A396098 G.f. A(x) satisfies Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} ( [x^k] 1/(1-x)^n )^2 for n >= 0.

1, 1, 5, 31, 235, 2033, 19220, 193438, 2038804, 22261830, 249955221, 2870522839, 33584727090, 399117777590, 4806462433605, 58548059001309, 720303118997053, 8939341693860181, 111801073862508091, 1407896478406885631, 17839002674912554079, 227290810261647023091, 2910570007328778112935

Offset

0,3

Comments

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

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^(2*n) = Sum_{k=0..2*n} [x^k] 1/(1-x)^n = (3*n)!/(n!*(2*n)!).

Links

Paul D. Hanna, Table of n, a(n) for n = 0..500

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} ( [x^k] 1/(1-x)^n )^2 for n >= 0.

(2) Sum_{k=0..n} [x^k] A(x)^n = A333592(n) where A333592(n) = Sum_{k=0..n} binomial(n+k-1, k)^2 for n >= 0.

(3) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) satisfies (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} A333592(n) * x^n.

(4) A(x) = (1-x) * (A(x) - x*A'(x)) * Sum_{n>=0} A333592(n) * x^n / A(x)^n.

Example

G.f.: A(x) = 1 + x + 5*x^2 + 31*x^3 + 235*x^4 + 2033*x^5 + 19220*x^6 + 193438*x^7 + 2038804*x^8 + 22261830*x^9 + ...
RELATED SERIES.
The series G(x) = A(x*G(x)) begins
G(x) = 1 + x + 6*x^2 + 47*x^3 + 440*x^4 + 4594*x^5 + 51597*x^6 + 610443*x^7 + 7508241*x^8 + 95169232*x^9 + ...
satisfies (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = D(x) where
D(x) = 1 + 2*x + 14*x^2 + 146*x^3 + 1742*x^4 + 22252*x^5 + 296438*x^6 + 4063866*x^7 + 56884430*x^8 + ... + A333592(n)*x^n + ...
Thus, G(x) = 1 + Integral G(x)*(D(x)-1)/x - D(x)*G(x)^2 dx
and A(x) = x/Series_Reversion(x*G(x)).
RELATED TABLES.
We shall compare the terms of the following tables to illustrate the defining property of the g.f. of this sequence.
Given the g.f. A(x), the table of coefficients of x^k in A(x)^n begins
  n = 1: [1, 1,  5,  31,  235,  2033,  19220, ...];
  n = 2: [1, 2, 11,  72,  557,  4846,  45817, ...];
  n = 3: [1, 3, 18, 124,  981,  8607,  81551, ...];
  n = 4: [1, 4, 26, 188, 1523, 13504, 128456, ...];
  n = 5: [1, 5, 35, 265, 2200, 19746, 188865, ...];
  n = 6: [1, 6, 45, 356, 3030, 27564, 265436, ...];
  ...
The table of binomial coefficients of x^k in 1/(1-x)^n begins
  n = 1: [1, 1,  1,  1,  1,   1, ...];
  n = 2: [1, 2,  3,  4,  5,   6, ...];
  n = 3: [1, 3,  6, 10, 15,  21, ...];
  n = 4: [1, 4, 10, 20, 35,  56, ...];
  n = 5: [1, 5, 15, 35, 70, 126, ...];
  ...
From these two tables, we see that the following partial sums of the rows are equal
  n = 1: (1 + 1) = (1 + 1^2);
  n = 2: (1 + 2 + 11) = (1 + 2^2 + 3^2);
  n = 3: (1 + 3 + 18 + 124) = (1 + 3^2 + 6^2 + 10^2);
  n = 4: (1 + 4 + 26 + 188 + 1523) = (1 + 4^2 + 10^2 + 20^2 + 35^2);
  n = 5: (1 + 5 + 35 + 265 + 2200 + 19746) = (1 + 5^2 + 15^2 + 35^2 + 70^2 + 126^2);
  ...

Prog

(PARI) /* By Definition (slow): */
{d(n) = sum(k=0, n, binomial(n+k-1, k)^2 )}
{a(n) = if(n==0, 1, (d(n) - sum(k=0, n, polcoef(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0, 12, print1(a(n), ", "))
(PARI) /* Faster, using series reversion: */
{d(n) = sum(k=0, n, binomial(n+k-1, k)^2 )}
{a(n) = my(D = sum(k=0, n, d(k)*x^k) +x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (D-1)*G/x - D*G^2));
polcoef(GF = x/serreverse(x*G +x^2*O(x^n)), n)}
{upto(n) = a(n); Vec(GF)}
upto(30)

Crossrefs

Cf. A232606, A396097, A333592.

Sequence in context: A213048 A349535 A069321 * A211179 A177797 A293717

Adjacent sequences: A396095 A396096 A396097 * A396099 A396100 A396102

Keyword

nonn

Author

Paul D. Hanna, May 29 2026

Status

approved