OFFSET
0,3
COMMENTS
Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n,k) * ([x^k] G(x)^n) is satisfied by G(x) = 1/(1-x) for n >= 1.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..200
FORMULA
a(n) ~ c * d^n * n!^2, where d = 0.691660276122579707675... = 4/BesselJZero(0,1)^2 = 4/A115368^2 and c = 3.8999463598998648630203... - Vaclav Kotesovec, Sep 10 2024
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 264*x^4 + 5100*x^5 + 138595*x^6 + 5021209*x^7 + 233863116*x^8 + ...
The table of coefficients of x^k in A(x)^n begins:
n=1: [1, 1, 3, 21, 264, 5100, 138595, ...];
n=2: [1, 2, 7, 48, 579, 10854, 289415, ...];
n=3: [1, 3, 12, 82, 954, 17352, 453657, ...];
n=4: [1, 4, 18, 124, 1399, 24696, 632656, ...];
n=5: [1, 5, 25, 175, 1925, 33001, 827900, ...];
n=6: [1, 6, 33, 236, 2544, 42396, 1041046, ...];
...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(n,k)^2 * ([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 - 9*3 + 9*12 - 1*82;
n=4: 0 = 1*1 - 16*4 + 36*18 - 16*124 + 1*1399;
n=5: 0 = 1*1 - 25*5 + 100*25 - 100*175 + 25*1925 - 1*33001;
n=6: 0 = 1*1 - 36*6 + 225*33 - 400*236 + 225*2544 - 36*42396 + 1*1041046;
...
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(#A-1, k)^2 * 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 10 2024
STATUS
approved