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 1-9 of 9 results.

A342573 The number of ordered n-tuples consisting of n permutations (not necessarily distinct) such that the first element of each of them is the same.

Original entry on oeis.org

1, 2, 24, 5184, 39813120, 17915904000000, 702142910300160000000, 3330690501757390081228800000000, 2534703826002712645182542460223488000000000, 395940866122425193243875570782668457763038822400000000000
Offset: 1

Views

Author

Tanya Khovanova and MIT PRIMES STEP Senior group, Mar 27 2021

Keywords

Comments

This is related to the stable marriage problem, as this counts the preference profiles for n men trying to marry n women when all of them prefer the same woman.
This sequence also counts the sets of n permutations of size n such that the i-th element of each of them is the same.
a(n) is a subsequence of A001013: products of factorial numbers.

Examples

			When n=3, we have 3 ways to fix the first element, and the remaining elements in each permutation can be in any order, yielding (3 - 1)! possible ways of ordering the rest of each permutation, so there are 3 * (2!)^3 = 24 sets of permutations.
		

Crossrefs

Programs

  • Mathematica
    Table[n (n - 1)!^n, {n, 10}]

Formula

a(n) = n*(n-1)!^n = n*A091868(n-1).

A343900 a(n) = Sum_{k=0..n} (k!)^(k+1) * binomial(n,k).

Original entry on oeis.org

1, 2, 11, 1324, 7967861, 2986023826166, 100306147958903465407, 416336313421816733159702737376, 281633758448076539969292901914477101456489, 39594086612245054028213574779019294652734771094507399786
Offset: 0

Views

Author

Seiichi Manyama, May 03 2021

Keywords

Comments

Binomial transform of (n!)^(n+1).

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[(k!)^(k+1) * Binomial[n, k], {k, 0, n}]; Array[a, 10, 0] (* Amiram Eldar, May 05 2021 *)
  • PARI
    a(n) = sum(k=0, n, k!^(k+1)*binomial(n, k));
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, k!^(k+1)*x^k/(1-x)^(k+1)))
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(x)*sum(k=0, N, (k!*x)^k)))

Formula

G.f.: Sum_{k>=0} (k!)^(k+1) * x^k/(1 - x)^(k+1).
E.g.f.: exp(x) * Sum_{k>=0} (k! * x)^k.

A165642 Number of ways to assemble an n-cube from 2n labeled (n-1)-cubes with labeled vertices, where left-handed and right-handed counterparts are considered distinct.

Original entry on oeis.org

2, 96, 7864320, 5917648890101760, 131757106216193173905620336640, 213712134396364787165698675955466240000000000000, 52598387159216969439004693376714880262603303706802782208000000000000000
Offset: 1

Views

Author

Andrew Weimholt, Sep 23 2009

Keywords

Examples

			For n=2, we are constructing a square from 4 labeled line-segments with labeled endpoints. Solutions which differ by a rotation are considered equivalent, but solutions which are a reflection of each other are considered distinct (assume the square we are constructing is embedded in a plane, so we cannot flip it over to convert a left-handed solution to right-handed solution). There are 6 ways to order the line-segments, and each line-segment can be oriented in 2 ways, so the total number of solutions is 6 * 2^4 = 96. For n=3, we are constructing a cube from 6 labeled squares with labeled vertices (assume we are confined to 3-space, so we consider reflections of the cube to be distinct). Without loss of generality, we can pick one labeled square to serve as our face of reference. For this face, we must decide which side of the square will face the interior of the cube, but we do not care about which rotation we pick as these just translate into rotations of the cube. From this reference square, there are 5! ways to assign the remaining squares to the faces of the cube, and each square can be oriented in 8 ways (we can pick which side of the square will face the interior of the cube, and we can pick from 4 rotations). This gives 2 * 8^5 * 5! solutions. The factor of "2" comes from the choice of which side of the reference square will face the interior of the cube (a choice which would go away if we considered reflections to be equivalent).
		

Crossrefs

Cf. A165643 (same idea, but reflections are equivalent). A165644 and A091868 are the corresponding sequences for simplices instead of cubes.

Formula

a(n) = 2 * ((2n-2)!!)^(2n-1) * (2n-1)!

Extensions

Example reformatted by Andrew Weimholt, Sep 25 2009

A165643 Number of ways to assemble an n-cube from 2n labeled (n-1)-cubes with labeled vertices, where left-handed and right-handed counterparts are considered equivalent.

Original entry on oeis.org

1, 48, 3932160, 2958824445050880, 65878553108096586952810168320, 106856067198182393582849337977733120000000000000, 26299193579608484719502346688357440131301651853401391104000000000000000
Offset: 1

Views

Author

Andrew Weimholt, Sep 23 2009

Keywords

Examples

			For n=2, we are constructing a square from 4 labeled line-segments with labeled endpoints. Solutions which differ by a rotation or a reflection are considered equivalent. There are 3 ways to order the line-segments, and each line-segment can be oriented in 2 ways, so the total number of solutions is 3 * 2^4 = 96. (equivalently, 6 * 2^3, which more closely resembles the formula provided.) For n=3, we are constructing a cube from 6 labeled squares with labeled vertices. Without loss of generality, we can pick one labeled square to serve as our face of reference. For this face, we do not care which side of the square will face the interior of the cube as this just translates into a reflection of the cube, nor we do not care about which rotation we pick as these just translate into rotations of the cube. From this reference square, there are 5! ways to assign the remaining squares to the faces of the cube, and each square can be oriented in 8 ways (we can pick which side of the square will face the interior of the cube, and we can pick from 4 rotations). This gives 8^5 * 5! solutions.
		

Crossrefs

Cf. A165642 (same idea, but reflections are distinct). A165644 and A091868 are the corresponding sequences for simplices instead of cubes.

Formula

a(n) = ((2n-2)!!)^(2n-1) * (2n-1)!

Extensions

Example reformatted by Andrew Weimholt, Sep 25 2009

A165644 Number of ways to assemble an n-simplex from n+1 labeled (n-1)-simplices with labeled vertices, where left-handed and right-handed counterparts are considered distinct.

Original entry on oeis.org

2, 16, 2592, 15925248, 5971968000000, 200612260085760000000, 832672625439347520307200000000, 563267516889491698929453880049664000000000
Offset: 1

Views

Author

Andrew Weimholt, Sep 23 2009, Sep 25 2009

Keywords

Examples

			For n=2, construct a triangle from 3 labeled line segments with labeled endpoints. Solutions which differ by a rotation are considered equivalent, but solutions which differ by a reflection are considered distinct (assume the triangle we are constructing will be embedded in a plane, so we cannot flip it over to convert a left-handed solution to a right-handed solution). There are 2 ways to order the line segments, and each line segment can be oriented in 2 ways, so the total number of solutions is 2 * 2^3 = 16.
For n=3, construct a tetrahedron from 4 labeled triangles with labeled vertices. Assume the tetrahedron is confined to 3-space where we cannot flip it around in 4-space to convert left-handed solutions to right-handed solutions, so we consider reflections to be distinct. Without loss of generality, we can pick one labeled triangle to serve as our face of reference. For this face, we must pick which side of the triangle will face the interior of the tetrahedron, but we do not care about which rotation we pick as these just translate into rotations of the tetrahedron. From this reference triangle, there are 6 (=3!) ways to assign the remaining triangles to the faces of the tetrahedron, and each triangle can be oriented in 6 (=3!) ways (we can pick which side of the triangle will face the interior of the tetrahedron, and we can pick from 3 rotations). This gives 2 * 6^4 solutions. The factor of "2" comes from the choice of which side of the reference triangle faces the interior (a choice that goes away if we consider reflections to be equivalent).
		

Crossrefs

Cf. A091868 (same idea, but reflections are equivalent). A165642 and A165643 are the corresponding sequences for cubes instead of simplices.

Formula

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

A176113 Determinant of n X n matrix whose (i,j)-th element is (i+j)^(n-1).

Original entry on oeis.org

1, -1, -8, 1296, 7962624, -2985984000000, -100306130042880000000, 416336312719673760153600000000, 281633758444745849464726940024832000000000
Offset: 1

Views

Author

Michel Lagneau, Apr 08 2010

Keywords

Comments

a(n) = A057077(n)*A091868(n-1), signed variant of A091868.
Concerning the general case det((u(i) + v(j))^(n-1) for 1<=i,j<=n and u(1), ...,u(n), v(1), ..., v(n) integers, the reference give the proof that det((u(i) + v(j))^(n-1) = (1/x) *(((n-1)!)^n)*y*z with the following results :
x = 1! * 2!*...(n-1)! ;
y = (u(1) - u(2))*((u(1) - u(3))*....*(u(n-1) - u(n)) ;
z = (v(n) - v(n-1))*((v(n) - v(n-2))*....*(v(2) - v(1)).
If the (i,j)-th element is (i+j)^(n-1), then u(i) = i and v(j) = n+1-j. Finally, det(n X n) = ((-1)^p)* (n!)^(n+1) with n = 2p + 1 or n=2p.

Examples

			a(3) = determinant(M_3) = -8 where M_3 is the matrix
[4  9   16]
[9  16  25]
[16 25  36]
		

References

  • J. M. Monier, Algebre & geometrie, Dunod (1996), p.216.

Crossrefs

Cf. A091868.

Programs

  • Mathematica
    a[n_] = ((-1)^((n - Mod[n, 2])/2))*((n - 1)!)^n; Table[a[n], {n, 9}] (* from Jean-François Alcover, Aug 31 2011 *)

Formula

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

A351781 a(n) = (n-1)^n*(n-1)!^n.

Original entry on oeis.org

0, 1, 64, 104976, 8153726976, 46656000000000000, 28079296819683655680000000, 2400095991902688012207233433600000000, 37800243186554601452585666030525214621696000000000
Offset: 1

Views

Author

Dan Eilers, Feb 19 2022

Keywords

Comments

a(n) is the number of women's ranking tables in the stable marriage problem that can be paired with a men's ranking table having no two men with the same first choice, without forming any mutual first choices. It has two terms: (n-1)^n from A065440(n), and (n-1)!^n from A091868(n-1). Such men's ranking tables having no two men with the same first choice arise in A343694, A343475, and A344663.
a(n)*A123234 is a useful alternative to A343696 which combines a Latin men's ranking table with an arbitrary women's table, since it gives fewer instances to consider.

Crossrefs

Programs

  • Mathematica
    Table[(n-1)^n*(n-1)!^n,{n,1,9}]

Formula

a(n) = (n-1)^n*(n-1)!^n.
a(n) = A065440(n)*A091868(n-1).

A358648 Number of preference profiles of the stable roommates problem with 2n participants.

Original entry on oeis.org

1, 1296, 2985984000000, 416336312719673760153600000000, 39594086612242519324387557078266845776303882240000000000, 16363214235219603423192858350259453436046713251360764276842772299776000000000000000000000000
Offset: 1

Views

Author

Dan Eilers, Nov 24 2022

Keywords

Comments

Each participant ranks all participants other than themselves in strict order, giving (2n-1)! orderings for each of 2n participants.

Crossrefs

Cf. A356584 (up to isomorphism), A185141 (Stable Marriage profiles), A001147 (possible roommate pairings).
Even bisection of A091868.

Programs

  • Mathematica
    Table[(2n-1)!^(2n), {n, 1, 6}]

Formula

a(n) = (2n-1)!^(2n).

A368806 a(n) = Product_{i=1..j, j=1..k, k=1..n} i*j*k.

Original entry on oeis.org

1, 1, 64, 60466176, 504857282956046106624, 46005119909369701466112000000000000000000000, 101230154592156481700985865260692304243040378536591360000000000000000000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 06 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Product[Product[Product[i*j*k, {i,1,j}], {j,1,k}], {k,1,n}], {n, 0, 6}]
    Table[Product[k^(k*(k+1)/2) * k!^(k+1), {k, 1, n}], {n, 0, 6}]

Formula

a(n) = Product_{k=1..n} k^(k*(k+1)/2) * k!^(k+1).
a(n) ~ (2*Pi)^(n^2/4 + 3*n/4 + 1/2) * n^(n^3/2 + 7*n^2/4 + 7*n/4 + 1/2) / exp(n^3/2 + 3*n^2/2 + 23*n/24 - 1/8).
Showing 1-9 of 9 results.