A375450 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(2*n-k,k) * ([x^k] A(x)^n) for n >= 1.

1, 1, 2, 9, 75, 960, 17056, 398023, 11785624, 431999096, 19225521917, 1022238356603, 64053139874787, 4673871388801269, 393051019651091208, 37747728969729478069, 4106730533743416782815, 502513668471090178354603, 68716238916399477889072499, 10440633447359638146139853297

Offset: 0

Author

Paul D. Hanna, Sep 11 2024

Keywords

• nonn

Comments

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k,2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 75*x^4 + 960*x^5 + 17056*x^6 + 398023*x^7 + 11785624*x^8 + 431999096*x^9 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  2,   9,  75,  960,  17056, ...];
  n=2: [1, 2,  5,  22, 172, 2106,  36413, ...];
  n=3: [1, 3,  9,  40, 297, 3477,  58412, ...];
  n=4: [1, 4, 14,  64, 457, 5120,  83454, ...];
  n=5: [1, 5, 20,  95, 660, 7091, 112010, ...];
  n=6: [1, 6, 27, 134, 915, 9456, 144632, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(2*n-k,k) * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 3*2 + 1*5;
  n=3: 0 = 1*1 - 5*3 + 6*9 - 1*40;
  n=4: 0 = 1*1 - 7*4 + 15*14 - 10*64 + 1*457;
  n=5: 0 = 1*1 - 9*5 + 28*20 - 35*95 + 15*660 - 1*7091;
  n=6: 0 = 1*1 - 11*6 + 45*27 - 84*134 + 70*915 - 21*9456 + 1*144632;
  ...
The triangle A054142(n,k) = binomial(2*n-k,k) begins:
  n=0: 1;
  n=1: 1,  1;
  n=2: 1,  3,  1;
  n=3: 1,  5,  6,  1;
  n=4: 1,  7, 15, 10,  1;
  n=5: 1,  9, 28, 35, 15,  1;
  n=6: 1, 11, 45, 84, 70, 21, 1;
  ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A054142, A375451.

Programs

PARI

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0);
A[#A] = sum(k=0, #A-1, (-1)^(#A-k) * binomial(2*(#A-1)-k, 1*k) * polcoef(Ser(A)^(#A-1), k) )/(#A-1) ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ c * 2^(2*n) * n^(2*n + 3/2) / (exp(2*n) * Pi^(2*n)), where c = 35.20725926431251936515... - Vaclav Kotesovec, Sep 11 2024