A115008 a(n) = a(n-1) + a(n-3) + a(n-4).

1, 0, 2, 4, 5, 7, 13, 22, 34, 54, 89, 145, 233, 376, 610, 988, 1597, 2583, 4181, 6766, 10946, 17710, 28657, 46369, 75025, 121392, 196418, 317812, 514229, 832039, 1346269, 2178310, 3524578, 5702886, 9227465, 14930353, 24157817, 39088168

Offset: 0

Author

Creighton Dement, Feb 23 2006

Keywords

• easy
• nonn
• changed

Comments

a(n+2) - a(n+1) - a(n) gives match to A000034, apart from signs.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Crossrefs

Cf. A000034, A000045, A001519, A001906, A006498, A116697, A116698, A116699.

Programs

Magma

A115008:= func< n | Fibonacci(n+1) - (n mod 2) + 2*0^((n+1) mod 4) >;
[A115008(n): n in [0..50]]; // G. C. Greubel, Aug 24 2025

Mathematica

Table[Fibonacci[n+1] -I^(n-1)*Mod[n,2], {n,0,50}] (* G. C. Greubel, Aug 24 2025 *)

SageMath

def A115008(n): return fibonacci(n+1) -i**(n-1)*(n%2)
print([A115008(n) for n in range(51)]) # G. C. Greubel, Aug 24 2025

Formula

a(2*n) = A000045(2*n+1) = A001519(n).

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

a(2*n+1) = (-1)^(n+1) + A001906(n+1) (compare with a similar property for A116697) - Creighton Dement, Mar 31 2006

From G. C. Greubel, Aug 24 2025: (Start)

a(n) = A000045(n+1) - i^(n-1)*(n mod 2).

E.g.f.: exp(x/2)*(cosh(p*x) + (1/(2*p))*sinh(p*x)) - sin(x), where 2*p = sqrt(5). (End)