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A156100 G.f.: A(x) = exp( Sum_{n>=1} (1 + 2^n*x)^n * x^n/n ). 5
1, 1, 3, 7, 25, 113, 741, 7181, 101139, 2089283, 61683087, 2600572391, 156100460443, 13231060891179, 1594932996895155, 270715422001769667, 65209448673400087945, 22130613779988110245993, 10631829612570393072035829, 7207580557759524950136903565, 6902254922678483464065364019049, 9340558535943272871301176019398265, 17831418294195720284498112713266643601 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Compare to g.f. exp( Sum_{m>=1} (1 + x)^m * x^m/m ) of the Fibonacci sequence.

Conjecture: a(n)^(1/n^2) tends to 2^(1/4). - Vaclav Kotesovec, Oct 17 2020

LINKS

Paul D. Hanna, Table of n, a(n) for n = 0..150

FORMULA

G.f.: A(x) = exp(F(x)) where F(x) is the l.g.f. of A156101.

EXAMPLE

G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 25*x^4 + 113*x^5 + 741*x^6 +...

log(A(x)) = (1 + 2*x)*x + (1 + 2^2*x)^2*x^2/2 + (1 + 2^3*x)^3*x^3/3 +...

log(A(x)) = x + 5*x^2/2 + 13*x^3/3 + 65*x^4/4 + 401*x^5/5 + 3521*x^6/6 +...

PROG

(PARI) {a(n)=polcoeff(exp(sum(m=1, n+1, (1+2^m*x)^m*x^m/m)+x*O(x^n)), n)}

for(n=0, 25, print1(a(n), ", "))

CROSSREFS

Cf. A156101 (log), A155810.

Sequence in context: A133206 A054092 A096648 * A245115 A215772 A019056

Adjacent sequences:  A156097 A156098 A156099 * A156101 A156102 A156103

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Feb 04 2009

STATUS

approved

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Last modified April 10 07:36 EDT 2021. Contains 342843 sequences. (Running on oeis4.)