A232687 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.

1, 1, 3, 7, 20, 66, 244, 980, 4182, 18674, 86353, 410541, 1996214, 9888844, 49760925, 253767097, 1309154825, 6822023553, 35865392690, 190038440422, 1014015337209, 5444707218851, 29401289997403, 159584901816255, 870267544114291, 4766246752344215, 26206635040151511

Offset: 0

Author

Paul D. Hanna, Dec 05 2013

Keywords

• nonn

Comments

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

a(n+1)/a(n) tends to 6.0295... - Vaclav Kotesovec, Jan 22 2014

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 20*x^4 + 66*x^5 + 244*x^6 + 980*x^7 +...
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, 1], 3,   7,   20,   66,   244,    980,   4182, ...;
A^2: [1, 2,  7], 20,   63,  214,   789,   3124,  13112, ...;
A^3: [1, 3, 12,  40], 138,  492,  1848,   7326,  30531, ...;
A^4: [1, 4, 18,  68,  255], 960,  3716,  14920,  62295, ...;
A^5: [1, 5, 25, 105,  425, 1691], 6785,  27805, 117165, ...;
A^6: [1, 6, 33, 152,  660, 2772, 11560], 48588, 207774, ...;
A^7: [1, 7, 42, 210,  973, 4305, 18676,  80746],351792, ...;
A^8: [1, 8, 52, 280, 1378, 6408, 28916, 128808, 573311], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals Sum_{k=0..n} C(n,k)^3 = A000172(n):
A000172(0) = 1 = 1;
A000172(1) = 1 + 1 = 2;
A000172(2) = 1 + 2 +  7 = 10;
A000172(3) = 1 + 3 + 12 +  40 = 56;
A000172(4) = 1 + 4 + 18 +  68 + 255 = 346;
A000172(5) = 1 + 5 + 25 + 105 + 425 + 1691 = 2252;
A000172(6) = 1 + 6 + 33 + 152 + 660 + 2772 + 11560 = 15184; ...
		

Links

• Paul D. Hanna and Vaclav Kotesovec, Table of n, a(n) for n = 0..460 (first 200 terms from Paul D. Hanna)

Crossrefs

Cf. A232606, A232683, A232690, A000172.

Programs

Mathematica

Franel[n_] := Sum[Binomial[n, k]^3, {k, 0, n}];
a[0] = 1; a[n_] := Module[{B, G}, B = Sum[Franel[k]*x^k, {k, 0, n+1}] + x^3*O[x]^n; G = 1+x*O[x]^n; For[i=1, i <= n, i++, G = 1+Integrate[(B-1)* (G/x)-B*G^2, x]]; SeriesCoefficient[x/InverseSeries[x*G, x], {x, 0, n}]];
Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jan 15 2018, translated from 2nd PARI program *)

PARI

/* By Definition (slow): */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=if(n==0, 1, (Franel(n) - 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), ", "))

PARI

/* Faster, using series reversion: */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=local(B=sum(k=0, n+1, Franel(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), ", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} C(n,k)^3 = A000172(n).

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} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1) = Sum_{n>=0} A000172(n)*x^n.