A191236 -id:A191236 - cp's OEIS frontend

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

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

Offset: 0

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

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

			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

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.

Alois P. Heinz, Table of n, a(n) for n = 0..345 (terms n = 1..200 from Vincenzo Librandi)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
O-Y. Chan and D. V. Manna, Divisibility properties of Stirling numbers of the second kind [From _Jason Kimberley_, Sep 14 2009]

Cf. A002465, A008277, A187646, A191236, A217913, A217914, A217915, A247238.

Cf. A350376, A383862, A384060.

A007820 := proc(n) Stirling2(2*n,n) ; end proc:
seq(A007820(n),n=0..20) ; # R. J. Mathar, Mar 15 2011

Table[StirlingS2[2n, n], {n, 1, 12}] (* Emanuele Munarini, Mar 12 2011 *)

makelist(stirling2(2*n,n),n,0,12); /* Emanuele Munarini, Mar 12 2011 */

a(n)=stirling(2*n,n,2); /* Joerg Arndt, Jul 01 2011 */

{a(n)=polcoeff(1/prod(k=1, n, 1-k*x +x*O(x^(2*n))), n)} \\ Paul D. Hanna, Oct 17 2012

{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

from sympy.functions.combinatorial.numbers import stirling
def A007820(n): return stirling(n<<1,n) # Chai Wah Wu, Jun 09 2025

[stirling_number2(2*i,i) for i in range(1,20)] # Zerinvary Lajos, Jun 26 2008

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

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

Paul D. Hanna, Oct 14 2012

sign

Compare the g.f. to the LambertW identity:

1 = Sum_{n>=0} -(n-1)^(n-1) * exp(-(n-1)*x) * x^n/n!.

			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.

G. C. Greubel, Table of n, a(n) for n = 0..345

Cf. A191236, A217906, A217899, A217900, A217901, A217902, A217903, A217904.

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 *)

{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)}

{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)}

{a(n)=1/n!*sum(k=0,n, -(-1)^(n-k)*binomial(n,k)*k^n*(k-1)^(n-1))}

{a(n)=polcoeff(1-x*(1-x)^(n-1)/prod(k=0, n, 1-k*x +x*O(x^n)), n)}

{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),", "))

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

A256500 Decimal expansion of the positive solution to x = 2*(1-exp(-x)).

Original entry on oeis.org

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

Offset: 1

Stanislav Sykora, Mar 31 2015

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 = 2.

The constant appears in asymptotic formula for A007820. - Vladimir Reshetnikov, Oct 10 2016

			1.5936242600400400923230418758751602417890024248188593649995...

Stanislav Sykora, Table of n, a(n) for n = 1..2000
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 249.
SpectralCalc, Calculation of Blackbody Radiance, Appendix C.
Wikipedia, Planck's law
Index entries for transcendental numbers

Cf. A194567 (q=3), A256501 (q=4), A256502 (q=5).

Cf. A106533, A191236, A217905, A229553.

RealDigits[2 + LambertW[-2 Exp[-2]], 10, 100][[1]] (* Vladimir Reshetnikov, Oct 10 2016 *)

a2=solve(x=0.1,10,x-2*(1-exp(-x))) \\ Use real precision in excess

Equals 2*(1-A106533). - Miko Labalan, Dec 18 2024

Equals log(A229553). - Hugo Pfoertner, Dec 19 2024

A199033 Number of ways to place n non-attacking bishops on a 2 X 2n board.

Original entry on oeis.org

1, 4, 22, 128, 771, 4744, 29618, 186880, 1188679, 7608764, 48953224, 316283264, 2050706932, 13336273528, 86953633242, 568221290496, 3720529001823, 24403423540348, 160314652983158, 1054635453261568, 6946703172803003, 45809043607167328, 302395650703501688

Offset: 0

Vaclav Kotesovec, Nov 02 2011

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
V. Kotesovec, Non-attacking chess pieces

Cf. A002465, A191236, A219197, A183160.

[(&+[Binomial(2*n-j+1,j)*Binomial(n+j+1,n-j): j in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 19 2019

Table[Sum[Binomial[2n-j+1,j]*Binomial[n+j+1,n-j],{j,0,n}],{n,0,25}]

A199033(n):=sum(binomial(n+k+1, n-k)*binomial(2*n-k+1,k),k,0,n)$ makelist(A199033(n),n,0,22); /* Martin Ettl, Nov 15 2012 */

{a(n)=sum(k=0, n, binomial(n+k+1, n-k)*binomial(2*n-k+1, k))}

{a(n)=local(G=1); for(i=0, n, G=1+x*G^3+O(x^(n+1))); polcoeff(G^2/(1-2*x*G^2-3*x^2*G^4), n)} \\ Paul D. Hanna, Nov 14 2012
for(n=0,25,print1(a(n),", "))

[sum(binomial(2*n-j+1,j)*binomial(n+j+1,n-j) for j in (0..n)) for n in (0..30)] # G. C. Greubel, Feb 19 2019

Recurrence: (112*n^4 + 968*n^3 + 3048*n^2 + 4136*n + 2040)*a(n+2) = (728*n^4 + 5914*n^3 + 17550*n^2 + 22510*n + 10530)*a(n+1) + (189*n^4 + 1539*n^3 + 4578*n^2 + 5886*n + 2760)*a(n). - Vaclav Kotesovec, Oct 30 2011
a(n) = Sum_{j=0..n} (binomial(2n-j+1,j)*binomial(n+j+1,n-j)).
a(n) ~ 3^(3n+4)/2^(2n+5)/sqrt(3*Pi*n).
Self-convolution of A219197. - Paul D. Hanna, Nov 14 2012
G.f.: A(x) = G(x)^2 / (1 - 2*x*G(x)^2 - 3*x^2*G(x)^4), where G(x) = 1 + x*G(x)^3 = g.f. of A001764. - Paul D. Hanna, Nov 14 2012
a(n) = [x^n] 1/((1 - x^2)*(1 - x)^(2*n+2)). - Ilya Gutkovskiy, Oct 25 2017

Offset changed to 0 and a(0)=1 added by Paul D. Hanna, Nov 14 2012

A352241 Maximal number of nonattacking black-square queens on an n X n chessboard.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10, 11, 12, 13, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 29, 30, 31, 32, 33, 33, 34, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 43, 44, 44, 45, 46, 47

Offset: 1

George Baloglou, Mar 09 2022

Andy Huchala, Illustration for n = 52.
Andy Huchala, Python program.
Math StackExchange, Black queens on n X n board, 2022.

Cf. A030978, A053757, A190394, A191236, A274616, A274933.

Cf. this sequence (maximal number for black-squares), A352325 (black-squares counts), A352426 (maximal number for white-squares), A352599 (white-squares counts).

Conjecture: a(5k)=4k-1, a(5k+1)=4k, a(5k+2)=4k+1, a(5k+3)=4k+1, a(5k+4)=4k+2. [This does not hold for n = 52 and n = 57. - Andy Huchala, Apr 02 2024]
a(n) = A053757(n-1), at least for 1 <= n <= 12. [This is unlikely to continue. - N. J. A. Sloane, Mar 11 2022] [Indeed the equality does not hold for n=13. - Martin Ehrenstein, Mar 11 2022]
a(n+1) >= a(n); a(2n) = A352426(2n). - Martin Ehrenstein, Mar 23 2022

a(13)-a(26) from Martin Ehrenstein, Mar 11 2022
a(27)-a(28) from Martin Ehrenstein, Mar 15 2022
a(29)-a(30) from Martin Ehrenstein, Mar 23 2022
a(31)-a(60) from Andy Huchala, Mar 27 2024

A274106 Triangle read by rows: T(n,k) = total number of configurations of k nonattacking bishops on the white squares of an n X n chessboard (0 <= k <= n-1+[n=0]).

Original entry on oeis.org

1, 1, 1, 2, 1, 4, 2, 1, 8, 14, 4, 1, 12, 38, 32, 4, 1, 18, 98, 184, 100, 8, 1, 24, 188, 576, 652, 208, 8, 1, 32, 356, 1704, 3532, 2816, 632, 16, 1, 40, 580, 3840, 12052, 16944, 9080, 1280, 16, 1, 50, 940, 8480, 38932, 89256, 93800, 37600, 3856, 32, 1, 60, 1390, 16000, 98292, 322848, 540080, 412800, 116656, 7744, 32

Offset: 0

N. J. A. Sloane, Jun 14 2016

From Eder G. Santos, Dec 16 2024: (Start)
The sequence counts every possible nonattacking configuration of k bishops on the white squares of an n X n chess board.
It is assumed that the n X n chess board has a black square in the upper left corner.
(End)

			Triangle begins:
  1;
  1;
  1,  2;
  1,  4,    2;
  1,  8,   14,     4;
  1, 12,   38,    32,     4;
  1, 18,   98,   184,   100,      8;
  1, 24,  188,   576,   652,    208,      8;
  1, 32,  356,  1704,  3532,   2816,    632,     16;
  1, 40,  580,  3840, 12052,  16944,   9080,   1280,     16;
  1, 50,  940,  8480, 38932,  89256,  93800,  37600,   3856,   32;
  1, 60, 1390, 16000, 98292, 322848, 540080, 412800, 116656, 7744, 32;
  ...
From _Eder G. Santos_, Dec 16 2024: (Start)
For example, for n = 3, k = 2, the T(3,2) = 2 nonattacking configurations are:
  +---+---+---+   +---+---+---+
  |   | B |   |   |   |   |   |
  +---+---+---+   +---+---+---+
  |   |   |   | , | B |   | B |
  +---+---+---+   +---+---+---+
  |   | B |   |   |   |   |   |
  +---+---+---+   +---+---+---+
(End)

Irving Kaplansky and John Riordan, The problem of the rooks and its applications, Duke Mathematical Journal 13.2 (1946): 259-268. See Section 9.
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, in Combinatorics, Duke Mathematical Journal, 13.2 (1946): 259-268. See Section 9. [Annotated scanned copy]
J. Perott, Sur le problème des fous, Bulletin de la S. M. F., tome 11 (1883), pp. 173-186.
Eder G. Santos, Counting non-attacking chess pieces placements: Bishops and Anassas. arXiv:2411.16492 [math.CO], 2024. (considered as black board).
Eric Weisstein's World of Mathematics, White Bishop Graph.

Columns k=0-1 give: A000012, A007590.
Alternate rows give A088960.
Row sums are A216078(n+1).
T(2n,n) gives A191236.
T(2n+1,n) gives A217900(n+1).
T(n+1,n) gives A060546.
Cf. A274105 (black squares), A288182, A201862, A002465.

with(combinat): with(gfun):
T := n -> add(stirling2(n+1,n+1-k)*x^k, k=0..n):
# bishops on white squares
bish := proc(n) local m,k,i,j,t1,t2; global T;
    if n=0 then return [1] fi;
    if (n mod 2) = 0 then m:=n/2;
        t1:=add(binomial(m,k)*T(2*m-1-k)*x^k, k=0..m);
    else
        m:=(n-1)/2;
        t1:=add(binomial(m,k)*T(2*m-k)*x^k, k=0..m+1);
    fi;
    seriestolist(series(t1,x,2*n+1));
end:
for n from 0 to 12 do lprint(bish(n)); od:

T[n_] := Sum[StirlingS2[n+1, n+1-k]*x^k, {k, 0, n}];
bish[n_] := Module[{m, t1, t2}, If[Mod[n, 2] == 0,
   m = n/2;     t1 = Sum[Binomial[m, k]*T[2*m-1-k]*x^k, {k, 0, m}],
   m = (n-1)/2; t1 = Sum[Binomial[m, k]*T[2*m - k]*x^k, {k, 0, m+1}]];
CoefficientList[t1 + O[x]^(2*n+1), x]];
Table[bish[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jul 25 2022, after Maple code *)

def stirling2_negativek(n, k):
  if k < 0: return 0
  else: return stirling_number2(n, k)
print([sum([binomial(floor(n/2), j)*stirling2_negativek(n-j, n-k) for j in [0..k]]) for n in [0..10] for k in [0..n-1+kronecker_delta(n,0)]]) # Eder G. Santos, Dec 01 2024

From Eder G. Santos, Dec 01 2024: (Start)
T(n,k) = Sum_{j=0..k} binomial(floor(n/2),j) * Stirling2(n-j,n-k).
T(n,k) = T(n-1,k) + (n-k+1-A000035(n)) * T(n-1,k-1), T(n,0) = 1, T(0,k) = delta(k,0). (End)

T(0,0) prepended by Eder G. Santos, Dec 01 2024

cp's OEIS Frontend

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

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

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A256500 Decimal expansion of the positive solution to x = 2*(1-exp(-x)).

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Mathematica

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A199033 Number of ways to place n non-attacking bishops on a 2 X 2n board.

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Magma

Mathematica

Maxima

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PARI

Sage

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A352241 Maximal number of nonattacking black-square queens on an n X n chessboard.

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A274106 Triangle read by rows: T(n,k) = total number of configurations of k nonattacking bishops on the white squares of an n X n chessboard (0 <= k <= n-1+[n=0]).

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