A336214 -id:A336214 - cp's OEIS frontend

A037966 a(n) = n^2binomial(2n-2, n-1).

0, 1, 8, 54, 320, 1750, 9072, 45276, 219648, 1042470, 4862000, 22355476, 101582208, 457002364, 2038517600, 9026235000, 39710085120, 173712232710, 756088415280, 3276123843300, 14138105520000, 60790319209620, 260516811228960, 1113068351807880, 4742456099097600, 20154752301937500, 85453569951920352

Offset: 0

N. J. A. Sloane

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
Han Mao Kiah, Alexander Vardy, and Hanwen Yao, Efficient Algorithms for the Bee-Identification Problem, arXiv:2212.09952 [cs.IT], 2022.

Cf. A000108, A000984, A037965, A336214.

[0] cat [n^3*Catalan(n-1): n in [1..30]]; // G. C. Greubel, Jun 19 2022

Array[#^2*Binomial[2#-2, #-1] &, 27, 0] (* Michael De Vlieger, Jul 15 2020 *)

{a(n) = n^2*binomial(2*n-2, n-1)} \\ Seiichi Manyama, Jul 15 2020

[n^3*catalan_number(n-1) for n in (0..30)] # G. C. Greubel, Jun 19 2022

a(n) = Sum_{k=0..n} k^2*binomial(n,k)^2. - Paul Barry, Mar 04 2003
a(n) = n^2*A000984(n-1). - Zerinvary Lajos, Jan 18 2007, corrected Jul 26 2015
a(n) = n*A037965(n). - Zerinvary Lajos, Jan 18 2007, corrected Jul 26 2015
(n-1)^3*a(n) = 2*n^2*(2*n-3)*a(n-1). - R. J. Mathar, Jul 26 2015
E.g.f.: x*exp(2*x)*((1 + 2*x)*BesselI(0,2*x) + 2*x*BesselI(1,2*x)). - Ilya Gutkovskiy, Mar 04 2021

More terms from Seiichi Manyama, Jul 15 2020

A336188 a(n) = Sum_{k=0..n} n^k * binomial(n,k)^n.

Original entry on oeis.org

1, 2, 13, 352, 38401, 16971876, 29359436149, 207003074670848, 5679112509686022145, 636468045901197095750500, 277939985126193076692203962501, 494649880078824954885176565423811200, 3447375085398645453825889951638344722092289, 97424105704407389799712313421357308088296084669504

Offset: 0

Seiichi Manyama, Jul 11 2020

Seiichi Manyama, Table of n, a(n) for n = 0..57
Vaclav Kotesovec, Plot of a(n) / (2^((n+1)*(2*n-1)/2) * n^(log(n)/8) / Pi^((n-1)/2)) for n = 1..10000000

Main diagonal of A336187.

Cf. A167010, A187021, A234971, A241247, A336202, A336213, A336214.

[(&+[n^j*Binomial(n,j)^n: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Aug 26 2022

Unprotect[Power]; 0^0 = 1; a[n_] := Sum[n^k * Binomial[n, k]^n, {k, 0, n} ]; Array[a, 14, 0] (* Amiram Eldar, Jul 11 2020 *)

{a(n) = sum(k=0, n, n^k*binomial(n, k)^n)}

[sum(n^j*binomial(n,j)^n for j in (0..n)) for n in (0..20)] # G. C. Greubel, Aug 26 2022

Let f(n) = 2^((n+1)*(2*n-1)/2) * n^(log(n)/8) / Pi^((n-1)/2). For sufficiently large n 0.7675... < a(n)/f(n) < 0.7900... - Vaclav Kotesovec, Jul 11 2020
The above bounds of Vaclav Kotesovec can be recast as: |a(n)/f(n) - exp(-1/4)| <= (3*Pi)^(-2) for sufficiently large n. - Peter Luschny, Jul 12 2020
a(n) ~ exp(-1/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-2), exp(-4)) * 2^(n^2 + n/2) / Pi^(n/2) if n is even and a(n) ~ exp(-3/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-4), exp(-4)) * 2^(n^2 + n/2) * sqrt(n)  / Pi^(n/2) if n is odd. - Vaclav Kotesovec, Jul 13 2020

A336828 a(n) = Sum_{k=0..n} binomial(n,k)^2 * k^n.

Original entry on oeis.org

1, 1, 8, 108, 2144, 56250, 1836792, 71799504, 3269445888, 169974711630, 9934458411800, 644825382429096, 46022332032100800, 3582265183110626740, 302002255041807372080, 27413749834141448520000, 2665789990569658618398720, 276477318687585566522176470

Offset: 0

Ilya Gutkovskiy, Aug 05 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..343

Cf. A000984, A002457, A037966, A037972, A072034, A074334, A187021, A329444, A329913, A336214, A341815.

Join[{1}, Table[Sum[Binomial[n, k]^2 k^n, {k, 0, n}], {n, 1, 17}]]

a(n) = sum(k=0, n, binomial(n, k)^2*k^n); \\ Michel Marcus, Aug 05 2020

a(n) ~ c * d^n * (n-1)!, where d = (1 + 2*LambertW(exp(-1/2)/2)) / (4*LambertW(exp(-1/2)/2)^2) = 6.476217542109791521947605963458797355564... and c = 0.21617818094152997942246965143216887599763501682724844713834495... - Vaclav Kotesovec, Feb 20 2021

A336213 a(n) = Sum_{k=0..n} k^k * binomial(n,k)^n, with a(0)=1.

Original entry on oeis.org

1, 2, 9, 163, 12609, 3906251, 4835455813, 23882051929709, 470073929716006913, 36867039626275056203923, 11562789460238169439667262501, 14393917436542502296957220221339601, 72060131612303615870363237649174605005057, 1424448870088911493303605765206905153730451241313

Offset: 0

Vaclav Kotesovec, Jul 12 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..58
Vaclav Kotesovec, Plot of a(n)/f(n) for n = 1..10000000

Cf. A072034, A086331, A167008, A167010, A336188, A336204, A336212, A336214.

Table[1 + Sum[k^k * Binomial[n, k]^n, {k, 1, n}], {n, 0, 15}]

a(n) = if (n==0, 1, sum(k=0, n, k^k * binomial(n,k)^n)); \\ Michel Marcus, Jul 13 2020

Let f(n) = exp(-1/4) * QPochhammer(exp(-4)) * 2^(n^2 - 1/4) * exp((3*log(n)^2 + 3*log(2)^2 + Pi^2 - 1)/24) * n^((1 - log(2))/4) / Pi^(n/2). For sufficiently large n 0.985... < a(n)/f(n) < 1.015...

a(n) ~ exp(-1/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-1)/2, exp(-4)) * 2^(n^2) / Pi^(n/2) if n is even and a(n) ~ exp(-1/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-3)/2, exp(-4)) * sqrt(n) * 2^(n^2 - 1/2) / Pi^(n/2) if n is odd.

cp's OEIS Frontend

A037966 a(n) = n^2*binomial(2*n-2, n-1).

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A336188 a(n) = Sum_{k=0..n} n^k * binomial(n,k)^n.

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Magma

Mathematica

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A336828 a(n) = Sum_{k=0..n} binomial(n,k)^2 * k^n.

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Mathematica

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Formula

A336213 a(n) = Sum_{k=0..n} k^k * binomial(n,k)^n, with a(0)=1.

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Mathematica

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Formula

A037966 a(n) = n^2binomial(2n-2, n-1).