A293411 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.

1, 2, 3, 4, 7, 12, 19, 30, 49, 80, 129, 208, 337, 546, 883, 1428, 2311, 3740, 6051, 9790, 15841, 25632, 41473, 67104, 108577, 175682, 284259, 459940, 744199, 1204140, 1948339, 3152478, 5100817, 8253296, 13354113, 21607408, 34961521, 56568930, 91530451

Offset: 0

Author

Clark Kimberling, Nov 25 2017

Keywords

• nonn
• easy

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that this sequence has the growth rate of the Fibonacci numbers (A000045).

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1999
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Crossrefs

Cf. A001622, A000045, A295619, A295620, A061646, A079472.

Programs

Mathematica

LinearRecurrence[{1, 0, 1, 1}, {1, 2, 3, 4}, 100]

Formula

a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4.

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

From Greg Dresden, Aug 25 2021: (Start)

a(2*n) = a(2*n - 1) + a(2*n - 2),

a(2*n) = 2*F(n+1)^2 - (-1)^n = A061646(n+1),

a(2*n+1) = 2*F(n+1)*F(n+2) = A079472(n+2), for F(n) the Fibonacci numbers A000045. (End)

a(n) = round (2*A000032(n+2)/5). - R. J. Mathar, Jul 20 2025