A350253 a(n) = A000045(n)^A000045(n+1) mod A000045(n+2).

0, 1, 1, 3, 3, 1, 8, 13, 1, 55, 55, 1, 144, 233, 1, 987, 987, 1, 2584, 4181, 1, 17711, 17711, 1, 46368, 75025, 1, 317811, 317811, 1, 832040, 1346269, 1, 5702887, 5702887, 1, 14930352, 24157817, 1, 102334155, 102334155, 1, 267914296, 433494437, 1, 1836311903, 1836311903, 1, 4807526976, 7778742049

Offset

0,4

Links

Robert Israel, Table of n, a(n) for n = 0..4763

Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,18,0,0,-18,0,0,-1,0,0,1).

Formula

a(n) = 1 if n == 2 or 5 (mod 6),

= A000045(n) if n == 0, 1 or 4 (mod 6),

= A000045(n+1) if n == 3 (mod 6).

G.f.: (z + z^2 + 3*z^3 + 2*z^4 + 5*z^6 - 8*z^7 - 18*z^8 - 7*z^9 + 6*z^10 - z^12 - z^13 + z^14)/(1 - z^3 - 18*z^6 + 18*z^9 + z^12 - z^15).

Example

a(3) = 2^3 mod 5 = 3.

Maple

f:= proc(n)
    if n mod 6 = 3 then combinat:-fibonacci(n+1)
  elif n mod 3 = 2 then 1
  else combinat:-fibonacci(n)
  fi
end proc:
map(f, [$0..100]);

Mathematica

Table[PowerMod @@ Fibonacci[n + {0, 1, 2}], {n, 0, 50}] (* Amiram Eldar, Dec 22 2021 *)

Prog

(PARI) a(n) = lift(Mod(fibonacci(n), fibonacci(n+2))^fibonacci(n+1)); \\ Michel Marcus, Dec 22 2021
(Python)
from sympy import fibonacci
def A350253(n): return 1 if (m := n % 6) == 2 or m == 5 else (fibonacci(n+1) if m == 3 else fibonacci(n)) # from formula Chai Wah Wu, Dec 22 2021

Crossrefs

Cf. A000045.

Sequence in context: A293294 A200342 A181330 * A262143 A284554 A078033

Adjacent sequences: A350250 A350251 A350252 * A350254 A350255 A350256

Keyword

nonn,easy

Author

J. M. Bergot and Robert Israel, Dec 21 2021

Status

approved