A113435 a(n) = a(n-1) + Sum_{k=0..n/3} a(n-3k) with a(0)=1.

1, 1, 1, 2, 3, 4, 7, 11, 16, 26, 41, 62, 98, 154, 237, 371, 581, 901, 1406, 2197, 3418, 5329, 8317, 12956, 20196, 31501, 49096, 76532, 119338, 186029, 289997, 452141, 704861, 1098826, 1713111, 2670692, 4163483, 6490879, 10119152, 15775426

Offset: 0

Author

Floor van Lamoen, Nov 04 2005

Keywords

• nonn
• easy

Comments

If presented in three rows a(3n), a(3n+1) and a(3n+2) each term is the sum of the previous term in the sequence and the partial sum of its row.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 19.
• Index entries for linear recurrences with constant coefficients, signature (1,0,2,-1).

Crossrefs

Cf. A113439, A113444, A028495.

Partial sums of A176848.

Programs

Mathematica

CoefficientList[Series[(1 - x^3)/(1 - x - 2*x^3 + x^4), {x,0,50}], x] (* G. C. Greubel, Mar 10 2017 *)
LinearRecurrence[{1,0,2,-1},{1,1,1,2},40] (* Harvey P. Dale, Dec 17 2023 *)

PARI

x='x+O(x^50); Vec((1 - x^3)/(1 - x - 2*x^3 + x^4)) \\ G. C. Greubel, Mar 10 2017

Formula

a(n) = a(n-1) + 2*a(n-3) - a(n-4) = 7*a(n-3) - 5*a(n-6) + 11*a(n-9) - a(n-12).

G.f.: (1-x^3)/(1-x-2*x^3+x^4).

G.f.: 1/(1-x) + x^3*Q(0)/(2-2*x) , where Q(k) = 1 + 1/(1 - x*(4*k+1 + 2*x^2 - x^3)/( x*(4*k+3 + 2*x^2 - x^3 ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 11 2013