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.

A054654 Triangle of Stirling numbers of 1st kind, S(n, n-k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -3, 2, 0, 1, -6, 11, -6, 0, 1, -10, 35, -50, 24, 0, 1, -15, 85, -225, 274, -120, 0, 1, -21, 175, -735, 1624, -1764, 720, 0, 1, -28, 322, -1960, 6769, -13132, 13068, -5040, 0
Offset: 0

Views

Author

N. J. A. Sloane, Apr 18 2000

Keywords

Comments

Triangle is the matrix product of the binomial coefficients with the Stirling numbers of the first kind.
Triangle T(n,k) giving coefficients in expansion of n!*C(x,n) in powers of x. E.g., 3!*C(x,3) = x^3-3*x^2+2*x.
The matrix product of binomial coefficients with the Stirling numbers of the first kind results in the Stirling numbers of the first kind again, but the triangle is shifted by (1,1).
Essentially [1,0,1,0,1,0,1,0,...] DELTA [0,-1,-1,-2,-2,-3,-3,-4,-4,...] where DELTA is the operator defined in A084938; mirror image of the Stirling-1 triangle A048994. - Philippe Deléham, Dec 30 2006
From Doudou Kisabaka, Dec 18 2009: (Start)
The sum of the entries on each row of the triangle, starting on the 3rd row, equals 0. E.g., 1+(-3)+2+0 = 0.
The entries on the triangle can be computed as follows. T(n,r) = T(n-1,r) - (n-1)*T(n-1,r-1). T(n,r) = 0 when r equals 0 or r > n. T(n,r) = 1 if n==1. (End)

Examples

			Matrix begins:
  1, 0,  0,  0,  0,   0,    0,     0,      0, ...
  0, 1, -1,  2, -6,  24, -120,   720,  -5040, ...
  0, 0,  1, -3, 11, -50,  274, -1764,  13068, ...
  0, 0,  0,  1, -6,  35, -225,  1624, -13132, ...
  0, 0,  0,  0,  1, -10,   85,  -735,   6769, ...
  0, 0,  0,  0,  0,   1,  -15,   175,  -1960, ...
  0, 0,  0,  0,  0,   0,    1,   -21,    322, ...
  0, 0,  0,  0,  0,   0,    0,     1,    -28, ...
  0, 0,  0,  0,  0,   0,    0,     0,      1, ...
  ...
Triangle begins:
  1;
  1,   0;
  1,  -1,   0;
  1,  -3,   2,    0;
  1,  -6,  11,   -6,    0;
  1, -10,  35,  -50,   24,     0;
  1, -15,  85, -225,  274,  -120,   0;
  1, -21, 175, -735, 1624, -1764, 720, 0;
  ...
		

References

  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 18, table 18:6:1 at page 152.

Crossrefs

Essentially Stirling numbers of first kind, multiplied by factorials - see A008276.
The Stirling2 counterpart is A106800.

Programs

  • Haskell
    a054654 n k = a054654_tabl !! n !! k
    a054654_row n = a054654_tabl !! n
    a054654_tabl = map reverse a048994_tabl
    -- Reinhard Zumkeller, Mar 18 2014
  • Maple
    a054654_row := proc(n) local k; seq(coeff(expand((-1)^n*pochhammer (-x,n)),x,n-k),k=0..n) end: # Peter Luschny, Nov 28 2010
    seq(seq(Stirling1(n, n-k), k=0..n), n=0..8); # Peter Luschny, Feb 21 2021
  • Mathematica
    row[n_] := Reverse[ CoefficientList[ (-1)^n*Pochhammer[-x, n], x] ]; Flatten[ Table[ row[n], {n, 0, 8}]] (* Jean-François Alcover, Feb 16 2012, after Maple *)
    Table[StirlingS1[n,n-k],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Jun 17 2023 *)
  • PARI
    T(n,k)=polcoeff(n!*binomial(x,n), n-k)
    

Formula

n!*binomial(x, n) = Sum_{k=0..n} T(n, k)*x^(n-k).
(In Maple notation:) Matrix product A*B of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling1(j,i) with i from 1 to d, j from 1 to d, d=9.
T(n, k) = (-1)^k*Sum_{j=0..k} E2(k, j)*binomial(n+j-1, 2*k), where E2(k, j) are the second-order Eulerian numbers A340556. - Peter Luschny, Feb 21 2021

Extensions

Additional comments from Thomas Wieder, Dec 29 2006
Edited by N. J. A. Sloane at the suggestion of Eric W. Weisstein, Jan 20 2008

A094791 Triangle read by rows giving coefficients of polynomials arising in successive differences of (n!)_{n>=0}.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 3, 5, 2, 1, 6, 17, 20, 9, 1, 10, 45, 100, 109, 44, 1, 15, 100, 355, 694, 689, 265, 1, 21, 196, 1015, 3094, 5453, 5053, 1854, 1, 28, 350, 2492, 10899, 29596, 48082, 42048, 14833, 1, 36, 582, 5460, 32403, 124908, 309602, 470328, 391641, 133496
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Let D_0(n)=n! and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n), then D_{k}(n)=n!*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.
The horizontal reversal of this triangle arises as a binomial convolution of the derangements coefficients der(n,i) (numbers of permutations of size n with i derangements = A098825(n,i) = number of permutations of size n with n-i rencontres = A008290(n,n-i), see formula section). - Olivier Gérard, Jul 31 2011

Examples

			D_3(n) = n!*(n^3 + 3*n^2 + 5*n + 2).
D_4(n) = n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9).
Table begins:
  1
  1  0
  1  1   1
  1  3   5   2
  1  6  17  20    9
  1 10  45 100  109   44
  1 15 100 355  694  689  265
  ...
		

Crossrefs

Successive differences of factorial numbers: A001563, A001564, A001565, A001688, A001689, A023043.
Rencontres numbers A008290. Partial derangements A098825.
Row sum is A000255. Signed version in A126353.

Programs

  • Maple
    with(LREtools): A094791_row := proc(n)
    delta(x!,x,n); simplify(%/x!); seq(coeff(%,x,n-j),j=0..n) end:
    seq(print(A094791_row(n)),n=0..9); # Peter Luschny, Jan 09 2015
  • Mathematica
    d[0][n_] := n!; d[k_][n_] := d[k][n] = d[k - 1][n + 1] - d[k - 1][n] // FullSimplify;
    row[k_] := d[k][n]/n! // FullSimplify // CoefficientList[#, n]& // Reverse;
    Array[row, 10, 0] // Flatten (* Jean-François Alcover, Aug 02 2019 *)

Formula

T(n, n) = A000166(n).
T(2, k) = A000217(k).
Sum_{k=0..n} T(n,n-k)*x^k = Sum_{i=0..n} der(n,i)*binomial( n+x, i) (an analog of Worpitzky's identity). - Olivier Gérard, Jul 31 2011

Extensions

Edited and T(0,0) corrected according to the author's definition by Olivier Gérard, Jul 31 2011

A126351 Triangle read by rows: matrix product of the Stirling numbers of the second kind with the binomial coefficients.

Original entry on oeis.org

1, 1, 2, 1, 5, 4, 1, 9, 19, 8, 1, 14, 55, 65, 16, 1, 20, 125, 285, 211, 32, 1, 27, 245, 910, 1351, 665, 64, 1, 35, 434, 2380, 5901, 6069, 2059, 128, 1, 44, 714, 5418, 20181, 35574, 26335, 6305, 256, 1, 54, 1110, 11130, 58107, 156660, 204205, 111645, 19171, 512
Offset: 1

Views

Author

Thomas Wieder, Dec 29 2006

Keywords

Comments

Many well-known integer sequences arise from such a matrix product of combinatorial coefficients. In the present case we have as the first row A000079 = the powers of two = 2^n. As the second row we have A001047 = 3^n - 2^n. As the column sums we have 1,3,10,37,151,674,3263,17007,94828 we have A005493 = number of partitions of [n+1] with a distinguished block.

Examples

			Matrix begins:
1, 2, 4,  8, 16,  32,   64,  128,   256, ... A000079
0, 1, 5, 19, 65, 211,  665, 2059,  6305, ... A001047
0, 0, 1,  9, 55, 285, 1351, 6069, 26335, ... A016269
0, 0, 0,  1, 14, 125,  910, 5901, 35574, ... A025211
0, 0, 0,  0,  1,  20,  245, 2380, 20181, ...
0, 0, 0,  0,  0,   1,   27,  434,  5418, ...
0, 0, 0,  0,  0,   0,    1,   35,   714, ...
0, 0, 0,  0,  0,   0,    0,    1,    44, ...
0, 0, 0,  0,  0,   0,    0,    0,     1, ...
Triangle begins:
1;
1,  2;
1,  5,  4;
1,  9, 19,  8;
1, 14, 55, 65, 16;
		

Crossrefs

Programs

  • Maple
    T:= (n, k)-> add(binomial(n-1, i-1) *Stirling2(i, n+1-k), i=1..n):
    seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Sep 29 2011
  • Mathematica
    T[n_, k_] := Sum[Binomial[n-1, i-1]*StirlingS2[i, n+1-k], {i, 1, n}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 08 2016, after Alois P. Heinz *)

Formula

(In Maple notation:) Matrix product B.A of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling2(j,i) with i from 1 to d, j from 1 to d, d=9.
T(n,k) = Sum_{i=1..n} C(n-1,i-1) * Stirling2(i, n+1-k). - Alois P. Heinz, Sep 29 2011

A126350 Triangle read by rows: matrix product of the binomial coefficients with the Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 9, 22, 15, 1, 14, 61, 99, 52, 1, 20, 135, 385, 471, 203, 1, 27, 260, 1140, 2416, 2386, 877, 1, 35, 455, 2835, 9156, 15470, 12867, 4140, 1, 44, 742, 6230, 28441, 72590, 102215, 73681, 21147
Offset: 1

Views

Author

Thomas Wieder, Dec 29 2006

Keywords

Comments

Many well-known integer sequences arise from such a matrix product of combinatorial coefficients. In the present case we have as the first row (not surprisingly) A000110 = Bell or exponential numbers: ways of placing n labeled balls into n indistinguishable boxes . As second row we have A033452 = "STIRLING" transform of squares A000290. As the column sums we have 1, 3, 11, 47, 227, 1215, 7107, 44959, 305091 which is A035009 = STIRLING transform of [1,1,2,4,8,16,32, ...].

Examples

			Matrix begins:
1 2 5 15 52 203  877  4140  21147
0 1 5 22 99 471 2386 12867  73681
0 0 1  9 61 385 2416 15470 102215
0 0 0  1 14 135 1140  9156  72590
0 0 0  0 1   20  260  2835  28441
0 0 0  0 0    1   27   455   6230
0 0 0  0 0    0    1    35    742
0 0 0  0 0    0    0     1     44
0 0 0  0 0    0    0     0      1
		

Crossrefs

Programs

  • Maple
    T:= (n, k)-> add(Stirling2(n, j)*binomial(j-1, n-k), j=n-k+1..n):
    seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Sep 03 2019
  • Mathematica
    T[dim_] := T[dim] = Module[{M}, M[n_, n_] = 1; M[, ] = 0; Do[M[n, k] = M[n-1, k-1] + (k+2) M[n-1, k] + (k+1) M[n-1, k+1], {n, 0, dim-1}, {k, 0, n-1}]; Array[M, {dim, dim}, {0, 0}]];
    dim = 9;
    Table[T[dim][[n]][[1 ;; n]] // Reverse, {n, 1, dim}] (* Jean-François Alcover, Jun 27 2019, from Sage *)
  • Sage
    def A126350_triangle(dim): # rows in reversed order
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+(k+2)*M[n-1,k]+(k+1)*M[n-1,k+1]
        return M
    A126350_triangle(9) # Peter Luschny, Sep 19 2012

Formula

(In Maple notation:) Matrix product A.B of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling2(j,i) with i from 1 to d, j from 1 to d, d=9.
Showing 1-4 of 4 results.