0,3

Comments from Miklos Kristof (kristmikl(AT)freemail.hu), Nov 04 2007: (Start) This is also the sequence of numerators associated with expansion of (1+x)^(1/x).

(1 + x)^(1/x)=exp(1)*(1 - 1/2*x + 11/24*x^2 - 7/16*x^3 + 2447/5760*x^4 - 959/2304*x^5 + 238043/580608*x^6 - ...).

(1+x)^(1/x)=exp(log(1+x)/x)=exp(1)*exp(-x/2)*exp(x^2/3)*exp(x^3/4)*...

Let a(n) be this sequence, let b(n) be A055535. Then (1+x)^(1/x)=exp(1)*a(n)/b(n) x^n.

a(n)/b(n) = sum(s(i,i-n)/(i !), i=n,...,infinity),... where s(n,m) is a Stirling number of the first kind.

exp(1)=1+sum(s(i,i)/i !, i=1,... infinity), for the n=1 case.

a(1)/b(1)=1/1 because 1+1/1!+1/2!+1/3!+1/4!+...=exp(1)

a(2)/b(2)=1/2 because 1/2!+3/3!+6/4!+10/5!+...=1/2*exp(1)

a(3)/b(3)=11/24 because 2/3!+11/4!+35/5!+85/6!+...=11/24*exp(1)

a(4)/b(4)=7/16 because 6/4!+50/5!+225/6!+735/7!+...=7/16*exp(1) (End)

M. Brede, On the convergence of the sequence defining Euler's number, Math. Intelligencer, 20 (1998), 25-29.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.1.

See Maple line for formula.

1+1/2*x+11/24*x^2+7/16*x^3+2447/5760*x^4+...

1, -1/2, 11/24, -7/16, 2447/5760, -959/2304, 238043/580608, -67223/165888, ...

T:=proc(u) local k, l; add( stirling1(u+k, k)*((u+k)!)^(-1)* add( (-1)^l/l!, l=0..u-k), k=0..u); end;

Cf. A055535, A094638, A130534, A055535.

Sequence in context: A060954 A038321 A002749 this_sequence A090841 A085757 A003567

Adjacent sequences: A055502 A055503 A055504 this_sequence A055506 A055507 A055508

nonn,frac,new

njas, Jul 11 2000

Edited by njas, Jul 01 2008 at the suggestion of R. J. Mathar