A036740 -id:A036740 - cp's OEIS frontend

A127488 a(n) = (n^2)!/(2*(n!)).

6, 30240, 435891456000, 64630041847212441600000, 258328699159653623241666283438080000000

Offset: 2

Artur Jasinski, Jan 16 2007

nonn

Cf. A000142, A001044, A000442, A036740, A010050, A009445, A134366, A134367, A134368, A134369, A134371, A134372, A134373, A134374, A134374

Mathematica
```
Table[(n^2)!/(2(n!)), {n, 2, 6}]
```

a(n) ~ n^(2*n^2 - n + 1/2) / (2 * exp(n*(n-1))). - Vaclav Kotesovec, Oct 26 2017

A181760 a(n) = (n!)(n!-1)(n!-2)...(n!-n+1).

Original entry on oeis.org

1, 1, 2, 120, 255024, 22869362880, 136434451994755200, 82262786502445667337542400, 6980114960816118346901632738195814400, 109099864394915605737486658299863377337267988480000, 395935956167605557454071116707328675502625329271836386079338496000

Offset: 0

Geoffrey Critzer

nonn

a(n) is the number of n X n matrices such that each row of the matrix is a different permutation of {1,2,...n}.

Cf. A036740.

a:= n-> mul(n!-k, k=0..n-1):
seq(a(n), n=0..10);  # Alois P. Heinz, Jan 17 2011

Mathematica
```
Table[FactorialPower[n!,n],{n,0,10}]
```

a(n) ~ (2*Pi)^(n/2) * n^(n*(2*n+1)/2) / exp(n^2-1/12). - Vaclav Kotesovec, Oct 26 2017

A213065 a(n) = n^n! - n!^n.

Original entry on oeis.org

-1, 0, 0, 513, 281474976378880, 752316384526264005099991383822237233803945956334136013765601092018187046026142190625

Offset: 0

Pratik Jain, Jun 04 2012

sign

a(6) has 561 digits: 1.8573779... * 10^560.

Cf. A036740, A053986.

Table[n^n! - n!^n, {n, 0, 4}] (* T. D. Noe, Jun 04 2012 *)

a(n) = n^n! - n!^n \\ Stefano Spezia, Aug 02 2025

a(5) from Stefano Spezia, Aug 02 2025

A336249 a(n) = (n!)^n * Sum_{k=0..n} 1 / ((k!)^n * (n-k)!).

Original entry on oeis.org

1, 2, 7, 172, 79745, 1375363126, 1445639634946657, 136511607703654177490168, 1597074319746489837872943936307201, 3049096207067719868011671739966873049880826186, 1209808678412717193052533393657339738066086793611743000000001

Offset: 0

Ilya Gutkovskiy, Jul 14 2020

nonn

Cf. A000079, A002720, A036740, A119400.

Table[(n!)^n Sum[1/((k!)^n (n - k)!), {k, 0, n}], {n, 0, 10}]
Table[(n!)^n SeriesCoefficient[Exp[x] Sum[x^k/(k!)^n, {k, 0, n}], {x, 0, n}], {n, 0, 10}]

a(n) = (n!)^n * sum(k=0, n, 1 / ((k!)^n * (n-k)!)); \\ Michel Marcus, Jul 14 2020

a(n) = (n!)^n * [x^n] exp(x) * Sum_{k>=0} x^k / (k!)^n.

a(n) ~ (2*Pi)^((n-1)/2) * n^(n^2 - n/2 + 1/2) / exp(n*(n-1) - 1/12). - Vaclav Kotesovec, Jul 14 2020

A336250 a(n) = (n!)^n * Sum_{k=1..n} (-1)^(k+1) / k^n.

Original entry on oeis.org

0, 1, 3, 197, 313840, 24191662624, 137300308036448256, 81994640912971156525105152, 6958651785463110878359050928999366656, 108902755985567407887534498777329973193771818418176, 395560567918154447056086270973712023435510589158871531520000000000

Offset: 0

Ilya Gutkovskiy, Jul 14 2020

nonn

Cf. A024167, A036740, A060943, A142999.

Table[(n!)^n Sum[(-1)^(k + 1)/k^n, {k, 1, n}], {n, 0, 10}]
Table[(n!)^n SeriesCoefficient[-PolyLog[n, -x]/(1 - x), {x, 0, n}], {n, 0, 10}]

a(n) = (n!)^n * sum(k=1, n, (-1)^(k+1) / k^n); \\ Michel Marcus, Jul 14 2020

a(n) = (n!)^n * [x^n] -polylog(n,-x) / (1 - x).

A360213 Number of distinct stable marriage problem instances up to gender exchange.

Original entry on oeis.org

1, 10, 23436, 55037822976, 309586821132441600000, 9704204980882671472665034752000000, 3411909590124519376908837990487929799751761920000000, 24394862766922609598505096548473341484170343775734092352694570188800000000

Offset: 1

Dan Eilers, Jan 29 2023

nonn

In the Stable Marriage Problem, the men's and women's preference lists can be swapped without affecting the number of blocking pairs or stable matchings, because the definitions of blocking pairs and stable matchings are symmetrical with respect to gender. a(n) is the number of instances in a canonical form where the men's preferences are lexicographically less than or equal to the women's preferences.
The A185141(n) instances of order n can be arranged in a square table with rows and columns indexed respectively by all possible men's and women's preferences in lexical order. The main diagonal of the square would be instances with men's preferences equal to women's preferences. The upper triangular region above the diagonal would contain instances with men's preferences less than women's preferences. The number of rows and columns in the table is given by A036740. The number of elements in the upper triangular region of a square, including the diagonal, is given by A000217. So a(n) composes A000217 with A036740 (performing A036740 first).
This sequence is like A351409 and A343700 in that they all involve means of reducing the search space, applied either individually or in combination, when searching for instances that maximize the number of stable matchings.

			For order 2 we have A185141(2) = 16 instances that can be arranged in a 4 X 4 square with A000217(4) = (4 * 5) / 2 = 10 distinct instances up to gender exchange in the upper triangular region including the diagonal. So a(2) = 10.

Wikipedia, Stable Marriage Problem

Cf. A185141, A036740, A000217, A351409, A343700.

Table[((n!)^n) * ((n!)^n + 1) / 2, {n, 1, 8}]

a(n) = A000217(A036740(n)).

a(n) = ((n!)^n) * ((n!)^n + 1) / 2.

A366672 a(n) = A002720(n)^n.

Original entry on oeis.org

1, 2, 49, 39304, 1908029761, 8831763846882976, 5602661527604399327549089, 659308109505417338723017914068713088, 18666765602783048904120522995911258148623099215361, 159740893387079678500933964995221201596055121928224632284394525184

Offset: 0

Paulius Dilkas, Oct 15 2023

This is the model count of the following sentence in first-order logic:
(forall w, x, y, z. P(x, y, z) /\ P(w, y, z) => x = w) /\
(forall w, x, y, z. P(x, y, z) /\ P(x, w, z) => y = w).

			When n = 2, i.e., the domain is [2] = {1, 2}, both P(x, y, 1) and P(x, y, 2) represent partial injective functions from [2] to [2]. Since there are seven such functions, a(n) = 7^2 = 49.

Paulius Dilkas and Vaishak Belle, Synthesising Recursive Functions for First-Order Model Counting: Challenges, Progress, and Conjectures, KR 2023, 198-207; arXiv:2306.04189 [cs.LO], 2023.

Cf. A002720, A036740.

Table[(n!*LaguerreL[n, -1])^n, {n, 0, 10}] (* Vaclav Kotesovec, Oct 20 2023 *)

a(n) ~ n^(n*(n + 1/4)) / (2^(n/2) * exp(n^2 - 2*n^(3/2) + n/2 - 31*sqrt(n)/48 + 17/192)) * (1 - 281/(5120*sqrt(n)) + 3074161/(52428800*n)). - Vaclav Kotesovec, Oct 20 2023

A379577 a(n) = (n!)^n + n^n.

Original entry on oeis.org

2, 2, 8, 243, 332032, 24883203125, 139314069504046656, 82606411253903523840823543, 6984964247141514123629140377616777216, 109110688415571316480344899355894085582848387420489, 395940866122425193243875570782668457763038822400000000010000000000, 409933016554924328182440935903164918932547530146724293451448320000000000285311670611

Offset: 0

Ctibor O. Zizka, Dec 26 2024

The equation (k!)^n + k^n = (n!)^k + n^k holds if and only if n = k. See the Proof of the Theorem 2.2 (p.182-183) in the Alzer and Luca article in Links section.

			n = 3: a(3) = (3!)^3 + 3^3 = 243.

Horst Alzer and Florian Luca, Diophantine equations involving factorials, Mathematica Bohemica 142.2 (2017), 181-184.

Cf. A000142, A000312, A036740.

seq((n!)^n + n^n, n=0..12); # Georg Fischer, Dec 30 2024

a[n_]:=(n!)^n + n^n; Array[a,12,0] (* Stefano Spezia, Dec 26 2024 *)

a(n) = A036740(n) + A000312(n).

a(9)-a(11) corrected by Georg Fischer, Dec 30 2024

cp's OEIS Frontend

A127488 a(n) = (n^2)!/(2*(n!)).

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Mathematica

Formula

A181760 a(n) = (n!)(n!-1)(n!-2)...(n!-n+1).

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Maple

Mathematica

Formula

A213065 a(n) = n^n! - n!^n.

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PARI

Extensions

A336249 a(n) = (n!)^n * Sum_{k=0..n} 1 / ((k!)^n * (n-k)!).

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Mathematica

PARI

Formula

A336250 a(n) = (n!)^n * Sum_{k=1..n} (-1)^(k+1) / k^n.

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Mathematica

PARI

Formula

A360213 Number of distinct stable marriage problem instances up to gender exchange.

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Mathematica

Formula

A366672 a(n) = A002720(n)^n.

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Programs

Mathematica

Formula

A379577 a(n) = (n!)^n + n^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica