A214645 E.g.f. A(x) satisfies: A'(x) = exp(A(A(x))).

1, 1, 3, 16, 126, 1333, 17895, 293461, 5721390, 129948787, 3384796695, 99848190706, 3301868304168, 121369298328835, 4923587573624940, 219090125559917698, 10637377855875861600, 560928617456424367993, 31993928581562975604588, 1966682218962058310721178

Offset

1,3

Comments

The e.g.f A(x) is divergent according to the answer by Pietro Majer to the MathOverflow question linked below. - Tom Copeland, Jan 16 2017

Links

Paul D. Hanna, Table of n, a(n) for n = 1..160

MathOverflow, How do I solve this: df/dx = exp[f^{(-1)}(x)], answer by Tom Copeland to a MathOverflow question by Zeraoulia Rafik, 2017.

MathOverflow, df/dx = exp[f^{(-1)}(x)] again, answer by Pietro Majer to a MathOverflow question by Fan Zheng, 2017.

Formula

E.g.f. A(x) satisfies:

(1) A''(x) = exp( 2*A(A(x)) + A(A(A(x))) ).

(2) exp(-A(x)) = d/dx Series_Reversion(A(x)).

(3) A(x) = Series_Reversion( Integral exp(-A(x)) dx ).

(4) A(x) = log(F(x)) where F(x) satisfies F( Integral 1/F(x) dx ) = exp(x) and equals the e.g.f. of A233335.

Example

E.g.f.: A(x) = x + x^2/2! + 3*x^3/3! + 16*x^4/4! + 126*x^5/5! + 1333*x^6/6! + ...
Related expansions:
A'(x) = 1 + x + 3*x^2/2! + 16*x^3/3! + 126*x^4/4! + 1333*x^5/5! + ...
A(A(x)) = log(A'(x)) = x + 2*x^2/2! + 9*x^3/3! + 65*x^4/4! + 657*x^5/5! + 8627*x^6/6! + 140433*x^7/7! + 2744360*x^8/8! + 62894577*x^9/9! + ...
The exponential of e.g.f. A(x) equals the e.g.f. of A233335:
exp(A(x)) = 1 + x + 2*x^2/2! + 7*x^3/3! + 38*x^4/4! + 292*x^5/5! + 2975*x^6/6! + 38350*x^7/7! + 604433*x^8/8! + 11351659*x^9/9! + ... + A233335(n)*x^n/n! + ...

Prog

(PARI) {a(n)=local(A=x+x^2); for(i=0, n, A=intformal(exp(subst(A, x, A+x*O(x^n))))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Crossrefs

Cf. A233335 (exp), A259267.

Sequence in context: A193242 A247591 A188805 * A296535 A088358 A082161

Adjacent sequences: A214642 A214643 A214644 * A214646 A214647 A214648

Keyword

nonn

Author

Paul D. Hanna, Jul 24 2012

Status

approved