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 1-7 of 7 results.

A007820 Stirling numbers of second kind S(2n,n).

Original entry on oeis.org

1, 1, 7, 90, 1701, 42525, 1323652, 49329280, 2141764053, 106175395755, 5917584964655, 366282500870286, 24930204590758260, 1850568574253550060, 148782988064375309400, 12879868072770626040000, 1194461517469807833782085, 118144018577011378596484455
Offset: 0

Views

Author

kemp(AT)sads.informatik.uni-frankfurt.de (Rainer Kemp)

Keywords

Comments

Chan and Manna prove that a(n) is odd if and only if n is in A003714. - Jason Kimberley, Sep 14 2009
The number of ways to partition a set of 2*n elements into n disjoint subsets. - Vladimir Reshetnikov, Oct 10 2016
Conjecture: a(2*n+1) is divisible by (2*n + 1)^2. - Peter Bala, Mar 30 2025

Examples

			G.f.: A(x) = 1 + x + 7*x^2 + 90*x^3 + 1701*x^4 + 42525*x^5 +...,
where A(x) = 1 + 1^2*x*exp(-1*x) + 2^4*exp(-2^2*x)*x^2/2! + 3^6*exp(-3^2*x)*x^3/3! + 4^8*exp(-4^2*x)*x^4/4! + 5^10*exp(-5^2*x)*x^5/5! + ... - _Paul D. Hanna_, Oct 17 2012
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.

Crossrefs

Programs

  • Maple
    A007820 := proc(n) Stirling2(2*n,n) ; end proc:
    seq(A007820(n),n=0..20) ; # R. J. Mathar, Mar 15 2011
  • Mathematica
    Table[StirlingS2[2n, n], {n, 1, 12}] (* Emanuele Munarini, Mar 12 2011 *)
  • Maxima
    makelist(stirling2(2*n,n),n,0,12); /* Emanuele Munarini, Mar 12 2011 */
    
  • PARI
    a(n)=stirling(2*n,n,2); /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    {a(n)=polcoeff(1/prod(k=1, n, 1-k*x +x*O(x^(2*n))), n)} \\ Paul D. Hanna, Oct 17 2012
    
  • PARI
    {a(n)=polcoeff(sum(m=1,n,(m^2)^m*exp(-m^2*x+x*O(x^n))*x^m/m!),n)} \\ Paul D. Hanna, Oct 17 2012
    
  • Python
    from sympy.functions.combinatorial.numbers import stirling
    def A007820(n): return stirling(n<<1,n) # Chai Wah Wu, Jun 09 2025
  • Sage
    [stirling_number2(2*i,i) for i in range(1,20)] # Zerinvary Lajos, Jun 26 2008
    

Formula

a(n) = A048993(2n,n). - R. J. Mathar, Mar 15 2011
Asymptotic: a(n) ~ (4*n/(e*z*(2-z)))^n/sqrt(2*Pi*n*(z-1)), where z = A256500 = 1.59362426... is a root of the equation exp(z)*(2-z)=2. - Vaclav Kotesovec, May 30 2011
a(n) = 1/n! * Sum_{k = 0..n} binomial(n,k)*(-1)^k*(n-k)^(2*n). - Emanuele Munarini, Jul 01 2011
a(n) = [x^n] 1 / Product_{k=1..n} (1-k*x). - Paul D. Hanna, Oct 17 2012
O.g.f.: Sum_{n>=1} (n^2)^n * exp(-n^2*x) * x^n/n! = Sum_{n>=1} S2(2*n,n)*x^n. - Paul D. Hanna, Oct 17 2012
G.f.: Sum_{n > 0} (a(n)*n!/(2*n)!)*x^n = x*B'(x)/B(x)-1, where B(x) satisfies B(x)^2 = x*(exp(B(x))-1). - Vladimir Kruchinin, Mar 13 2013
a(n) = Sum_{j = 0..n} (-1)^(n-j)*n^j*binomial(2*n,j)*stirling2(2*n-j,n). - Vladimir Kruchinin, Jun 14 2013

Extensions

Typo in Mathematica program fixed by Vincenzo Librandi, May 04 2013
a(0)=1 prepended by Alois P. Heinz, Feb 01 2018

A217905 O.g.f.: Sum_{n>=0} -n^n*(n-1)^(n-1) * exp(-n*(n-1)*x) * x^n / n!.

Original entry on oeis.org

1, -1, -2, -14, -184, -3532, -89256, -2800016, -104967808, -4578528464, -227816059360, -12735645181536, -790296855912576, -53905019035510528, -4008716449677965312, -322807879692969879552, -27983800239966141382656, -2598368754552749176202496, -257284990746988090769530368
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2012

Keywords

Comments

Compare the g.f. to the LambertW identity:
1 = Sum_{n>=0} -(n-1)^(n-1) * exp(-(n-1)*x) * x^n/n!.

Examples

			O.g.f.: A(x) = 1 - x - 2*x^2 - 14*x^3 - 184*x^4 - 3532*x^5 - 89256*x^6 +...
where
A(x) = 1 - 1^1*0^0*x*exp(-1*0*x) - 2^2*1^1*exp(-2*1*x)*x^2/2! - 3^3*2^2*exp(-3*2*x)*x^3/3! - 4^4*3^3*exp(-4*3*x)*x^4/4! - 5^5*4^4*exp(-5*4*x)*x^5/5! +...
simplifies to a power series in x with integer coefficients.
		

Crossrefs

Programs

  • Mathematica
    Join[{1, -1}, Table[(1/n!)*Sum[(-1)^(n - k + 1)*Binomial[n, k]*k^n*(k - 1)^(n - 1), {k, 0, n}], {n, 2, 50}]] (* G. C. Greubel, Nov 16 2017 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,-m^m*(m-1)^(m-1)*x^m*exp(-m*(m-1)*x+x*O(x^n))/m!),n)}
    
  • PARI
    {a(n)=(1/n!)*polcoeff(sum(k=0, n, -k^k*(k-1)^(k-1)*x^k/(1+k*(k-1)*x +x*O(x^n))^(k+1)), n)}
    
  • PARI
    {a(n)=1/n!*sum(k=0,n, -(-1)^(n-k)*binomial(n,k)*k^n*(k-1)^(n-1))}
    
  • PARI
    {a(n)=polcoeff(1-x*(1-x)^(n-1)/prod(k=0, n, 1-k*x +x*O(x^n)), n)}
    
  • PARI
    {a(n)=polcoeff(1-x*(1+x)^n/prod(k=0, n, 1-(k-1)*x +x*O(x^n)), n)}
    for(n=0,30,print1(a(n),", "))

Formula

a(n) = -A191236(n-1) for n>=1. [corrected by Vaclav Kotesovec, Aug 22 2018]
a(n) = 1/n! * Sum_{k=0..n} -(-1)^(n-k)*binomial(n,k) * k^n * (k-1)^(n-1) for n>=0.
a(n) = 1/n! * [x^n] Sum_{k>=0} -k^k*(k-1)^(k-1)*x^k / (1 + k*(k-1)*x)^(k+1).
a(n) = [x^n] 1 - x*(1-x)^(n-1) / Product_{k=1..n} (1-k*x).
a(n) = [x^n] 1 - x*(1+x)^(n-1) / Product_{k=1..n} (1-(k-1)*x).
a(n) ~ -2^(n-1) * exp(n*(r-1)-r) * n^(n - 3/2) / (sqrt(Pi*(r-1)*(2-r)) * r^(n-1)), where r = 2 + LambertW(-2*exp(-2)) = A256500 = 1.5936242600400400923230418... - Vaclav Kotesovec, Aug 22 2018

A106533 The rumor constant: decimal expansion of the number x defined by x*e^(2 - 2*x) = 1.

Original entry on oeis.org

2, 0, 3, 1, 8, 7, 8, 6, 9, 9, 7, 9, 9, 7, 9, 9, 5, 3, 8, 3, 8, 4, 7, 9, 0, 6, 2, 0, 6, 2, 4, 1, 9, 8, 7, 9, 1, 0, 5, 4, 9, 8, 7, 8, 7, 5, 9, 0, 5, 7, 0, 3, 1, 7, 5, 0, 0, 2, 4, 7, 7, 4, 4, 1, 5, 1, 9, 5, 7, 5, 0, 7, 5, 9, 1, 9, 0, 6, 0, 2, 4, 8, 8, 3, 6, 2, 5, 0, 3, 6, 1, 6, 9, 0, 7, 7, 9, 6, 4, 2, 9, 1, 4, 6, 9
Offset: 0

Views

Author

Robert G. Wilson v, May 03 2005

Keywords

Examples

			c = 0.20318786997997995383847906206241987910549878759057031750024774...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[ -ProductLog[ -2/E^2]/2, 10, 111][[1]]
    RealDigits[x/.FindRoot[x E^(2-2x)==1,{x,2},WorkingPrecision->120]][[1]] (* Harvey P. Dale, Jul 05 2025 *)
  • PARI
    solve(x=0, 0.5, x*exp(2-2*x)-1) \\ Michel Marcus, Mar 13 2016

Formula

Solution to x*exp(2 - 2*x) = 1 with x not equal to 1.
Equals -1/2*LambertW(-2*exp(-2)). - Vladeta Jovovic, May 30 2005
Constant c satisfies: exp(c*x)/(1-2*c) = Sum_{n>=0} (x + 2*n)^n * exp(-2*n)/n!. - Paul D. Hanna, Mar 12 2016
Equals (2-A256500)/2. - Miko Labalan, Dec 18 2024

A191236 Number of ways to place n nonattacking bishops on black squares of a 2n X 2n board.

Original entry on oeis.org

1, 2, 14, 184, 3532, 89256, 2800016, 104967808, 4578528464, 227816059360, 12735645181536, 790296855912576, 53905019035510528, 4008716449677965312, 322807879692969879552, 27983800239966141382656
Offset: 0

Views

Author

Vaclav Kotesovec, May 27 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Join[{1}, Table[(1/n!)*Sum[(-1)^(n - k)*Binomial[n, k]*(k*(k + 1))^n, {k, 0, n}], {n,1,50}]] (* G. C. Greubel, Feb 03 2017 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m^m*(m+1)^m*x^m*exp(-m*(m+1)*x+x*O(x^n))/m!),n)} \\ Paul D. Hanna, Oct 15 2012
    
  • PARI
    {a(n)=sum(k=0,n, binomial(n,k) * stirling(2*n-k,n,2))} \\ Paul D. Hanna, Nov 13 2012

Formula

a(n) = 1/n! * Sum_{j=0..n} (-1)^(n-j) * binomial(n,j) * (j*(j+1))^n.
Asymptotic: a(n) ~ 1/sqrt(Pi*(z-1)*(2-z)*n)*(2*n*exp(z-1)/z)^n or a(n) ~ exp(z/2)*Stirling2(2*n,n) where z = A256500 = 1.59362426... is a root of the equation exp(z)*(2-z)=2.
O.g.f.: Sum_{n>=0} n^n*(n+1)^n * exp(-n*(n+1)*x) * x^n/n! = Sum_{n>=0} a(n)*x^n. - Paul D. Hanna, Oct 15 2012
a(n) = Sum_{k=0..n} binomial(n,k) * Stirling2(2*n-k,n), where Stirling2(n,k) = A008277(n,k). - Paul D. Hanna, Nov 13 2012

Extensions

Offset changed to 0 and a(0)=1 added by Paul D. Hanna, Nov 13 2012

A094090 Decimal expansion of positive solution to 5*(1-exp(u)) + u*exp(u) = 0.

Original entry on oeis.org

4, 9, 6, 5, 1, 1, 4, 2, 3, 1, 7, 4, 4, 2, 7, 6, 3, 0, 3, 6, 9, 8, 7, 5, 9, 1, 3, 1, 3, 2, 2, 8, 9, 3, 9, 4, 4, 0, 5, 5, 5, 8, 4, 9, 8, 6, 7, 9, 7, 2, 5, 0, 9, 7, 2, 8, 1, 4, 4, 4, 6, 1, 4, 4, 7, 8, 0, 4, 6, 3, 9, 8, 7, 9, 5, 7, 4, 5, 2, 9, 7, 2, 2, 3, 8, 2, 7, 0, 4, 5, 0, 6, 6, 0, 0, 0, 9, 6, 0, 8, 2, 9, 7, 7, 6
Offset: 1

Views

Author

Jeppe Stig Nielsen, May 01 2004

Keywords

Comments

This purely mathematical constant turns up when in physics one derives Wien's displacement law from the Planck black-body radiation law (see link).
Positive solution to x = 5*(1-exp(-x)). More comments in A256500. - Stanislav Sykora, Apr 01 2015

Examples

			u=4.965114231744276...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[5 + ProductLog[ -5/E^5], 10, 120][[1]] (* Robert G. Wilson v, May 04 2004 *)
  • PARI
    a5=solve(x=0.1, 10, x-5*(1-exp(-x))) \\ Use real precision in excess

Formula

u = 5 + W(-5*exp(-5)), where W() is Lambert's W-function.

Extensions

More terms from Robert G. Wilson v, May 04 2004

A194567 Decimal expansion of the positive solution to x = 3*(1-exp(-x)).

Original entry on oeis.org

2, 8, 2, 1, 4, 3, 9, 3, 7, 2, 1, 2, 2, 0, 7, 8, 8, 9, 3, 4, 0, 3, 1, 9, 1, 3, 3, 0, 2, 9, 4, 4, 8, 5, 1, 9, 5, 3, 4, 5, 8, 8, 1, 7, 4, 4, 0, 7, 3, 1, 1, 4, 0, 9, 2, 2, 7, 9, 8, 5, 7, 6, 9, 3, 9, 4, 1, 2, 1, 4, 3, 0, 4, 5, 0, 5, 5, 1, 7, 3, 9, 1, 2, 4, 5, 6, 8, 6, 5, 6, 5, 3, 4, 7, 8, 3, 9, 6, 4, 4, 3, 8, 9, 5, 9
Offset: 1

Views

Author

Jean-François Alcover, Aug 29 2011

Keywords

Comments

The positive solution to x=3*(1-exp(-x)) is the dimensionless coefficient corresponding to the maximum brightness in Planck's law of radiation.
It can be symbolically expressed as 3+W(-3/e^3), where W stands for Lambert (a.k.a. "ProductLog") function.

Examples

			2.821439372...
		

Crossrefs

Programs

  • Maple
    evalf(3+LambertW(-3/exp(3)), 130);  # Alois P. Heinz, May 08 2024
  • Mathematica
    RealDigits[ N[x /. ToRules[ Reduce[x > 0 && x == 3*(1 - E^-x), x, Reals]], 100]][[1]]
    RealDigits[3 + ProductLog[-3/E^3], 10, 111][[1]] (* Robert G. Wilson v, Oct 16 2013 *)
    RealDigits[x/.FindRoot[x==3(1-Exp[-x]),{x,2},WorkingPrecision->120]][[1]] (* Harvey P. Dale, Aug 09 2023 *)
  • PARI
    a3=solve(x=0.1,10,x-3*(1-exp(-x))) \\ Use real precision in excess
    
  • PARI
    3+lambertw(-3/exp(3)) \\ Charles R Greathouse IV, Sep 13 2022

A256501 Decimal expansion of the positive solution to x = 4*(1-exp(-x)).

Original entry on oeis.org

3, 9, 2, 0, 6, 9, 0, 3, 9, 4, 8, 7, 2, 8, 8, 6, 3, 4, 3, 5, 6, 0, 8, 9, 1, 3, 5, 2, 6, 1, 3, 5, 3, 6, 2, 2, 0, 5, 2, 5, 6, 2, 7, 3, 7, 1, 2, 0, 7, 9, 8, 4, 5, 3, 0, 4, 0, 1, 1, 7, 5, 0, 0, 5, 7, 9, 0, 5, 0, 5, 6, 4, 8, 3, 6, 6, 7, 0, 5, 7, 5, 7, 4, 3, 3, 6, 5, 6, 6, 0, 1, 8, 9, 4, 8, 3, 6, 5, 8, 9, 0, 4, 7, 3, 0
Offset: 1

Views

Author

Stanislav Sykora, Apr 01 2015

Keywords

Comments

Each of the positive solutions to x = q*(1-exp(-x)) obtained for q = 2, 3, 4, and 5, appears in several formulas pertinent to Planck's black-body radiation law. For a given q, the solution can be also written as q+W(-q/exp(q)), where W is the Lambert function. Here q = 4.

Examples

			3.9206903948728863435608913526135362205256273712079845304011750...
		

Crossrefs

Cf. A094090 (q=5), A194567 (q=3), A256500 (q=2).

Programs

  • Mathematica
    RealDigits[x/.FindRoot[x==4(1-Exp[-x]),{x,3},WorkingPrecision->120]] [[1]] (* Harvey P. Dale, May 08 2017 *)
  • PARI
    a4=solve(x=0.1, 10, x-4*(1-exp(-x))) \\ Use real precision in excess
Showing 1-7 of 7 results.