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EXAMPLE
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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; ...
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PROG
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(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), ", "))
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