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%I #42 Sep 30 2023 16:38:26
%S 1,1,6,96,2976,151416,11449296,1204566336,168233625216,30110372009856,
%T 6719377991060736,1829013279998846976,596449130341224185856,
%U 229556544889929225117696,102956750031135241952280576,53228316147100497514507862016,31423560379886826670772937424896
%N E.g.f.: ( (1 + cos(sqrt(6)*x))/2 )^(-1/3), showing coefficients of only the even powers of x.
%C Previous name was: Column 0 and row sums of symmetric triangle A126150.
%H Vaclav Kotesovec, <a href="/A126151/b126151.txt">Table of n, a(n) for n = 0..232</a>
%H E. Norton, <a href="http://arxiv.org/abs/1302.5411">Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions</a>, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
%F a(n) = Sum_{0<=k<=n} A087736(n,k)*2^(n-k). - _Philippe Deléham_, Jul 17 2007
%F G.f.: 1/(1-x/(1-5*x/(1-12*x/(1-22*x/(1-35*x/(1-51*x/(1-70*x/(1-...- (n*(3*n-1)/2)*x/(1-...))))))))), a continued fraction involving pentagonal numbers A000326. - _Paul D. Hanna_, Feb 15 2012
%F E.g.f. satisfies: A(x) = exp( Integral Integral A(x)^3 dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)! and the constant of integration is zero. - _Paul D. Hanna_, May 29 2015
%F E.g.f. satisfies: A(x) = exp( Integral A(x)^(3/2) * Integral 1/A(x)^(3/2) dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)! and the constant of integration is zero. - _Paul D. Hanna_, Jun 02 2015
%F a(n) ~ Gamma(1/3) * 2^(3*n+4/3) * 3^(n+1/2) * n^(2*n+1/6) / (exp(2*n) * Pi^(2*n+7/6)). - _Vaclav Kotesovec_, May 30 2015
%F The computation can be based on the pentagonal numbers, a(n) = T(n, n) where T(n, k) = A000326(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, n-1) if n > 0. This is equivalent to _Paul D. Hanna_'s continued fraction 2012. - _Peter Luschny_, Sep 30 2023
%e E.g.f.: A(x) = 1 + x^2/2! + 6*x^4/4! + 96*x^6/6! + 2976*x^8/8! + 151416*x^10/10! +...
%e where the logarithm begins:
%e log(A(x)) = x^2/2! + 3*x^4/4! + 36*x^6/6! + 918*x^8/8! + 40176*x^10/10! + 2686608*x^12/12! +...
%e compare the logarithm to
%e A(x)^3 = 1 + 3*x^2/! + 36*x^4/4! + 918*x^6/6! + 40176*x^8/8! + 2686608*x^10/10! +...
%e where A(x)^3 = 2/(1 + cos(sqrt(6)*x)).
%p A000326 := n -> n * (3 * n - 1) / 2;
%p T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else A000326(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
%p a := n -> T(n, n): seq(a(n), n = 0..16); # _Peter Luschny_, Sep 30 2023
%t terms = 18;
%t CoefficientList[((1 + Cos[Sqrt[6] x])/2)^(-3^(-1)) + O[x]^(2 terms), x] Range[0, 2 terms - 2]! // DeleteCases[#, 0]& (* _Jean-François Alcover_, Jul 26 2018 *)
%o (PARI) /* Continued Fraction involving pentagonal numbers A000326: */
%o {a(n)=local(CF=1+x*O(x),m,P); for(k=1, n,m=n-k+1;P=m*(3*m-1)/2; CF=1/(1-P*x*CF)); polcoeff(CF, n, x)}
%o for(n=0,20,print1(a(n),","))
%o (PARI) /* E.g.f. A(x) = exp( Integral^2 A(x)^3 dx^2 ): */
%o {a(n)=local(A=1+x*O(x)); for(i=1, n, A=exp(intformal(intformal(A^3 + x*O(x^(2*n))))) ); (2*n)!*polcoeff(A, 2*n, x)}
%o for(n=0,20,print1(a(n),", "))
%o (PARI) /* E.g.f. A(x) = exp( Integral A(x)^(3/2) * Integral 1/A(x)^(3/2) dx dx ) */
%o {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( A^(3/2) * intformal( 1/A^(3/2) + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
%o for(n=0,20,print1(a(2*n),", "))
%Y Cf. A126150; diagonals: A126152, A126153.
%Y Cf. A000326.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 19 2006
%E New name from _Paul D. Hanna_, May 30 2015
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