A244651 G.f. A(x) satisfies: Sum_{k=0..n} [x^k] A(x)^n = binomial(6*n,2*n).

1, 14, 135, 1148, 9325, 74634, 596083, 4775288, 38447961, 311305350, 2534757855, 20749571316, 170705908421, 1410874891522, 11710273480395, 97573698950384, 815919118022833, 6845174820882174, 57601263531202871, 486057767175907180, 4112073577799441181, 34871360280503319674

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 + 14*x + 135*x^2 + 1148*x^3 + 9325*x^4 + 74634*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, 14], 135,   1148,    9325,    74634,     596083, ...;
A^2: [1, 28,  466],  6076,   69019,   720328,    7117572, ...;
A^3: [1, 42,  993,  17528], 258462,  3377556,   40526262, ...;
A^4: [1, 56, 1716,  38248,  695450, 10968552,  155816996, ...;
A^5: [1, 70, 2635,  70980, 1536195, 28435134,  467948465, ...;
A^6: [1, 84, 3750, 118468, 2975325, 63276528, 1185303544],...; ...
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,2*n):
C( 0, 0) = 1 = 1;
C( 6, 2) = 1 + 14 = 15;
C(12, 4) = 1 + 28 +  466 = 495;
C(18, 6) = 1 + 42 +  993 +  17528 = 18564;
C(24, 8) = 1 + 56 + 1716 +  38248 +  695450 = 735471;
C(30,10) = 1 + 70 + 2635 +  70980 + 1536195 + 28435134 = 30045015;
C(36,12) = 1 + 84 + 3750 + 118468 + 2975325 + 63276528 + 1185303544 = 1251677700; ...
		

Links

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

Crossrefs

Cf. A232606, A232683, A232687, A244650, A244652, A244653, A244654.

Programs

PARI

/* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(6*n,2*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,2*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+3)*(4*n+1)*a(n) = (44*n^3 + 91*n^2 + 99*n + 46)*a(n-1) - (76*n^3 + 159*n^2 + 53*n - 60)*a(n-2) + 9*(n-1)*(n+2)*(4*n+5)*a(n-3). - Vaclav Kotesovec, Jul 04 2014

a(n) ~ 9^(n+4) / (64*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 04 2014