cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A239897 Bisection of A055505.

Original entry on oeis.org

1, 11, 2447, 238043, 559440199, 29128857391, 9447595434410813, 225037938358318573, 3651003047854884043877, 104388909491649724435759747, 1372557084260440289321615059133, 107881945709178295095123859185817, 98682616643700175634367947900986085893
Offset: 0

Views

Author

N. J. A. Sloane, Apr 05 2014

Keywords

Comments

Based on a very good approximation to e.

Examples

			Numerators of the fractions 1, 11/24, 2447/5760, 238043/580608, ... (see A055505/A055535).
		

Crossrefs

Cf. A055505/A055535, A239898 (denominators).

Extensions

More terms from Amiram Eldar, May 08 2024

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

Views

Author

N. J. A. Sloane, Jul 11 2000

Keywords

Comments

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

Examples

			(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 + ...).
		

References

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

Crossrefs

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

Programs

  • Maple
    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
  • Mathematica
    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 *)

Formula

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)

Extensions

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

A276977 Duplicate of A106827.

Original entry on oeis.org

1, 1, 11, 315, 17129, 1510425, 196385475, 35327367075, 8399994587985, 2550903574364145, 963207568455370875, 442613044315692124875, 243195136160954426677305, 157442856285298191126143625, 118607799383105394973766029875, 102867257381973743111023517821875
Offset: 0

Views

Author

Vladimir Reshetnikov, Sep 23 2016

Keywords

Crossrefs

Programs

  • Mathematica
    Table[(-1)^n Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}], {n, 0, 20}]

Formula

a(n) = (-1)^n * Sum_{k=0..n} Stirling1(n+k, k) * !(n-k) * C(2*n, n+k), where !n = A000166(n) is the subfactorial, C(n,k) are binomial coefficients.

A106827 Numerators in expansion of (1 - x)^(-1/x) / e.

Original entry on oeis.org

1, 1, 11, 315, 17129, 1510425, 196385475, 35327367075, 8399994587985, 2550903574364145, 963207568455370875, 442613044315692124875, 243195136160954426677305, 157442856285298191126143625, 118607799383105394973766029875, 102867257381973743111023517821875
Offset: 0

Views

Author

Philippe Deléham, May 21 2005

Keywords

Examples

			G.f. = 1 + 1*x/2! + 11*x^2/4! + 315*x^3/6! + 17129*x^4/8! + 503475*x^5/10! + ...
		

References

  • L. Comtet, Analyse Combinatoire, P. U. F., 1970, tome second, p. 140, #12.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1 . 3 . 1.

Crossrefs

Programs

  • Magma
    m:=31; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (&*[Exp(x^(j-1)/j): j in [2..40]]) )); [Factorial(2*n-2)*b[n]: n in [1..m]]; // G. C. Greubel, Sep 14 2021
    
  • Mathematica
    Table[(-1)^n Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 23 2016 *)
    With[{m=30}, CoefficientList[Series[(1-x)^(-1/x)/E, {x,0,m}], x]*(2*Range[0,m])!] (* G. C. Greubel, Sep 14 2021 *)
  • Sage
    def A_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( product(exp(x^(j-1)/j) for j in (2..41)) ).list()
    A=A_list(40)
    [factorial(2*n)*A[n] for n in (0..31)] # G. C. Greubel, Sep 14 2021

Formula

Sum_{n>=0} a(n)/(2n)!*x^n = (1 - x)^(-1/x) / e.
a(n) = A055505(n)*(2n)! / A055535(n).
a(n) = (-1)^n * Sum_{k=0..n} Stirling1(n+k, k) * !(n-k) * C(2*n, n+k), where !n = A000166(n) is the subfactorial, C(n,k) are binomial coefficients. - Vladimir Reshetnikov, Sep 23 2016
a(n) = (2*n)! * coefficients of Product_{j >= 2} exp(x^(j-1)/j). - G. C. Greubel, Sep 14 2021

Extensions

a(5) corrected by G. C. Greubel, Sep 14 2021

A239898 Bisection of A055535.

Original entry on oeis.org

1, 24, 5760, 580608, 1393459200, 73574645760, 24103053950976000, 578473294823424000, 9440684171518279680000, 271211974879377138647040000, 3579998068407778230140928000000, 282308419108727654719684608000000, 258955866680053703121272297226240000000
Offset: 0

Views

Author

N. J. A. Sloane, Apr 05 2014

Keywords

Comments

Based on a very good approximation to e.

Examples

			Denominators of the fractions 1, 11/24, 2447/5760, 238043/580608, ... (see A055505/A055535).
		

Crossrefs

Cf. A055505/A055535, A239897 (numerators).

Extensions

More terms from Amiram Eldar, May 08 2024
Showing 1-5 of 5 results.