A276977 -id:A276977 - cp's OEIS frontend

A055505 Numerators in expansion of (1-x)^(-1/x)/e.

1, 1, 11, 7, 2447, 959, 238043, 67223, 559440199, 123377159, 29128857391, 5267725147, 9447595434410813, 1447646915836493, 225037938358318573, 29911565062525361, 3651003047854884043877, 38950782815463986767

Offset: 0

N. J. A. Sloane, Jul 11 2000

From Miklos Kristof, 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_{i>=n} s(i,i-n)/i! where s(n,m) is a Stirling number of the first kind.
exp(1) = 1 + Sum_{i>=1} s(i,i)/i!, 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)

			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, ...

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.1.

G. C. Greubel, Table of n, a(n) for n = 0..250
Markus Brede, On the convergence of the sequence defining Euler's number, Math. Intelligencer, 27, no. 3 (2005), 6-7.
Chao-Ping Chen and Junesang Choi, An Asymptotic Formula for (1+1/x)^x Based on the Partition Function, Amer. Math. Monthly 121 (2014), no. 4, 338--343. MR3183017.
Branko Malesevic, Yue Hu, and Cristinel Mortici, Accurate Estimates of (1+x)^{1/x} Involved in Carleman Inequality and Keller Limit, arXiv:1801.04963 [math.CA], 2018.

Cf. A094638, A130534, A055535 (denominators).
See also A239897/A239898.
Cf. A276977.

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;

a[n_] := Sum[StirlingS1[n+k, k]/(n+k)!*Sum[(-1)^j/j!, {j, 0, n-k}], {k, 0, n}]; Table[a[n] // Numerator // Abs, {n, 0, 17}] (* Jean-François Alcover, Mar 04 2014, after Maple *)
Numerator[((1-x)^(-1/x)/E + O[x]^20)[[3]]] (* or *)
Numerator[Table[Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}] (-1)^n/(2n)!, {n, 0, 10}]] (* Vladimir Reshetnikov, Sep 23 2016 *)

See Maple line for formula.

Edited by N. J. A. Sloane, Jul 01 2008 at the suggestion of R. J. Mathar

A055535 Denominators in expansion of (1-x)^(-1/x)/e.

Original entry on oeis.org

1, 2, 24, 16, 5760, 2304, 580608, 165888, 1393459200, 309657600, 73574645760, 13377208320, 24103053950976000, 3708162146304000, 578473294823424000, 77129772643123200, 9440684171518279680000, 100969884187361280000

Offset: 0

N. J. A. Sloane, Jul 11 2000

Or, equally, denominators in expansion of (1+x)^(1/x)/e.

			(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 + ...).

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.1.

Robert Israel, Table of n, a(n) for n = 0..377
Chao-Ping Chen and Junesang Choi, An Asymptotic Formula for (1+1/x)^x Based on the Partition Function, Amer. Math. Monthly 121 (2014), no. 4, 338--343. MR3183017.

Cf. A094638, A130534, A055505 (numerators), A276977.

G:= (1-x)^(-1/x)/exp(1):
S:= series(G,x,32):
seq(denom(coeff(S,x,j)),j=0..30); # Robert Israel, Sep 23 2016

a[n_] := Sum[StirlingS1[n+k, k]/(n+k)!*Sum[(-1)^j/j!, {j, 0, n-k}], {k, 0, n}]; Table[a[n] // Denominator, {n, 0, 17}] (* Jean-François Alcover, Mar 04 2014 *)
Denominator[((1+x)^(1/x)/E + O[x]^20)[[3]]] (* or *)
Denominator[Table[Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}]/(2n)!, {n, 0, 10}]] (* Vladimir Reshetnikov, Sep 23 2016 *)

From Miklos Kristof, Nov 04 2007 (Start):
(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 A055505, let b(n) be this sequence. Then (1+x)^(1/x) = exp(1)*a(n)/b(n) x^n.
a(n)/b(n) = Sum_{i>=n} s(i,i-n)/i! where s(n,m) is a Stirling number of the first kind.
exp(1) = 1 + Sum_{i>=1} s(i,i)/i! 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)

Edited by N. J. A. Sloane, Jul 25 2008 at the suggestion of R. J. Mathar and Eric Rowland

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A055505 Numerators in expansion of (1-x)^(-1/x)/e.

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