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 1-4 of 4 results.

A322093 Square array A(n,k), n >= 1, k >= 1, read by antidiagonals, where A(n,k) is the number of permutations of n copies of 1..k with no element equal to another within a distance of 1.

Original entry on oeis.org

1, 2, 0, 6, 2, 0, 24, 30, 2, 0, 120, 864, 174, 2, 0, 720, 39480, 41304, 1092, 2, 0, 5040, 2631600, 19606320, 2265024, 7188, 2, 0, 40320, 241133760, 16438575600, 11804626080, 134631576, 48852, 2, 0, 362880, 29083420800, 22278418248240, 131402141197200, 7946203275000, 8437796016, 339720, 2, 0
Offset: 1

Views

Author

Seiichi Manyama, Nov 26 2018

Keywords

Examples

			Square array begins:
   1, 2,    6,        24,           120,                 720, ...
   0, 2,   30,       864,         39480,             2631600, ...
   0, 2,  174,     41304,      19606320,         16438575600, ...
   0, 2, 1092,   2265024,   11804626080,     131402141197200, ...
   0, 2, 7188, 134631576, 7946203275000, 1210527140790855600, ...
		

Crossrefs

Columns k=3 gives A110706.
Main diagonal gives A321634.
Cf. A322013.

Programs

  • Mathematica
    Table[Table[SeriesCoefficient[1/(1 - Sum[x[i]/(1 + x[i]), {i, 1, n}]), Sequence @@ Table[{x[i], 0, k}, {i, 1, n}]],{n, 1, 6}], {k, 1, 5}] (* Zlatko Damijanic, Nov 03 2024 *)
  • PARI
    q(n,x) = sum(i=1, n, (-1)^(n-i) * binomial(n-1, n-i) * x^i/i!)
    T(n,k) = subst(serlaplace(q(n,x)^k), x, 1) \\ Andrew Howroyd, Feb 03 2024

Formula

A(n,k) = k! * A322013(n,k).
Let q_n(x) = Sum_{i=1..n} (-1)^(n-i) * binomial(n-1, n-i) * x^i/i!.
A(n,k) = Integral_{0..infinity} (q_n(x))^k * exp(-x) dx.

A190830 Number of permutations of 4 copies of 1..n introduced in order 1..n with no element equal to another within a distance of 1.

Original entry on oeis.org

1, 0, 1, 182, 94376, 98371884, 182502973885, 551248360550999, 2536823683737613858, 16904301142107043464659, 156690501089429126239232946, 1955972150994131850032960933480, 32016987304767134806200915633253966, 672058204939482014438623912695190927357
Offset: 0

Views

Author

R. H. Hardin, May 21 2011

Keywords

Examples

			Some solutions for n=3:
  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2
  3  1  3  3  1  3  3  3  3  3  3  1  1  3  1  1
  1  2  2  2  2  2  2  1  1  2  1  3  3  1  3  2
  2  3  3  3  1  1  3  2  2  3  2  1  2  3  1  1
  3  2  1  2  3  2  1  3  1  1  3  2  3  2  2  3
  1  3  2  3  2  3  3  1  3  2  2  3  2  1  3  1
  3  2  3  1  3  1  2  3  1  1  1  2  1  3  2  3
  1  1  1  3  1  3  1  2  2  3  3  3  2  2  3  2
  2  3  2  1  3  2  3  1  3  1  1  2  3  3  1  3
  3  1  1  2  2  3  1  3  2  2  2  3  1  1  3  2
  2  3  3  1  3  1  2  2  3  3  3  1  3  2  2  3
		

Crossrefs

Row n=4 of A322013.

Formula

From Vaclav Kotesovec, Nov 24 2018: (Start)
Recurrence: 3*(64*n^3 - 280*n^2 + 414*n - 245)*a(n) = (2048*n^6 - 12032*n^5 + 30400*n^4 - 42608*n^3 + 32484*n^2 - 14624*n + 1731)*a(n-1) + 3*(3840*n^5 - 20640*n^4 + 40104*n^3 - 36340*n^2 + 23378*n - 13429)*a(n-2) - 18*(384*n^4 - 1488*n^3 + 1556*n^2 - 986*n + 649)*a(n-3) - 27*(64*n^3 - 88*n^2 + 46*n - 47)*a(n-4).
a(n) ~ 2^(5*n+1) * n^(3*n) / (3^n * exp(3*n + 3)). (End)

Extensions

a(0)=1 prepended by Seiichi Manyama, Nov 16 2018

A321634 Number of arrangements of n 1's, n 2's, ..., n n's avoiding equal consecutive terms.

Original entry on oeis.org

1, 1, 2, 174, 2265024, 7946203275000, 12229789732207993835280, 12202002913678756821228939869239920, 10937192762438008527903830198163831816546577931520, 11655577382287102750765311537460065620507094071664576111302628243840
Offset: 0

Views

Author

Seiichi Manyama, Nov 15 2018

Keywords

Crossrefs

Programs

  • PARI
    {a(n) = sum(i=n, n^2, i!*polcoef(sum(j=1, n, (-1)^(n-j)*binomial(n-1, j-1)*x^j/j!)^n, i))} \\ Seiichi Manyama, May 27 2019

Formula

a(n) ~ n^(n^2 - n/2 + 1) / ((2*Pi)^((n-1)/2) * exp(n - 5/12)). - Vaclav Kotesovec, Nov 24 2018

A330266 Number of ways to shuffle a deck of 4n cards, with 4 cards in each of n ranks, so that adjacent cards have different ranks.

Original entry on oeis.org

1, 0, 1152, 15095808, 751480602624, 93995798935633920, 25111340235557122867200, 12742555660097789273088983040, 11259023892340311657074592904642560, 16205462460428776872054787528078739374080, 36051066700209244649349258741114804984663244800, 118807003903158552156678227915553602167323425243136000
Offset: 0

Views

Author

David Radcliffe, Dec 07 2019

Keywords

Examples

			a(13) = 3668033946384704437729512814619767610579526911188666362431432294400 is the number of ways to shuffle a standard 52-card deck of playing cards so that no two cards of the same rank are adjacent.
		

Crossrefs

Cf. A007060 (2n cards), A193624 (3n cards).

Programs

  • Mathematica
    Table[Integrate[(x^4 - 12x^3 + 36x^2 - 24x)^n *Exp[-x],{x,0,Infinity}],{n,0,10}] (* Stefano Spezia, Dec 09 2019 *)

Formula

a(n) = Integral_{x=0..oo} (x^4 - 12x^3 + 36x^2 - 24x)^n*exp(-x) dx.
a(n) = 24^n * A321633(n).
Conjecture: Limit_{n->oo} a(n)/(4n)! = 1/e^3. The conjecture is based on the observation of the asymptotic behavior of A007060 and A193624; it seems that it can be generalized in the following way. Let b(n) be the number of ways to shuffle a deck of k*n cards, with k cards in each of n ranks, so that adjacent cards have different ranks. Then, lim_{n->oo} b(n)/(kn)! = 1/e^(k-1); maybe we could prove it with the help of rook polynomials theory or in some other way. - Sergey Kirgizov, Sep 29 2023
Showing 1-4 of 4 results.