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A369265 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^3) ).

%I A369265 #18 Feb 15 2024 04:20:52
%S A369265 1,2,7,31,153,806,4439,25250,147193,874732,5279635,32276245,199439761,
%T A369265 1243633652,7815804351,49455190791,314807497953,2014530780524,
%U A369265 12952334769203,83628832755779,542022781854953,3525150296312984,22998642171764363,150478455899387966
%N A369265 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^3) ).
%H A369265 P. Bala, <a href="/A251592/a251592.pdf">Fractional iteration of a series inversion operator</a>
%H A369265 <a href="/index/Res#revert">Index entries for reversions of series</a>
%F A369265 a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(n+1,k) * binomial(3*n-3*k+1,n-3*k).
%F A369265 D-finite with recurrence 16*(n+1)*(2*n+1)*a(n) +4*(-89*n^2+15*n+2)*a(n-1) +3*(345*n^2-603*n+274)*a(n-2) +18*(-41*n^2+45*n+94)*a(n-3) +54*(-4*n^2+57*n-137)*a(n-4) +486*(n-4)*(n-5)*a(n-5) -243*(n-4)*(n-5)*a(n-6)=0. - _R. J. Mathar_, Jan 25 2024
%F A369265 a(n) = (1/(n+1)) * [x^n] ( 1/(1-x)^2 * (1+x^3) )^(n+1). - _Seiichi Manyama_, Feb 14 2024
%p A369265 A369265 := proc(n)
%p A369265     add(binomial(n+1,k) * binomial(3*n-3*k+1,n-3*k),k=0..floor(n/3)) ;
%p A369265     %/(n+1) ;
%p A369265 end proc;
%p A369265 seq(A369265(n),n=0..70) ; # _R. J. Mathar_, Jan 25 2024
%o A369265 (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2/(1+x^3))/x)
%o A369265 (PARI) a(n, s=3, t=1, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);
%Y A369265 Cf. A369267, A369269.
%Y A369265 Cf. A071969, A370247.
%K A369265 nonn
%O A369265 0,2
%A A369265 _Seiichi Manyama_, Jan 18 2024