OFFSET
0,3
COMMENTS
In general, we have the following identity:
given A(x) = Sum_{n>=0} a(n)*x^n satisfies
A(x) = 1 + x*Sum_{n>=0} p^n * log( A(q^n*x) )^n / n!,
then a(n+1) = [x^n] A(x)^(p*q^n) for n >= 0, with a(0)=1,
for arbitrary fixed parameters p and q.
Here, p = 2 and q = 3.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..50
FORMULA
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = 1 + x*Sum_{n>=0} 2^n*log( A(3^n*x) )^n / n!.
(2) a(n+1) = [x^n] A(x)^(2*3^n) for n >= 0, with a(0)=1.
EXAMPLE
G.f.: A(x) = 1 + x + 6*x^2 + 261*x^3 + 56070*x^4 + 56526498*x^5 + 334429044030*x^6 + 15777272891508021*x^7 + 6500948711591606135796*x^8 + ...
where
A(x) = 1 + x*[1 + 2*log(A(3*x)) + 2^2*log(A(3^2*x))^2/2! + 2^3*log(A(3^2*x))^3/3! + ... + 2^n*log(A(3^n*x))^n/n! + ...].
RELATED SERIES.
log(A(x)) = x + 11*x^2/2 + 766*x^3/3 + 223187*x^4/4 + 282345766*x^5/5 + 2006233236098*x^6/6 + 110438567161208518*x^7/7 + ...
RELATED TABLE.
The table of coefficients of x^k in A(x)^(2*3^n) begins:
n=0: [1, 2, 13, 534, 112698, 113168268, ...];
n=1: [1, 6, 51, 1766, 345165, 340906254, ...];
n=2: [1, 18, 261, 7350, 1112382, 1035922644, ...];
n=3: [1, 54, 1755, 56070, 4589001, 3250238022, ...];
n=4: [1, 162, 14013, 894294, 56526498, 12817431900, ...];
n=5: [1, 486, 120771, 20555046, 2731197285, 334429044030, ...]; ...
in which the main diagonal equals this sequence shift left,
illustrating that a(n+1) = [x^n] A(x)^(2*3^n) for n >= 0.
PROG
(PARI) {a(n) = my(A=[1, 1]); for(i=1, n, A=concat(A, Vec(Ser(A)^(2*3^(#A-1)))[ #A])); A[n+1]}
for(n=0, 15, print1(a(n), ", "))
(PARI) {a(n) = my(A=1+x); for(i=1, n, A = 1 + x*sum(m=0, #A, 2^m*log( subst(Ser(A), x, 3^m*x +x*O(x^n)))^m/m!) ); polcoeff(A, n)}
for(n=0, 15, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 16 2023
STATUS
approved