A007820 -id:A007820 - cp's OEIS frontend

A350366 a(n) = [x^n] Product_{k=1..n} (1 + kx)/(1 - kx).

1, 2, 18, 312, 8000, 271770, 11502162, 583036832, 34437042432, 2322677883330, 176137593178250, 14835018315726312, 1373972097646792800, 138787120025382437882, 15184417945878202716450, 1788809909368939651651200, 225755544056485027686459392

Offset: 0

Seiichi Manyama, Dec 27 2021

nonn

Cf. A007820, A008275, A008277, A298851, A384043, A384044.

a[n_] := Coefficient[Series[Product[(1 + k*x)/(1 - k*x), {k, 1, n}], {x, 0, n}], x, n]; Array[a, 17, 0] (* Amiram Eldar, Dec 27 2021 *)
Table[Sum[(-1)^(n - k)*StirlingS1[n + 1, k + 1] * StirlingS2[k + n, n], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Dec 29 2021 *)

a(n) = sum(k=0, n, (-1)^(n-k)*stirling(n+1, k+1, 1)*stirling(k+n, n, 2));

a(n) = Sum_{k=0..n} (-1)^(n-k) * Stirling1(n+1, k+1) * Stirling2(k+n, n).

a(n) ~ c * d^n * (n-1)!, where d = (1+r) / ((-1 + exp(r + LambertW(-1, -exp(-r)*r))) * LambertW(-exp(-1-r)*(1+r))) = 8.406107401279769476199925123910168..., r = 0.7545302104650497245839827141610818561001159135034... is the root of the equation r*(1 + r + LambertW(-exp(-1 - r)*(1 + r))) = -(1 + r)*(r + LambertW(-1, -exp(-r)*r)) and c = 0.281498742412700978029375818376931142913157133987685... - Vaclav Kotesovec, Dec 29 2021

A350376 a(n) = [x^n] Product_{k=1..n} 1/(1 - k*x)^2.

Original entry on oeis.org

1, 2, 23, 480, 14627, 587580, 29331038, 1750923328, 121673580435, 9648709656300, 859874920598850, 85078769750118144, 9254316901029412110, 1097635452798476278232, 140986468651523106196060, 19496446561112852736019200, 2887977880849714395963280515

Offset: 0

Seiichi Manyama, Dec 27 2021

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Cf. A007820, A008277, A129256, A298851, A350366, A383862, A384060.

a[n_] := Coefficient[Series[Product[1/(1 - k*x)^2, {k, 1, n}], {x, 0, n}], x, n]; Array[a, 17, 0] (* Amiram Eldar, Dec 28 2021 *)
Table[Sum[StirlingS2[n + k, n]*StirlingS2[2*n - k, n], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Dec 29 2021 *)

a(n) = sum(k=0, n, stirling(n+k, n, 2)*stirling(2*n-k, n, 2));

a(n) = Sum_{k=0..n} Stirling2(n+k, n) * Stirling2(2*n-k, n).

a(n) ~ c * d^n * (n-1)!, where d = 27 / (4*LambertW(-3*exp(-3/2)/2)^2 * (3 + 2*LambertW(-3*exp(-3/2)/2))) = 9.858422414446789720857925020919293523149... and c = sqrt(3/(-LambertW(-3*exp(-3/2)/2) * (1 + LambertW(-3*exp(-3/2)/2)))) / (4*Pi) = 0.28482428628793763109169664913715827647091747... - Vaclav Kotesovec, Dec 28 2021, updated May 14 2025

A187655 Self-convolution of the central Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 15, 194, 3631, 89712, 2764268, 102207394, 4411265695, 217707856946, 12092696127691, 746552539553152, 50708165735187572, 3757864633323765824, 301719332111553586612, 26089939284112306045362, 2417245528055399202851119

Offset: 0

Emanuele Munarini, Mar 12 2011

The sequence of the central Stirling numbers of the second kind is 1, 1, 7, 90, 1701,... with offset 0 (see A007820).

Cf. A187656.

seq( add(combinat[stirling2](2*k,k) *combinat[stirling2](2*(n-k),n-k) ,k=0..n), n=0..12);

Table[Sum[StirlingS2[2k, k]StirlingS2[2n - 2k, n - k], {k, 0, n}], {n, 0, 16}]

makelist(sum(stirling2(2*k,k)*stirling2(2*n-2*k,n-k),k,0,n),n,0,12);

a(n) = sum_{k=0..n} A048993(2k,k)*A048993(2n-2k,n-k).

a(n) ~ 2^(2*n+1/2) * n^(n-1/2) / (sqrt(Pi*(1-c)) * exp(n) * (c*(2-c))^n), where c = -LambertW(-2*exp(-2)) = 0.4063757399599599... . - Vaclav Kotesovec, May 20 2014

A242446 a(n) = Sum_{k=1..n} C(n,k) * k^(2*n).

Original entry on oeis.org

1, 18, 924, 93320, 15609240, 3903974592, 1364509038592, 635177480713344, 379867490829555840, 283825251434680651520, 259092157573229145859584, 283735986144895532781391872, 367138254141051794797009309696, 554136240038549806366753446051840

Offset: 1

Vaclav Kotesovec, May 14 2014

Generally, for p>=1, a(n) = Sum_{k=1..n} C(n,k) * k^(p*n) is asymptotic to sqrt(r/(p+r-p*r)) * r^(p*n) * n^(p*n) / (exp(p*n) * (1-r)^n), where r = p/(p+LambertW(p*exp(-p))).

Sum_{k=1..n} (-1)^(n-k) * C(n,k) * k^(p*n) = n! * stirling2(p*n,n).

Cf. A007820, A072034, A242449, A256016.

Table[Sum[Binomial[n,k]*k^(2*n),{k,1,n}],{n,1,20}]

a(n) ~ sqrt(r/(2-r)) * r^(2*n) * n^(2*n) / (exp(2*n) * (1-r)^n), where r = 2/(2+LambertW(2*exp(-2))).

A210029 Number of sequences over the alphabet of n symbols of length 2n which have n distinct symbols. Also number of placements of 2n balls into n cells where no cell is empty.

Original entry on oeis.org

1, 14, 540, 40824, 5103000, 953029440, 248619571200, 86355926616960, 38528927611574400, 21473732319740064000, 14620825330739032204800, 11941607887300551753216000, 11523529003703200697461248000, 12970646659082235068963297280000

Offset: 1

Washington Bomfim, Mar 16 2012

nonn

			a(2) = 14 because the 2^4 sequences on 2 symbols of length 4 can be represented by 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100,1110, and 1111. Only two of them do not have n distinct symbols.
a(10)= 21473732319740064000 since all digits appear in 21473732319740064000 nonnegative integers with 20 digits.
O.g.f.: A(x) = 1 + 14*x + 540*x^2 + 40824*x^3 + 5103000*x^4 + ... where
A(x) = x/(1+x)^2 + 2^4*x^2/(1+4*x)^3 + 3^6*x^3/(1+9*x)^4 + 4^8*x^4/(1+16*x)^5 + 5^10*x^5/(1+25*x)^6 +... - _Paul D. Hanna_, Feb 24 2013
E.g.f.: E(x) = 1 + 14*x + 540*x^2/2! + 40824*x^3/3! + 5103000*x^4/4! + ... where
E(x) = exp(-x)*x + 2^4*exp(-4*x)*x^2/2! + 3^6*exp(-9*x)*x^3/3! + 4^8*exp(-16*x)*x^4/4! + 5^10*exp(-25*x)*x^5/5! +... - _Paul D. Hanna_, Feb 24 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A131689, A210024, A209899, A209900, A007820, A008277.

P := proc(n) option remember; if n = 0 then return 1 fi;
expand(add(binomial(n,k)*P(n-k)*x, k=1..n)) end:
a := n -> coeff(P(2*n), x, n); # Peter Luschny, Sep 11 2019

Table[Sum[((-1)^v*Binomial[n, v]*(n - v)^(2 n)), {v, 0, n - 1}], {n, 20}] (* T. D. Noe, Mar 16 2012 *)

{a(n)=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 26 2012

{a(n)=polcoeff(sum(k=1, n, (k^2)^k*x^k/(1+k^2*x +x*O(x^n))^(k+1)), n)} for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 24 2013

a(n) = Sum_{v=0..n-1}( (-1)^v * binomial(n,v) * (n-v)^(2n) ).
a(n) = n! * S2(2*n,n), where S2(n,k) = A008277(n,k) are the Stirling numbers of the second kind. - Paul D. Hanna, Oct 26 2012 [Also the central column of A131689 (suggesting a(0) = 1). - Peter Luschny, Sep 11 2019]
E.g.f.: Sum_{n>=1} (n^2)^n * exp(-n^2*x) * x^n/n! = Sum_{n>=1} a(n)*x^n/n!. - Paul D. Hanna, Oct 26 2012
a(n) ~ n^(2*n)*(2/(exp(c)*(2-c)))^n / sqrt(1-c), where c = -LambertW(-2/exp(2)) = 0.406375739959959907676958... - Vaclav Kotesovec, Jan 02 2013
O.g.f.: Sum_{n>=1} n^(2*n) * x^n / (1 + n^2*x)^(n+1). - Paul D. Hanna, Feb 24 2013
a(n) = [x^n] P(2*n) where P(n) = Sum_{k=1..n} binomial(n, k)*P(n-k)*x based in P(0) = 1. - Peter Luschny, Sep 11 2019

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

A383862 a(n) = [x^n] Product_{k=0..n} 1/(1 - k*x)^3.

Original entry on oeis.org

1, 3, 48, 1386, 58278, 3225915, 221726711, 18216234288, 1741626159966, 189977753488050, 23285057201978520, 3168272346322892094, 473878954663846060735, 77281168674525142984020, 13647787698908399220563400, 2594721838238358445753776000, 528401900314147344955336365822

Offset: 0

Seiichi Manyama, May 17 2025

nonn

Cf. A007820, A350376, A384022, A384060.

Cf. A384012.

Table[SeriesCoefficient[Product[1/(1-k*x)^3, {k, 1, n}], {x, 0, n}], {n, 0, 16}] (* Vaclav Kotesovec, May 18 2025 *)
(* or *)
Table[Sum[StirlingS2[i + n, n] * StirlingS2[j + n, n] * StirlingS2[2*n - i - j, n], {i, 0, n}, {j, 0, n-i}], {n, 0, 16}] (* Vaclav Kotesovec, May 18 2025 *)

a(n) = sum(i=0, n, sum(j=0, n-i, stirling(i+n, n, 2)*stirling(j+n, n, 2)*stirling(2*n-i-j, n, 2)));

a(n) = Sum_{i, j, k>=0 and i+j+k=n} Stirling2(i+n,n) * Stirling2(j+n,n) * Stirling2(k+n,n).

a(n) ~ 2^(8*n + 3/2) * n^(n - 1/2) / (sqrt(Pi*(1-w)) * exp(n) * 3^(3*n + 3/2) * (4 - 3*w)^n * w^(3*n + 1)), where w = -LambertW(-4*exp(-4/3)/3) = 0.727473355414332993149219573314579663... - Vaclav Kotesovec, May 18 2025

A242449 a(n) = Sum_{k=0..n} C(n,k) * (2k+1)^(2n+1).

Original entry on oeis.org

1, 28, 3612, 1064480, 560632400, 462479403072, 550095467201728, 891290348282967040, 1887146395301619304704, 5058811707344107766328320, 16746136671945501439084657664, 67088193422344140016282100785152, 319900900946743851959321101768511488

Offset: 0

Vaclav Kotesovec, May 14 2014

Generally, for p>=1, a(n) = Sum_{k=0..n} C(n,k) * (p*k+1)^(p*n+1) is asymptotic to n^(p*n+1) * p^(p*n+1) * r^(p*n+3/2+1/p) / (sqrt(p+r-p*r) * exp(p*n) * (1-r)^(n+1/p)), where r = p/(p+LambertW(p*exp(-p))).

G. C. Greubel, Table of n, a(n) for n = 0..190

Cf. A242373, A220955, A242446, A072034, A007820, A221214, A213193, A195242.

Table[Sum[Binomial[n,k]*(2*k+1)^(2*n+1),{k,0,n}],{n,0,20}]

for(n=0,30, print1(sum(k=0,n, binomial(n,k)*(2*k+1)^(2*n+1)), ", ")) \\ G. C. Greubel, Nov 16 2017

a(n) ~ n^(2*n+1) * 2^(2*n+1) * r^(2*n+2) / (sqrt(2-r) * exp(2*n) * (1-r)^(n+1/2)), where r = 2/(2+LambertW(2*exp(-2))) = 0.901829091937052...

A384060 a(n) = [x^n] Product_{k=0..n} 1/(1 - k*x)^4.

Original entry on oeis.org

1, 4, 82, 3024, 162154, 11438280, 1001454024, 104777127616, 12755141675754, 1771354690734420, 276386332002204450, 47870892086756660064, 9113932961179205496744, 1891845220489637114281216, 425240943851497448491619600, 102899751348092720847554016000

Offset: 0

Vaclav Kotesovec, May 18 2025

nonn

In general, for m>=1, [x^n] Product_{k=0..n} 1/(1 - k*x)^m ~ (m+1)^((m+1)*n + (m-1)/2) * n^(n - 1/2) / (sqrt(2*Pi*(1-w)) * exp(n) * (m+1-m*w)^n * m^(m*(n + 1/2)) * w^(m*n + (m-1)/2)), where w = -LambertW(-(m+1)*exp(-(m+1)/m)/m).

The general formula is valid even for m=n, where after modifications we get the formula for A351508.

Cf. A007820 (m=1), A350376 (m=2), A383862 (m=3), A351508 (m=n).

Cf. A384031.

Table[SeriesCoefficient[Product[1/(1-k*x)^4, {k, 1, n}], {x, 0, n}], {n, 0, 15}]

a(n) = Sum_{i, j, k, l>=0 and i+j+k+l=n} Stirling2(i+n,n) * Stirling2(j+n,n) * Stirling2(k+n,n) * Stirling2(l+n,n). - Seiichi Manyama, May 18 2025

a(n) ~ 5^(5*n + 3/2) * n^(n - 1/2) / (sqrt(Pi*(1-w)) * 2^(8*n + 9/2) * exp(n) * (5 - 4*w)^n * w^(4*n + 3/2)), where w = -LambertW(-5*exp(-5/4)/4) = 0.7857872456211833502961937693700363613539172187... - Vaclav Kotesovec, May 18 2025

A218141 a(n) = Stirling2(n^2, n).

Original entry on oeis.org

1, 1, 7, 3025, 171798901, 2436684974110751, 14204422416132896951197888, 50789872166903636182659702516635946082, 155440114706926165785630654089245708839702615196926765, 541500903058656141876322139677626107784896646583041951351456223689104719

Offset: 0

Paul D. Hanna, Oct 21 2012

nonn

			O.g.f.: A(x) = 1 + x + 7*x^2 + 3025*x^3 + 171798901*x^4 + 2436684974110751*x^5 +...

Paul D. Hanna, Table of n, a(n) for n = 0..25

Cf. A008277, A218142, A218143, A007820, A217913, A217914, A217915.

Table[StirlingS2[n^2, n],{n,0,10}] (* Vaclav Kotesovec, May 11 2014 *)

makelist(stirling2(n^2,n),n,0,30 ); /* Martin Ettl, Oct 21 2012 */

{a(n)=polcoeff(sum(k=0,n,(k^n)^k*exp(-k^n*x +x*O(x^n))*x^k/k!),n)}

{a(n)=polcoeff(1/prod(k=1, n, 1-k*x +x*O(x^(n^2+1))), n^2-n)}

{Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
{a(n) = Stirling2(n^2, n)}
for(n=0, 10, print1(a(n), ", "))

a(n) = [x^n] Sum_{k>=0} k^(n*k) * exp(-k^n*x) * x^k / k!.
a(n) = [x^(n^2-n)] 1 / Product_{k=1..n} (1-k*x).
a(n) ~ n^(n^2)/n!. - Vaclav Kotesovec, May 11 2014

cp's OEIS Frontend

A350366 a(n) = [x^n] Product_{k=1..n} (1 + k*x)/(1 - k*x).

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A350376 a(n) = [x^n] Product_{k=1..n} 1/(1 - k*x)^2.

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A187655 Self-convolution of the central Stirling numbers of the second kind.

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A242446 a(n) = Sum_{k=1..n} C(n,k) * k^(2*n).

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A210029 Number of sequences over the alphabet of n symbols of length 2n which have n distinct symbols. Also number of placements of 2n balls into n cells where no cell is empty.

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

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A383862 a(n) = [x^n] Product_{k=0..n} 1/(1 - k*x)^3.

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A242449 a(n) = Sum_{k=0..n} C(n,k) * (2*k+1)^(2*n+1).

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A350366 a(n) = [x^n] Product_{k=1..n} (1 + kx)/(1 - kx).

A242449 a(n) = Sum_{k=0..n} C(n,k) * (2k+1)^(2n+1).