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%I #55 Jan 27 2024 10:08:29
%S 1,1,1,2,3,4,8,13,19,38,64,98,196,337,531,1062,1851,2974,5948,10468,
%T 17060,34120,60488,99658,199316,355369,590563,1181126,2115577,3540464,
%U 7080928,12731141,21430267,42860534,77306428,130771376,261542752,473018396,803538100
%N Expansion of f(x^3)/(1-x*f(x^3)), where f(x) is the g.f. of A001764, whose n-th term is binomial(3n,n)/(2n+1).
%C Row sums of number triangle A111373.
%C Interleaves T(3n,2n), T(3n+1,2n+1) and T(3n+2,2n+2) for T(n,k) = A047089(n,k).
%C One step forward and two steps back: number of nonnegative walks of n steps where the steps are size 1 forwards and size 2 backwards. - _David Scambler_, Mar 15 2011
%C Brown's criterion ensures that the sequence is complete (see formulae). - _Vladimir M. Zarubin_, Aug 05 2019
%H G. C. Greubel, <a href="/A126042/b126042.txt">Table of n, a(n) for n = 0..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BrownsCriterion.html">Brown's Criterion</a>
%F a(n) = Sum_{k=0..n} binomial(3*floor((n+2k)/3) - 2k, floor((n+2k)/3)-k)*(k+1)/(2*floor((n+2k)/3) - k + 1)(2*cos(2*Pi*(n-k)/3) + 1)/3.
%F G.f.: (1/x)*Series_Reversion( x*(1+x)^2/((1+x)^3+x^3) ). - _Paul D. Hanna_, Mar 15 2011
%F From _Vladimir M. Zarubin_, Aug 05 2019: (Start)
%F a(0) = 1, a(1) = 1, a(2) = 1 and for k>0
%F a(3*k) = 2*a(3*k-1),
%F a(3*k+1) = 2*a(3*k) - binomial(3*k,k)/(2*k+1),
%F a(3*k+2) = 2*a(3*k+1) - binomial(3*k+1,k)/(k+1),
%F where binomial(3*k,k)/(2*k+1) = A001764(k)
%F and binomial(3*k+1,k)/(k+1) = A006013(k). (End)
%F a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} (n-3*k+1) * binomial(n+1,k). - _Seiichi Manyama_, Jan 27 2024
%p a:= proc(n) option remember; `if`(n<4, [1$3, 2][n+1], (a(n-1)*
%p 2*(20*n^4-14*n^3-31*n^2-n+8)-6*(3*n-1)*(5*n-6)*a(n-2)
%p +9*(n-2)*(15*n^3-48*n^2+15*n+14)*a(n-3)-54*(n-2)*(n-3)*
%p (5*n^2-n-2)*a(n-4))/(2*(2*n+1)*(n+1)*(5*n^2-11*n+4)))
%p end:
%p seq(a(n), n=0..45); # _Alois P. Heinz_, Sep 07 2022
%t Table[Binomial[n, Floor[n/3]] -Sum[Binomial[n,i], {i,0,Floor[n/3] -1}], {n,0,40}] (* _David Callan_, Oct 26 2017 *)
%o (PARI) {a(n)=polcoeff((1/x)*serreverse(x*(1+x)^2/((1+x)^3+x^3+x*O(x^n))),n)}
%o (PARI)
%o n=30;
%o {a0=1;a1=1;a2=1;for(k=1, n/3,print1(a0,", ",a1,", ",a2,", ");
%o a0=2*a2;a1=2*a0-binomial(3*k,k)/(2*k+1);a2=2*a1-binomial(3*k+1,k)/(k+1))
%o } // _Vladimir M. Zarubin_, Aug 05 2019
%o (Magma) [n lt 3 select 1 else Binomial(n, Floor(n/3)) - (&+[Binomial(n,j): j in [0..Floor(n/3)-1]]): n in [0..40]]; // _G. C. Greubel_, Jul 30 2022
%o (SageMath) [binomial(n, (n//3)) - sum(binomial(n,j) for j in (0..(n//3)-1)) for n in (0..40)] # _G. C. Greubel_, Jul 30 2022
%Y Cf. A047089, A111373.
%K easy,nonn
%O 0,4
%A _Paul Barry_, Dec 16 2006
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