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EXAMPLE
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G.f.: A(x) = 1 + 20*x + 280*x^2 + 3521*x^3 + 42945*x^4 + 521913*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, 20], 280, 3521, 42945, 521913, 6379233, ...;
A^2: [1, 40, 960], 18242, 305130, 4733386, 70081627, ...;
A^3: [1, 60, 2040, 52163], 1122555, 21563619, 382898782, ...;
A^4: [1, 80, 3520, 113284, 2991220], 68901812, 1437454298, ...;
A^5: [1, 100, 5400, 209605, 6567125, 176797165], 4251203775, ...;
A^6: [1, 120, 7680, 349126, 12666270, 390658878, 10654434813],...; ...
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(7*n,2*n):
C( 0, 0) = 1 = 1;
C( 7, 2) = 1 + 20 = 21;
C(14, 4) = 1 + 40 + 960 = 1001;
C(21, 6) = 1 + 60 + 2040 + 52163 = 54264;
C(28, 8) = 1 + 80 + 3520 + 113284 + 2991220 = 3108105;
C(35,10) = 1 + 100 + 5400 + 209605 + 6567125 + 176797165 = 183579396;
C(42,12) = 1 + 120 + 7680 + 349126 + 12666270 + 390658878 + 10654434813 = 11058116888; ...
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PROG
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(PARI) /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(7*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(7*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), ", "))
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