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.

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A173882 Triangle T(n, k) = A090443(n-1)/(A090443(k-1)*A090443(n-k-1)) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 24, 24, 1, 1, 60, 240, 60, 1, 1, 120, 1200, 1200, 120, 1, 1, 210, 4200, 10500, 4200, 210, 1, 1, 336, 11760, 58800, 58800, 11760, 336, 1, 1, 504, 28224, 246960, 493920, 246960, 28224, 504, 1, 1, 720, 60480, 846720, 2963520, 2963520, 846720, 60480, 720, 1
Offset: 0

Views

Author

Roger L. Bagula, Mar 01 2010

Keywords

Comments

A090443 is defined as +1 at negative indices here, which keeps the definition valid in the range 0 <= k <= n.
Row sums are 1, 2, 8, 50, 362, 2642, 19322, 141794, 1045298, 7742882, ....

Examples

			Triangle begins as:
  1;
  1,   1;
  1,   6,      1;
  1,  24,     24,       1;
  1,  60,    240,      60,        1;
  1, 120,   1200,    1200,      120,        1;
  1, 210,   4200,   10500,     4200,      210,        1;
  1, 336,  11760,   58800,    58800,    11760,      336,       1;
  1, 504,  28224,  246960,   493920,   246960,    28224,     504,      1;
  1, 720,  60480,  846720,  2963520,  2963520,   846720,   60480,    720,   1;
  1, 990, 118800, 2494800, 13970880, 24449040, 13970880, 2494800, 118800, 990, 1;
  ...

Links

Crossrefs

Cf. A056939.

Programs

  • Magma
    T:= func< n,k | k eq 0 or k eq n select 1 else 2*Binomial(n-1,k-1)*Binomial(n,k)*Binomial(n+1,k+1)*(n-k)/(n-k+1) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 17 2021
  • Maple
    A090443 := proc(n) (n+2)!*(n+1)!*n!/2 ; end proc:
A173882 := proc(n,m) if m=0 or m= n then 1; else A090443(n-1)/A090443(m-1)/A090443(n-m-1) ; end if; end proc:
seq(seq(A173882(n,m),m=0..n),n=0..5) ; # R. J. Mathar, Mar 19 2011
  • Mathematica
    T[n_,k_]:= If[k==0||k==n, 1, 2*Binomial[n-1,k-1]*Binomial[n,k]*Binomial[n+1,k+1]*(n-k)/(n-k+1)];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 17 2021 *)
  • Sage
    def T(n,k): return 1 if (k==0 or k==n) else 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 17 2021

Formula

T(n, k) = 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1) with T(n, 0) = T(n, n) = 1.
T(n, k) = c(n)/(c(k)*c(n-k)) where c(n) = Product_{j=2..n} (j-1)*j*(j+1) = (n-1)!*n!*(n+1)!/2 and c(0) = c(1) = 1.

A090441 Symmetric triangle of certain normalized products of decreasing factorials.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 6, 12, 6, 1, 1, 24, 144, 144, 24, 1, 1, 120, 2880, 8640, 2880, 120, 1, 1, 720, 86400, 1036800, 1036800, 86400, 720, 1, 1, 5040, 3628800, 217728000, 870912000, 217728000, 3628800, 5040, 1, 1, 40320, 203212800, 73156608000
Offset: 0

Views

Author

Wolfdieter Lang, Dec 23 2003

Keywords

Comments

Similar to, but different from, superfactorial Pascal triangle A009963.
A009963(n,m) = (Product_{p=0..m-1} (n-p)!)/superfac(m) with n >= m >= 0, otherwise 0.
From Natalia L. Skirrow, Apr 13 2025 (Start)
Denoting this sequence as the superbinomial sb(n,k), the hook length formula for a j X k rectangular Young tableau states the number of configurations of j*k distinct numbers such that each row and column is strictly increasing is (j*k)!/sb(j+k,j), ie. 1/sb(j+k,j) is the probability that a random permutation is a Young tableau.
Meanwhile, if the numbers are placed into the array with repetition, but the columns are still strictly increasing, there are c(n,j,k) = sb(n+1,j+k)/(sb(n+1-j,k)*sb(n+1-k,j)) configurations.
If the strict criterion is relaxed to monotonic, this becomes C(n,j,k) = sb(n-1+j+k,j+k)/(sb(n-1+j,j)*sb(n-1+k,k)).
By proposition 13.2(i) of Stanley's PhD thesis, for fixed j,k, c(n,j,k) and C(n,j,k) are polynomials in n of degree j*k, and c(n,j,k) = (-1)^(j*k)*C(-n,j,k).
For example, c(n,1,k)=(n choose k) and C(n,1,k)=(n+k-1 choose k), while c(n,2,k) = N(n,k+1) and C(n,2,k) = N(n+k,k+1), so the binomial coefficients and Narayana numbers N=A001263 obey the dualities (under continuation as polynomials) (n choose k) = (-1)^k*(k-1-n choose k) and N(n,k) = N(k-1-n,k).
(End)

Examples

			Rows for n = 0, 1, 2, 3, ...:
  1;
  1,  1;
  1,  1,  1;
  1,  2,  2,  1;
  1,  6, 12,  6,  1;
  ...

Links

  • Donald Knuth, Two Notes on Notation, arXiv:math/9205211 [math.HO], 1992. (Page 16-17 explain and give examples; the case with Narayana numbers come from tying together the poset P_k's 'shoelaces' with inequalities, into a 2 X k rectangle.)
  • Wolfdieter Lang, First 9 rows
  • Richard P. Stanley, Ordered Structures and Partitions, 1971.

Crossrefs

Column sequences give: A000012 (powers of 1), A000142 (factorials), A010790, A090443-4, etc.
Cf. A090445 (row sums), A090446 (alternating row sums).

Programs

  • PARI
    spf(n) = prod(k=2, n, k!);
T(n,m) = spf(n-1)/spf(m-1)/spf(n-m-1);
row(n) = vector(n+1, k, T(n, k-1)); \\ Michel Marcus, Apr 13 2025

Formula

a(n, m) = 0 if n < m;
a(n, m) = 1 if m = 0 or m = n;
a(n, m) = (Product_{p=1..m} (n-p)!)/superfac(m-1) if n >= 0, 1 <= m <= n+1, where superfac(n) := A000178(n), n >= 0, (superfactorials).
a(n, m) = superfac(n-1)/superfac(m-1)/superfac(n-m-1)
With offset 1, equals ConvOffsStoT transform of the factorials, A000142: (1, 1, 2, 6, 24, ...); e.g., ConvOffs transform of (1, 1, 2, 6) = (1, 6, 12, 6, 1). - Gary W. Adamson, Apr 21 2008

Extensions

OFFSET changed from -1 to 0 by Natalia L. Skirrow, Apr 13 2025

A176037 a(n) = n!*(n+1)!*(n+2)!.

Original entry on oeis.org

2, 12, 288, 17280, 2073600, 435456000, 146313216000, 73741860864000, 53094139822080000, 52563198423859200000, 69383421919494144000000, 119061952013851951104000000, 260031303198252661211136000000, 709885457731229765106401280000000, 2385215137976932010757508300800000000
Offset: 0

Views

Author

Jonathan Vos Post, Apr 07 2010

Keywords

Examples

			a(2) = (2)!*(2+1)!*(2+2)! = (2)!*(3)!*(4)! = 2*6*24 = 288.

Crossrefs

Cf. A000142, A010790, A090443.

Programs

  • Mathematica
    Times@@@Partition[Range[0,15]!,3,1] (* Harvey P. Dale, Aug 29 2012 *)

Formula

a(n) = A000142(n)*A000142(n+1)*A000142(n+2) = A010790(n)*A000142(n+2).
a(n) = 2*A090443(n). - Pontus von Brömssen, Jun 14 2024
Showing 1-3 of 3 results.

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