A012855 - OEIS

%I #40 May 18 2019 02:09:52

%S 0,1,1,1,2,7,28,114,465,1897,7739,31572,128801,525456,2143648,8745217,

%T 35676949,145547525,593775046,2422362079,9882257736,40315615410,

%U 164471408185,670976837021,2737314167775,11167134898976

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

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

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

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5, -4, 1).

%F a(n) = A000931(5*n-12) for n >= 3. - _Alois P. Heinz_, Feb 04 2019

%F 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

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

%p 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

%t 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 *)

%o (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

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

%K nonn

%O 0,5

%A _N. J. A. Sloane_

%E 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