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 1-8 of 8 results.

A187783 De Bruijn's triangle, T(m,n) = (m*n)!/(n!^m) read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 20, 90, 24, 1, 1, 1, 70, 1680, 2520, 120, 1, 1, 1, 252, 34650, 369600, 113400, 720, 1, 1, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1
Offset: 0

Views

Author

Robert G. Wilson v, Jan 05 2013

Keywords

Comments

From Tilman Piesk, Oct 28 2014: (Start)
Number of permutations of a multiset that contains m different elements n times. These multisets have the signatures A249543(m,n-1) for m>=1 and n>=2.
In an m-dimensional Pascal tensor (the generalization of a symmetric Pascal matrix) P(x1,...,xn) = (x1+...+xn)!/(x1!*...*xn!), so the main diagonal of an m-dimensional Pascal tensor is D(n) = (m*n)!/(n!^m). These diagonals are the rows of this array (with m>0), which begins like this:
m\n:0 1 2 3 4 5
0: 1 1 1 1 1 1 ... A000012;
1: 1 1 1 1 1 1 ... A000012;
2: 1 2 6 20 70 252 ... A000984;
3: 1 6 90 1680 34650 756756 ... A006480;
4: 1 24 2520 369600 63063000 11732745024 ... A008977;
5: 1 120 113400 168168000 305540235000 623360743125120 ... A008978;
6: 1 720 7484400 137225088000 3246670537110000 88832646059788350720 ... A008979;
with columns: A000142 (n=1), A000680 (n=2), A014606 (n=3), A014608 (n=4), A014609 (n=5).
A089759 is the transpose of this matrix. A034841 is its diagonal. A141906 is its lower triangle. A120666 is the upper triangle of this matrix with indices starting from 1. A248827 are the diagonal sums (or the row sums of the triangle).
(End)

Examples

			T(3,5) = (3*5)!/(5!^3) = 756756 = A014609(3) = A006480(5) is the number of permutations of a multiset that contains 3 different elements 5 times, e.g., {1,1,1,1,1,2,2,2,2,2,3,3,3,3,3}.
		

Crossrefs

Cf. A089759 (transposed), A141906 (subtriangle), A120666 (subtriangle transposed), A060538 (1st row/column removed).
Main diagonal gives: A034841.
Row sums of the triangle: A248827.

Programs

  • Magma
    [Factorial(k*(n-k))/(Factorial(n-k))^k: k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 26 2022
    
  • Mathematica
    T[n_, k_]:= (k*n)!/(n!)^k; Table[T[n, k-n], {k, 9}, {n, 0, k-1}]//Flatten
  • SageMath
    def A187783(n,k): return gamma(k*(n-k)+1)/(factorial(n-k))^k
    flatten([[A187783(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Dec 26 2022

Formula

T(m,n) = (m*n)!/(n!)^m.
A060540(m,n) = T(m,n)/m! . - R. J. Mathar, Jun 21 2023

Extensions

Row m=0 prepended by Tilman Piesk, Oct 28 2014

A089759 Table T(n,k), 0<=k, 0<=n, read by antidiagonals, defined by T(n,k) = (k*n)! / (n!)^k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 24, 90, 20, 1, 1, 1, 120, 2520, 1680, 70, 1, 1, 1, 720, 113400, 369600, 34650, 252, 1, 1, 1, 5040, 7484400, 168168000, 63063000, 756756, 924, 1, 1, 1, 40320, 681080400, 137225088000, 305540235000, 11732745024, 17153136, 3432, 1, 1
Offset: 0

Views

Author

Philippe Deléham, Jan 08 2004; revised Jun 08 2005

Keywords

Comments

T(n,k) is the number of lattice paths from {n}^k to {0}^k using steps that decrement one component by 1. - Alois P. Heinz, May 06 2013

Examples

			Row n=0: 1, 1,   1,      1,           1,               1, ... A000012
Row n=1: 1, 1,   2,      6,          24,             120, ... A000142
Row n=2: 1, 1,   6,     90,        2520,          113400, ... A000680
Row n=3: 1, 1,  20,   1680,      369600,       168168000, ... A014606
Row n=4: 1, 1,  70,  34650,    63063000,    305540235000, ... A014608
Row n=5: 1, 1, 252, 756756, 11732745024, 623360743125120, ... A014609
		

Crossrefs

Cf. A000680, A014606, A014608, A014609, A000984, A187783 (transposed version).
Main diagonal gives A034841.

Programs

  • Maple
    T:= (n, k)-> (k*n)!/(n!)^k:
    seq(seq(T(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 16 2012
  • Mathematica
    T[n_, k_] := (k*n)!/(n!)^k; Table[T[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 19 2015 *)

Extensions

Corrected by Alois P. Heinz, Aug 16 2012

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5*n+1,i) * binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).

Original entry on oeis.org

1, 1, 25, 100, 100, 25, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275
Offset: 1

Views

Author

Yahia Kahloune, Feb 05 2014

Keywords

Comments

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,5,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 5 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 08 2020

Examples

			T(n,0) = 1;
T(n,1) = 6^n - (5*n+1);
T(n,2) = 21^n - (5*n+1)*6^n + C(5*n+1,2);
T(n,3) = 56^n - (5*n+1)*21^n + C(5*n+1,2)*6^n - C(5*n+1,3) ;
T(n,4) = 126^n - (5*n+1)*56^n + C(5*n+1,2)*21^n - C(5*n+1,3)*6^n  + C(5*n+1,4).
Triangle T(n,k) begins:
1;
1, 25, 100, 100, 25, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 125, 1;
1, 7750, 3882250, 447069750, 18746073375, 359033166276, 3575306548500, 20052364456500, 66640122159000, 135424590593500, 171219515211316, 135424590593500, 66640122159000, 20052364456500, 3575306548500, 359033166276, 18746073375, 447069750, 3882250, 7750, 1;
...
Example:
Sum_{i=1..n} C(4+i,5)^3 = C(n+5,16) + 200*C(n+6,16) + 5925*(n+7,16) + 52800*C(n+8,16) + 182700*C(n+9,16) + 273504*C(n+10,16) + 182700*C(n+11,16) + 52800*C(n+12,16) + 5925*C(n+13,16) + 200*C(n+14,16) + C(n+15,16).
C(n,5)^3 = C(n,15) + 200*C(n+1,15) + 5925*C(n+2,15) + 52800*C(n+3,15) + 182700*C(n+4,15) + 273504*C(n+5,15) + 182700*C(n+6,15) + 52800*C(n+7,15) + 5925*C(n+8,15) + 200*C(n+9,15) + C(n+10,15).
		

Crossrefs

Columns k=2..5 are A151647, A151648, A151649, A151650.
Row sums are A014609.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, A236463, this sequence, A237252.
Sum_{i=1..n} binomial(4+i,5)^p for p=2..3 gives: A086025, A086026.

Programs

  • Mathematica
    b[k_, 5, p_] := Sum[(-1)^i*Binomial[5*p+1, i]*Binomial[k-i, 5]^p /. k -> 5+i, {i, 0, k-5}]; row[p_] := Table[b[k, 5, p], {k, 5, 5*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)
  • PARI
    T(n,k)={sum(i=0, k, (-1)^i*binomial(5*n+1, i)*binomial(k+5-i, 5)^n)} \\ Andrew Howroyd, May 08 2020

Formula

Sum_{i=1..n} binomial(4+i,5)^p = Sum{k=0..5*(p-1)} T(p,k) * binomial(n+5+k, 5*p+1).
binomial(n,5)^p = Sum_{k=0..5*(p-1)} T(p,k) * binomial(n+k, 5*p).

Extensions

Edited by Andrew Howroyd, May 08 2020

A060538 Square array read by antidiagonals of number of ways of dividing n*k labeled items into n labeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 20, 90, 24, 1, 70, 1680, 2520, 120, 1, 252, 34650, 369600, 113400, 720, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1, 3432, 17153136, 11732745024, 305540235000, 137225088000, 681080400, 40320, 1, 12870
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Examples

			       1        1        1        1
       2        6       20       70
       6       90     1680    34650
      24     2520   369600 63063000
		

Crossrefs

Subtable of A187783.
Rows include A000012, A000984, A006480, A008977, A008978 etc.
Columns include A000142, A000680, A014606, A014608, A014609 etc.
Main diagonal is A034841.

Programs

  • PARI
    T(n,k)=(n*k)!/k!^n;
    for(n=1, 6, for(k=1, 6, print1(T(n,k), ", ")); print) \\ Harry J. Smith, Jul 06 2009

Formula

T(n, k) = (nk)!/k!^n = T(n-1, k)*binomial(nk, k) = T(n-1, k)*A060539(n, k) = A060540(n, k)*A000142(k).

A177301 Number of permutations of 5 copies of 1..n with all adjacent differences <= 1 in absolute value.

Original entry on oeis.org

1, 1, 252, 7002, 284002, 8956752, 276285002, 8039989002, 226901044252, 6232521421502, 167765555015002, 4441865811412752, 116042171156004002, 2998175631045512002, 76751828260736962252, 1949598912568702682502, 49197330547669842130002, 1234523426440511782592752
Offset: 0

Views

Author

R. H. Hardin, May 06 2010

Keywords

Comments

a(n) = (5n)!/120^n = A014609(n) for n<=2.

Crossrefs

Cf. A014609.
Column k=5 of A331562.

Extensions

a(10) from Alois P. Heinz, Jan 21 2020
Terms a(11) and beyond from Andrew Howroyd, May 15 2020

A269115 Number of sequences with 5 copies each of 1,2,...,n avoiding the pattern 12...n.

Original entry on oeis.org

0, 0, 1, 26945, 1474050783, 173996758190594, 41489887418194798439, 18675035107760751336633904, 15006398706095549205738572240519, 20584293980310480336378076914369061769, 46436573450646384263588550418764923934014699
Offset: 0

Views

Author

Alois P. Heinz, Feb 19 2016

Keywords

Crossrefs

Column k=5 of A269129.

Formula

a(n) = A014609(n) - A268848(n).

A177302 Number of permutations of 5 copies of 1..n with all adjacent differences <= 2 in absolute value.

Original entry on oeis.org

1, 1, 252, 756756, 457507512, 184619047020, 97087382008034, 49254897805981056, 22802419346927105088
Offset: 0

Views

Author

R. H. Hardin, May 06 2010

Keywords

Comments

a(n) = (5n)!/120^n for n<=3.

Crossrefs

Cf. A014609.

Extensions

a(0), a(8) from Alois P. Heinz, Jan 19 2016

A177303 Number of permutations of 5 copies of 1..n with all adjacent differences <= 3 in absolute value.

Original entry on oeis.org

1, 1, 252, 756756, 11732745024, 52977847995072, 131319696504676680, 302061042944500510428, 847432462636241316703042
Offset: 0

Views

Author

R. H. Hardin, May 06 2010

Keywords

Comments

a(n) = (5n)!/120^n for n<=4.

Crossrefs

Cf. A014609.

Extensions

a(0), a(8) from Alois P. Heinz, Jan 19 2016
Showing 1-8 of 8 results.