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%I #47 Aug 01 2023 08:06:12
%S 1,1,3,12,54,262,1337,7072,38426,213197,1202795,6879160,39794416,
%T 232429030,1368806610,8118934656,48458809586,290832756606,
%U 1754059333738,10625545472716,64620970743082,394409682103262,2415084675723048,14832185219521152,91339478577683664
%N G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
%H Robert Israel, <a href="/A186241/b186241.txt">Table of n, a(n) for n = 0..1108</a>
%H Nathan Gabriel, Katherine Peske, Lara Pudwell, and Samuel Tay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Pudwell/pudwell.html">Pattern avoidance in ternary trees</a> J. Integer Seq. 15 (2012), no. 1, Article 12.1.5, 20 pp.
%H Vladimir Kruchinin and D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.
%F a(n) = 1/(2*n-1)*Sum_{j=0..2*n-1} binomial(2*n-1,j)*Sum_{i=j..n+j-1} binomial(j,i-j)*binomial(2*n-j-1,3*j-3*n-i+1))), n>0.
%F From _Paul D. Hanna_, Nov 11 2011: (Start)
%F G.f. A(x) satisfies:
%F (1) A(x) = sqrt( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^2 ) ).
%F (2) A( x/(1 + x + x^2 + x^3)^2 ) = 1 + x + x^2 + x^3.
%F (3) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = g.f. of A036765 (number of rooted trees with a degree constraint).
%F (4) a(n) = [x^n] (1 + x + x^2 + x^3)^(2*n+1) / (2*n+1).
%F (5) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * [Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(2*k)] ).
%F (6) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * [(1-x*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(2*k) )] ).
%F (End)
%F From _Peter Bala_, Jun 21 2015: (Start)
%F a(n) = 1/(2*n + 1)*Sum_{k = 0..floor(n/2)} binomial(2*n + 1,k)*binomial(2*n + 1,n - 2*k).
%F More generally, the coefficient of x^n in A(x)^r equals r/(2*n + r)*Sum_{k = 0..floor(n/2)} binomial(2*n + r,k)*binomial(2*n + r,n - 2*k) by the Lagrange-Bürmann formula.
%F O.g.f. A(x) = exp(Sum_{n >= 1} 1/2*b(n)*x^n/n), where b(n) = Sum_{k = 0..floor(n/2)} binomial(2*n,k)*binomial(2*n,n - 2*k). Cf. A036765, A198951, A200731. (End)
%F Recurrence: 5*n*(5*n - 1)*(5*n + 1)*(5*n + 2)*(5*n + 3)*(13144*n^4 - 57784*n^3 + 90149*n^2 - 59354*n + 13980)*a(n) = 8*(2*n - 1)*(16259128*n^8 - 71478808*n^7 + 108653137*n^6 - 60530902*n^5 - 2811173*n^4 + 12694433*n^3 - 2398482*n^2 - 352503*n + 78570)*a(n-1) + 128*(n-1)*(2*n - 3)*(2*n - 1)*(52576*n^6 - 178560*n^5 + 136156*n^4 + 22938*n^3 - 16067*n^2 - 3138*n - 405)*a(n-2) + 2048*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(13144*n^4 - 5208*n^3 - 4339*n^2 + 168*n + 135)*a(n-3). - _Vaclav Kotesovec_, Nov 17 2017
%F A(x^2) = (1/x) * series reversion of x/(1 + x^2 + x^4 + x^6). - _Peter Bala_, Jul 27 2023
%e G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 54*x^4 + 262*x^5 + 1337*x^6 +...
%e where A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^4).
%e Related expansions:
%e A(x)^2 = 1 + 2*x + 7*x^2 + 30*x^3 + 141*x^4 + 704*x^5 + 3666*x^6 +...
%e A(x)^4 = 1 + 4*x + 18*x^2 + 88*x^3 + 451*x^4 + 2392*x^5 + 13022*x^6 +...
%e A(x)^6 = 1 + 6*x + 33*x^2 + 182*x^3 + 1014*x^4 + 5718*x^5 + 32623*x^6 +...
%e where A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
%e From _Paul D. Hanna_, Nov 11 2011: (Start)
%e The logarithm of the g.f. A = A(x) equals the series:
%e log(A(x)) = (1 + x*A^2)*x*A + (1 + 2^2*x*A^2 + x^2*A^4)*x^2*A^2/2 +
%e (1 + 3^2*x*A^2 + 3^2*x^2*A^4 + x^3*A^6)*x^3*A^3/3 +
%e (1 + 4^2*x*A^2 + 6^2*x^2*A^4 + 4^2*x^3*A^6 + x^4*A^8)*x^4*A^4/4 +
%e (1 + 5^2*x*A^2 + 10^2*x^2*A^4 + 10^2*x^3*A^6 + 5^2*x^4*A^8 + x^5*A^10)*x^5*A^5/5 + ...
%e which involves squares of binomial coefficients. (End)
%p F:= proc(n) if n::even then
%p simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
%p else
%p simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
%p fi
%p end proc:
%p map(F, [$0..30]); # _Robert Israel_, Jun 22 2015
%t a[n_] := 1/(2n + 1) Sum[Binomial[2n + 1, k] Binomial[2n + 1, n - 2k], {k, 0, n/2}];
%t (* or: *)
%t a[n_] := (Binomial[2n + 1, n] HypergeometricPFQ[{-2n - 1, 1/2 - n/2, -n/2}, {n/2 + 1, n/2 + 3/2}, -1])/(2n + 1);
%t Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Nov 17 2017 *)
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^4)+x*O(x^n));polcoeff(A,n)} /* _Paul D. Hanna_ */
%o (PARI) {a(n)=polcoeff(sqrt((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^2)), n)} /* _Paul D. Hanna_ */
%o (PARI) {a(n)=polcoeff( (1 + x + x^2 + x^3+x*O(x^n))^(2*n+1)/(2*n+1), n)} /* _Paul D. Hanna_ */
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (x*A+x*O(x^n))^m/m*sum(j=0, m, binomial(m, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* _Paul D. Hanna_ */
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,x^m*A^m/m*(1-x*A^2)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* _Paul D. Hanna_ */
%Y Cf. A199876, A199877, A198951, A198953, A198957, A192415, A198888, A036765.
%Y Cf. A200731.
%Y Cf. A199874, A200074, A200075, A200718, A200719, A215576.
%K nonn,easy
%O 0,3
%A _Vladimir Kruchinin_, Feb 15 2011
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