A012855 a(0) = 0, a(1) = 1, a(2) = 1; thereafter a(n) = 5*a(n-1) - 4*a(n-2) + a(n-3).

0, 1, 1, 1, 2, 7, 28, 114, 465, 1897, 7739, 31572, 128801, 525456, 2143648, 8745217, 35676949, 145547525, 593775046, 2422362079, 9882257736, 40315615410, 164471408185, 670976837021, 2737314167775, 11167134898976

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Old name was "Take every 5th term of Padovan sequence A000931".

Lim_{n -> infinity} a(n+1)/a(n) = p^5 = 4.0795956..., where p is the plastic constant (A060006). - Jianing Song, Feb 04 2019

Links

• Index entries for linear recurrences with constant coefficients, signature (5, -4, 1).

Crossrefs

Cf. A000931, A012814, A012864, A060006, A176476.

Programs

Maple

A012855 := proc(n,A,B,C) option remember; if n = 0 then A elif n = 1 then B elif n = 2 then C else 5*procname(n-1,A,B,C)-4*procname(n-2,A,B,C)+procname(n-3,A,B,C); fi; end; [ seq(A012855(i,0,1,1),i = 0..40) ]; # R. J. Mathar, Dec 30 2011

Mathematica

CoefficientList[Series[(4x^2-x)/(x^3-4x^2+5x-1),{x,0,40}],x] (* or *) LinearRecurrence[{5,-4,1},{0,1,1},40] (* Harvey P. Dale, Mar 28 2013 *)

PARI

a(n) = my(v=vector(n+1), u=[0,1,1]); for(k=1, n+1, v[k]=if(k<=3, u[k], 5*v[k-1] - 4*v[k-2] + v[k-3])); v[n+1] \\ Jianing Song, Feb 04 2019

Formula

a(n) = A000931(5*n-12) for n >= 3. - Alois P. Heinz, Feb 04 2019

G.f. (4x^2 - x)/(x^3 - 4x^2 + 5x - 1). For n > 2, a(n) = 1 + Sum_{k=0..n-3} A012814(k). - Ralf Stephan, Jan 15 2004

a(n) = 1 + A176476(n-3) = 1 + Sum_{k=0..n-3} A000931(5*k+2) for n >= 3. - Jianing Song, Feb 04 2019

Extensions

Edited by N. J. A. Sloane, Feb 06 2019 at the suggestion of Jianing Song, replacing imprecise definition with formula from Harvey P. Dale, Mar 28 2013