OFFSET
0,3
COMMENTS
Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+2*k, 3*k) * ([x^k] D(x)^n) for n >= 1 is satisfied by the function D(x) = 1 + x*D(x)^3 (g.f. of A001764), where coefficient [x^k] D(x)^n = binomial(n+3*k-1, k)*n/(n+2*k).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..200
FORMULA
a(n) ~ c * d^n * n!^3 * n^alpha, where d = 0.1579852929267375678916376580224..., alpha = 2.6601429516008505168108..., c = 0.86048778713891683578001... - Vaclav Kotesovec, Sep 12 2024
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 27*x^3 + 520*x^4 + 17461*x^5 + 924123*x^6 + 72565316*x^7 + 8092491188*x^8 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
n=1: [1, 1, 3, 27, 520, 17461, 924123, ...];
n=2: [1, 2, 7, 60, 1103, 36124, 1887017, ...];
n=3: [1, 3, 12, 100, 1758, 56097, 2890755, ...];
n=4: [1, 4, 18, 148, 2495, 77500, 3937572, ...];
n=5: [1, 5, 25, 205, 3325, 100466, 5029880, ...];
n=6: [1, 6, 33, 272, 4260, 125142, 6170284, ...];
...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(3*n-2*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 - 4*2 + 1*7;
n=3: 0 = 1*1 - 7*3 + 10*12 - 1*100;
n=4: 0 = 1*1 - 10*4 + 28*18 - 20*148 + 1*2495;
n=5: 0 = 1*1 - 13*5 + 55*25 - 84*205 + 35*3325 - 1*100466;
n=6: 0 = 1*1 - 16*6 + 91*33 - 220*272 + 210*4260 - 56*125142 + 1*6170284;
...
The triangle A193636(n,k) = binomial(3*n-2*k, k) begins:
n=0: 1;
n=1: 1, 1;
n=2: 1, 4, 1;
n=3: 1, 7, 10, 1;
n=4: 1, 10, 28, 20, 1;
n=5: 1, 13, 55, 84, 35, 1;
n=6: 1, 16, 91, 220, 210, 56, 1;
...
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A-1;
A[#A] = sum(k=0, m, (-1)^(m-k+1) * binomial(3*m-2*k, 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