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A186241
G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
11
1, 1, 3, 12, 54, 262, 1337, 7072, 38426, 213197, 1202795, 6879160, 39794416, 232429030, 1368806610, 8118934656, 48458809586, 290832756606, 1754059333738, 10625545472716, 64620970743082, 394409682103262, 2415084675723048, 14832185219521152, 91339478577683664
OFFSET
0,3
LINKS
Nathan Gabriel, Katherine Peske, Lara Pudwell, and Samuel Tay, Pattern avoidance in ternary trees J. Integer Seq. 15 (2012), no. 1, Article 12.1.5, 20 pp.
Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.
FORMULA
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.
From Paul D. Hanna, Nov 11 2011: (Start)
G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^2 ) ).
(2) A( x/(1 + x + x^2 + x^3)^2 ) = 1 + x + x^2 + x^3.
(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).
(4) a(n) = [x^n] (1 + x + x^2 + x^3)^(2*n+1) / (2*n+1).
(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)] ).
(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) )] ).
(End)
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(2*n + 1)*Sum_{k = 0..floor(n/2)} binomial(2*n + 1,k)*binomial(2*n + 1,n - 2*k).
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.
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)
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
A(x^2) = (1/x) * series reversion of x/(1 + x^2 + x^4 + x^6). - Peter Bala, Jul 27 2023
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 54*x^4 + 262*x^5 + 1337*x^6 +...
where A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^4).
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 30*x^3 + 141*x^4 + 704*x^5 + 3666*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 88*x^3 + 451*x^4 + 2392*x^5 + 13022*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 182*x^3 + 1014*x^4 + 5718*x^5 + 32623*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
From Paul D. Hanna, Nov 11 2011: (Start)
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^2)*x*A + (1 + 2^2*x*A^2 + x^2*A^4)*x^2*A^2/2 +
(1 + 3^2*x*A^2 + 3^2*x^2*A^4 + x^3*A^6)*x^3*A^3/3 +
(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 +
(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 + ...
which involves squares of binomial coefficients. (End)
MAPLE
F:= proc(n) if n::even then
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))
else
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))
fi
end proc:
map(F, [$0..30]); # Robert Israel, Jun 22 2015
MATHEMATICA
a[n_] := 1/(2n + 1) Sum[Binomial[2n + 1, k] Binomial[2n + 1, n - 2k], {k, 0, n/2}];
(* or: *)
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);
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Nov 17 2017 *)
PROG
(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 */
(PARI) {a(n)=polcoeff(sqrt((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^2)), n)} /* Paul D. Hanna */
(PARI) {a(n)=polcoeff( (1 + x + x^2 + x^3+x*O(x^n))^(2*n+1)/(2*n+1), n)} /* Paul D. Hanna */
(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 */
(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 */
KEYWORD
nonn,easy,changed
AUTHOR
Vladimir Kruchinin, Feb 15 2011
STATUS
approved