n) = (A070352(n+2)*A000032(n+2) + 3*(-1)^n)/5.
a(n) + 2*a(n + 1) + 3*a(n + 2) + 5*a(n + 3) + a(n + 4) - a(n + 5) - a(n + 7) = 0 for n >= 1.
G.f.: -3 + 3/(5*(1+x)) + (3+x)/(2*(1-x-x^2)) + (9-4*x+6*x^2-x^3)/(10*(1+3*x^2+x^4)).
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