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%I #30 Jul 20 2023 16:29:23
%S 1,1,3,9,28,92,319,1154,4302,16382,63391,248499,984867,3940121,
%T 15891386,64545971,263783729,1083883910,4475194635,18557356409,
%U 77251869363,322723617687,1352518263334,5684939482522,23959266771808,101226312702475,428650606083144,1818991203750774,7734098181837847
%N G.f. satisfies: x = A(x) * (1 - A(x)) / (1 - A(x) - A(x)^3) such that A(0) = 1.
%C Compare to A243157, the series reversion of x*(1 - x)/(1 - x - x^3).
%H Paul D. Hanna, <a href="/A243156/b243156.txt">Table of n, a(n) for n = 0..300</a>
%F G.f.: A(x) = x / Series_Reversion(x*(1 + Series_Reversion(x / (1 + 4*x + 3*x^2 + x^3)))).
%F G.f. satisfies: x = (1+x)*A(x) - A(x)^2 + x*A(x)^3 such that A(0) = 1.
%F a(n) ~ sqrt((s-1)*s/(3*r*s-1)) / (2*sqrt(Pi) * r^n * n^(3/2)), where r = 2/(3 + sqrt(13 + 16*sqrt(2))) = 0.22299351557517... and s = (1+sqrt(5+4*sqrt(2)))/2 = 2.1322418823119... . - _Vaclav Kotesovec_, May 31 2014
%F a(n) = floor(((6^(2*n)*(6^(-n)+1)^3+6^n)/(6*(1+6^n)))^n-6^n*floor((((6^(2*n)*(6^(-n)+1)^3+6^n)/(36*(1+6^n)))^n)))/n, for n>0. - _Tani Akinari_, May 20 2018
%F D-finite with recurrence n*(n+1)*a(n) -9*n*(n-1)*a(n-1) +6*(6*n^2 -22*n +17)*a(n-2) -27*(3*n-7) *(n-4)*a(n-3) +2*(14*n^2 -172*n +435)*a(n-4) +3*(53*n -192) *(n-5) *a(n-5) -341*(n-5)*(n-6) *a(n-6)=0. - _R. J. Mathar_, Jul 20 2023
%e G.f.: A(x) = 1 + x + 3*x^2 + 9*x^3 + 28*x^4 + 92*x^5 + 319*x^6 + 1154*x^7 + 4302*x^8 + 16382*x^9 + 63391*x^10 + ... where A(x) = x * (1 - A(x) - A(x)^3) / (1 - A(x)).
%t CoefficientList[x/InverseSeries[x*(1+InverseSeries[Series[x/(1 + 4*x + 3*x^2 + x^3),{x,0,20}],x]),x],x] (* _Vaclav Kotesovec_, May 31 2014 after _Paul D. Hanna_ *)
%o (PARI) {a(n)=polcoeff(x/serreverse(x*(1+serreverse(x/(1 + 4*x + 3*x^2 + x^3 +x*O(x^n))))),n)}
%o for(n=0,30,print1(a(n),", "))
%o (PARI) {a(n)=0^n+floor(((6^(2*n)*(6^(-n)+1)^3+6^n)/(6*(1+6^n)))^n-6^n*floor((((6^(2*n)*(6^(-n)+1)^3+6^n)/(36*(1+6^n)))^n)))/(n+0^n)}
%o for(n=0, 50, print1(a(n), ", ")) \\ _Tani Akinari_, May 20 2018
%o (PARI) {a(n)=0^n+sum(k=1,n,binomial(n,k)*binomial(3*k-n,k-1))/(n+0^n)} \\ _Tani Akinari_, May 20 2018
%Y Cf. A243157, A115344.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 31 2014
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