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a(n) = Sum_{k=0..floor(n/5)} C(n-4k,k+1).
2

%I #22 Jul 12 2018 02:36:50

%S 0,1,2,3,4,5,7,10,14,19,25,33,44,59,79,105,139,184,244,324,430,570,

%T 755,1000,1325,1756,2327,3083,4084,5410,7167,9495,12579,16664,22075,

%U 29243,38739,51319,67984,90060,119304,158044,209364,277349,367410,486715

%N a(n) = Sum_{k=0..floor(n/5)} C(n-4k,k+1).

%H Harvey P. Dale, <a href="/A099559/b099559.txt">Table of n, a(n) for n = 0..1000</a>

%H Minerva Catral, P. L. Ford, P. E. Harris, S. J. Miller, et al. <a href="http://arxiv.org/abs/1606.09312">Legal Decompositions Arising from Non-positive Linear Recurrences</a>, arXiv preprint arXiv:1606.09312 [math.CO], 2016. See Table 2.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,1,-1).

%F Partial sums of A003520 (with leading zero).

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

%F a(n) = 2a(n-1)-a(n-2)+a(n-5)-a(n-6).

%F 7*a(n) = A117373(n+2) -7 +10*b(n) +15*b(n-1) +9*b(n-2), where b(n) = A182097(n). - _R. J. Mathar_, Aug 07 2017

%F a(n) = A003520(n+4) -1. - _R. J. Mathar_, Aug 07 2017

%t LinearRecurrence[{2,-1,0,0,1,-1},{0,1,2,3,4,5},50] (* _Harvey P. Dale_, Feb 20 2017 *)

%o (PARI) a(n) = sum(k=0,n\5, binomial(n-4*k, k+1)); \\ _Michel Marcus_, Jul 11 2018

%Y Cf. A098578.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Oct 22 2004

%E Values from a(5) on corrected by _R. J. Mathar_, Jul 29 2008