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^(2*n) * n^(2*n + 3/2) / (exp(2*n) * Pi^(2*n)), where c = 35.20725926431251936515... - Vaclav Kotesovec, Sep 11 2024
EXAMPLE
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;
...
PROG
(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), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 11 2024
STATUS
approved