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%I #39 Feb 06 2019 02:33:55
%S 0,1,6,27,113,464,1896,7738,31571,128800,525455,2143647,8745216,
%T 35676948,145547524,593775045,2422362078,9882257735,40315615409,
%U 164471408184,670976837020,2737314167774,11167134898975,45557394660800,185855747875875,758216295635151
%N Partial sums of A012814.
%C Old name was "a(n) is the minimum integer that can be expressed as the sum of n Padovan numbers (see A000931)".
%C Lim_{n -> infinity} a(n+1)/a(n) = p^5 = 4.0795956..., where p is the plastic constant (A060006).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,5,-1).
%F a(n) = A012855(n+3) - 1. a(n) = 6*a(n-1) - 9*a(n-2) + 5*a(n-3) - a(n-4). - _R. J. Mathar_, Oct 18 2010
%F G.f.: x/(1 - 6*x + 9*x^2 - 5*x^3 + x^4). - _Colin Barker_, Feb 03 2012
%F From _Jianing Song_, Feb 04 2019: (Start)
%F a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n) + 1.
%F a(n) = Sum_{k=0..n} A012814(k) = Sum_{k=0..n} A000931(5*k+2). (End)
%e a(5) = A000931(2) + A000931(7) + A000931(12) + A000931(17) + A000931(22) + A000931(27) = 0 + 1 + 5 + 21 + 86 + 351 = 464.
%o (PARI) a(n) = my(v=vector(n+1), u=[0,1,6,27]); for(k=1, n+1, v[k]=if(k<=4, u[k], 5*v[k-1] - 4*v[k-2] + v[k-3] + 1)); v[n+1] \\ _Jianing Song_, Feb 04 2019
%Y Cf. A000931, A012814, A012855, A060006.
%K nonn,easy
%O 0,3
%A _Carmine Suriano_, Apr 18 2010
%E New name, more terms and a(0) = 0 prepended by _Jianing Song_, Feb 04 2019
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