A244652 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(6*n,3*n).

1, 19, 262, 3322, 41455, 520165, 6602716, 84860884, 1103478733, 14500102087, 192309166018, 2571407785918, 34631087423419, 469382779109305, 6398055968407480, 87653105740545976, 1206315271455768505, 16669999282643795899, 231219555870655381438, 3217973871571202211778

Offset: 0

Author

Paul D. Hanna, Jul 03 2014

Keywords

• nonn

Comments

Compare to Sum_{k=0..n} [x^k] 1/(1-x)^(5*n) = binomial(6*n,n).

Compare to Sum_{k=0..n} [x^k] 1/((1-x)*(1-2*x)^2)^n = binomial(4*n,2*n).

Examples

			G.f.: A(x) = 1 + 19*x + 262*x^2 + 3322*x^3 + 41455*x^4 + 520165*x^5 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],  0,    0,      0,        0,         0,          0, ...;
A^1: [1,  19], 262,   3322,    41455,    520165,    6602716, ...;
A^2: [1,  38,  885], 16600,   277790,   4356348,   65729806, ...;
A^3: [1,  57, 1869,  46693],  992751,  19018983,  339483259, ...;
A^4: [1,  76, 3214, 100460,  2600405], 59206736, 1229790360, ...;
A^5: [1,  95, 4920, 184760,  5645140, 149282604],3549124200, ...;
A^6: [1, 114, 6987, 306452, 10801665, 325750014, 8738270067],...; ...
then we can illustrate how the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals C(6*n,3*n):
C( 0, 0) = 1 = 1;
C( 6, 3) = 1 +  19 = 20;
C(12, 6) = 1 +  38 +  885 = 924;
C(18, 9) = 1 +  57 + 1869 +  46693 = 48620;
C(24,18) = 1 +  76 + 3214 + 100460 +  2600405 = 2704156;
C(30,21) = 1 +  95 + 4920 + 184760 +  5645140 + 149282604 = 155117520;
C(36,24) = 1 + 114 + 6987 + 306452 + 10801665 + 325750014 + 8738270067 = 9075135300; ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244651, A244653, A244654.

Programs

PARI

/* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(6*n,3*n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j/1!)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))

PARI

/* Faster, using series reversion: */
{a(n)=local(B=sum(k=0, n+1, binomial(6*k,3*k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

Recurrence: n*(n+2)*(n^2 + 5*n - 2)*a(n) = (18*n^4 + 103*n^3 + 39*n^2 + 56*n + 12)*a(n-1) - (49*n^4 + 264*n^3 - 169*n^2 - 372*n - 180)*a(n-2) + 8*(6*n^4 + 31*n^3 - 51*n^2 - 112*n - 48)*a(n-3) - 16*(n-3)*(n+1)*(n^2 + 7*n + 4)*a(n-4). - Vaclav Kotesovec, Jul 04 2014

a(n) ~ 64*sqrt(6*(26*sqrt(3)-45)) * (8+4*sqrt(3))^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 04 2014