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