A232606 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (2*n)!^2/n!^4, the square of the central binomial coefficients (A000984), for n>=0.

1, 3, 10, 42, 221, 1379, 9678, 73666, 594326, 5007958, 43641702, 390632678, 3573598539, 33289289533, 314871186248, 3017358158132, 29242725947318, 286209134234602, 2825613061237808, 28111283170770480, 281598654896870051, 2838309465080014489, 28767973963085929656, 293059625830028920012

Offset: 0

Author

Paul D. Hanna, Nov 26 2013

Keywords

• nonn

Comments

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).

a(n+1)/a(n) tends to 11.3035... - Vaclav Kotesovec, Jan 23 2014

Examples

			G.f.: A(x) = 1 + 3*x + 10*x^2 + 42*x^3 + 221*x^4 + 1379*x^5 + 9678*x^6 +...
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,        0,         0, ...;
A^1: [1,  3], 10,   42,   221,   1379,    9678,    73666,    594326, ...;
A^2: [1,  6,  29], 144,   794,   4924,   33814,   251544,   1988885, ...;
A^3: [1,  9,  57,  333], 1989,  12669,   86935,   639123,   4979499, ...;
A^4: [1, 12,  94,  636,  4157], 27728,  193504,  1423120,  11006058, ...;
A^5: [1, 15, 140, 1080,  7730,  54538], 391970,  2915490,  22558825, ...;
A^6: [1, 18, 195, 1692, 13221,  99102,  739547], 5612016,  43767477, ...;
A^7: [1, 21, 259, 2499, 21224, 169232, 1317722, 10267666], 81223912, ...;
A^8: [1, 24, 332, 3528, 32414, 274792, 2238492, 17990904, 145096413], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals the square of the central binomial coefficients:
1^1 = 1;
2^2 = 1 + 3;
6^2 = 1 +  6 +  29;
20^2 = 1 +  9 +  57 +  333;
70^2 = 1 + 12 +  94 +  636 +  4157;
252^2 = 1 + 15 + 140 + 1080 +  7730 +  54538;
924^2 = 1 + 18 + 195 + 1692 + 13221 +  99102 +  739547;
3432^2 = 1 + 21 + 259 + 2499 + 21224 + 169232 + 1317722 + 10267666; ...
RELATED SERIES.
From a main diagonal in the above array we can derive sequence A232607:
[1/1, 6/2, 57/3, 636/4, 7730/5, 99102/6, 1317722/7, 17990904/8, ...] =
[1, 3, 19, 159, 1546, 16517, 188246, 2248863, 27844369, 354576634, ...];
from which we can form the series G(x) = A(x*G(x)):
G(x) = 1 + 3*x + 19*x^2 + 159*x^3 + 1546*x^4 + 16517*x^5 + 188246*x^6 +...
such that
(G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = 1 + 2^2*x + 6^2*x^2 + 20^2*x^3 + 70^2*x^4 + 252^2*x^5 +...+ A000984(n)^2*x^n +...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..350

Crossrefs

Cf. A232683, A232607, A232687, A000984, A002894.

Programs

Mathematica

terms = 24; a[0] = 1; A[x_] = Sum[a[n]*x^n, {n, 0, terms - 1}];
c[n_] := Sum[Coefficient[B[x], x, k], {k, 0, n}] == (2*n)!^2/n!^4 // Solve // First;
Do[B[x_] = A[x]^n + O[x]^(n+1) // Normal; A[x_] = (A[x] /. c[n]) + O[x]^terms, {n, 0, terms-1}];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)

PARI

/* By Definition: */
{a(n)=if(n==0, 1, ((2*n)!^2/n!^4 - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0,20,print1(a(n)*1!,", "))

PARI

/* Faster, using series reversion: */
{a(n)=local(CB2=sum(k=0,n,binomial(2*k,k)^2*x^k)+x*O(x^n), G=1+x*O(x^n));
for(i=1,n,G = 1 + intformal( (CB2-1)*G/x - CB2*G^2));polcoeff(x/serreverse(x*G),n)}
for(n=0,30,print1(a(n),", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (2*n)!^2/n!^4 = A000984(n)^2.

Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (2*n)!^2/n!^4 * x^n.