A228078 a(n) = 2^n - Fibonacci(n) - 1.

0, 0, 2, 5, 12, 26, 55, 114, 234, 477, 968, 1958, 3951, 7958, 16006, 32157, 64548, 129474, 259559, 520106, 1041810, 2086205, 4176592, 8359950, 16730847, 33479406, 66987470, 134021309, 268117644, 536356682, 1072909783, 2146137378, 4292788986, 8586410013

Offset: 0

Author

Reinhard Zumkeller, Aug 15 2013

Keywords

• nonn
• easy

Comments

a(n+1) = sum of n-th row of the triangle in A228074.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See p. 30.
• Index entries for linear recurrences with constant coefficients, signature (4,-4,-1,2).

Programs

Haskell

a228078 = subtract 1 . a099036

Magma

[2^n - Fibonacci(n) - 1: n in [0..40]]; // Vincenzo Librandi, Aug 16 2013

Mathematica

Table[(2^n - Fibonacci[n] - 1), {n, 0, 40}] (* Vincenzo Librandi, Aug 16 2013 *)

PARI

concat([0,0], Vec(x^2*(3*x-2)/((x-1)*(2*x-1)*(x^2+x-1)) + O(x^100))) \\ Colin Barker, Mar 20 2015

Formula

a(n) = A000079(n) - A000045(n) - 1 = A000225(n) - A000045(n) = A000079(n) - A001611(n) = A099036(n) - 1.

From Colin Barker, Mar 20 2015: (Start)

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

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

a(n) = (-1+2^n+(((1-sqrt(5))/2)^n-((1+sqrt(5))/2)^n)/sqrt(5)). - Colin Barker, Nov 02 2016