OFFSET
0,3
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).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..200
FORMULA
a(n) ~ c * 2^(4*n) * n^(2*n + 1/2) / (Pi^(2*n) * exp(2*n)), where c = 7.23682343848882192289996... - Vaclav Kotesovec, Sep 12 2024
EXAMPLE
G.f.: A(x) = 1 + x + 5*x^2 + 65*x^3 + 1593*x^4 + 61953*x^5 + 3476813*x^6 + 265517441*x^7 + 26492540401*x^8 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
n=1: [1, 1, 5, 65, 1593, 61953, 3476813, ...];
n=2: [1, 2, 11, 140, 3341, 127742, 7097687, ...];
n=3: [1, 3, 18, 226, 5259, 197637, 10869476, ...];
n=4: [1, 4, 26, 324, 7363, 271928, 14799444, ...];
n=5: [1, 5, 35, 435, 9670, 350926, 18895290, ...];
n=6: [1, 6, 45, 560, 12198, 434964, 23165174, ...];
...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*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 - 6*2 + 1*11;
n=3: 0 = 1*1 - 15*3 + 15*18 - 1*226;
n=4: 0 = 1*1 - 28*4 + 70*26 - 28*324 + 1*7363;
n=5: 0 = 1*1 - 45*5 + 210*35 - 210*435 + 45*9670 - 1*350926;
n=6: 0 = 1*1 - 66*6 + 495*45 - 924*560 + 495*12198 - 66*434964 + 1*23165174;
...
The triangle A086645(n,k) = binomial(2*n, 2*k) begins:
n=0: 1;
n=1: 1, 1;
n=2: 1, 6, 1;
n=3: 1, 15, 15, 1;
n=4: 1, 28, 70, 28, 1;
n=5: 1, 45, 210, 210, 45, 1;
n=6: 1, 66, 495, 924, 495, 66, 1;
...
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A-1;
A[m+1] = sum(k=0, m, (-1)^(m-k+1) * binomial(2*m, 2*k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 11 2024
STATUS
approved