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EXAMPLE
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G.f.: A(x) = 1 + 9*x + 55*x^2 + 290*x^3 + 1430*x^4 + 6827*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, 9], 55, 290, 1430, 6827, 32083, ...;
A^2: [1, 18, 191], 1570, 11105, 71294, 428452, ...;
A^3: [1, 27, 408, 4569], 42390, 345546, 2564272, ...;
A^4: [1, 36, 706, 10016, 115211], 1142108, 10130498, ...;
A^5: [1, 45, 1085, 18640, 256055, 2992934], 30938150, ...;
A^6: [1, 54, 1545, 31170, 497970, 6708456, 79254029],...; ...
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(5*n,2*n):
C( 0, 0) = 1 = 1;
C( 5, 2) = 1 + 9 = 10;
C(10, 4) = 1 + 18 + 191 = 210;
C(15, 6) = 1 + 27 + 408 + 4569 = 5005;
C(20, 8) = 1 + 36 + 706 + 10016 + 115211 = 125970;
C(25,10) = 1 + 45 + 1085 + 18640 + 256055 + 2992934 = 3268760;
C(30,12) = 1 + 54 + 1545 + 31170 + 497970 + 6708456 + 79254029 = 86493225; ...
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PROG
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(PARI) /* By Definition (slow): */
{a(n)=if(n==0, 1, ( binomial(5*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(5*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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