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_Paul D. Hanna (pauldhanna(AT)juno.com), _, Sep 19 2008
proposed
approved
editing
proposed
E.g.f. satisfies: a(n+1) = [x^n/n!] exp(x)*A(x)^(3*n+3).
ILLUSTRATE FORMULA a(n+1) = [x^n/n!] exp(x)*A(x)^(3*n+3) as follows.
Form a table of coefficients of x^k/k! in exp(x)*A(x)^(3*n) for n>=1, k>=0:
exp(x)*A(x)^3: [(1), 4, 34, 685, 27256, 1747159, 159049000, ...];
exp(x)*A(x)^6: [1, (7), 85, 1909, 75193, 4654417, 410053357, ...];
exp(x)*A(x)^9: [1, 10,(154), 3835, 153208, 9284725, 795356632, ...];
exp(x)*A(x)^12:[1, 13, 241, (6625), 272641, 16385713, 1373165425, ...];
exp(x)*A(x)^15:[1, 16, 346, 10441,(446776), 26918851, 2221660936, ...];
exp(x)*A(x)^18:[1, 19, 469, 15445, 690841, (42088609), 3443635405, ...];
exp(x)*A(x)^21:[1, 22, 610, 21799, 1022008, 63371617,(5171653432), ...]; ...
then the terms along the main diagonal form this sequence shift left.
(PARI) {a(n)=local(A=1+x+sum(k=2, n-1, a(k)*x^k/k!)+x*O(x^n)); if(n==0, 1, (n-1)!*polcoeff(exp(x+x*O(x^n))*A^(3*n), n-1))}
approved
editing
E.g.f. satisfies: A(x/A(x)^3) = exp(x).
1, 1, 7, 154, 6625, 446776, 42088609, 5171653432, 792466370161, 146737621112464, 32079442602647521, 8134165156479090136, 2358873194743497162889, 773523517692799444058632, 284154419348084944647780289
0,3
E.g.f. satisfies: A(x) = exp( x*A(log A(x))^3 ).
E.g.f. A(x) = G(3x)^(1/3) where G(x/G(x)) = exp(x) and G(x) is the g.f. of A144681.
E.g.f. A(x) = 1 + x + 7*x^2/2! + 154*x^3/3! + 6625*x^4/4! +...
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(n=0, n, A=exp(serreverse(x/A^3))); n!*polcoeff(A, n)}
nonn
Paul D. Hanna (pauldhanna(AT)juno.com), Sep 19 2008
approved