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 1-10 of 261 results. Next

A131689 Triangle of numbers T(n,k) = k!*Stirling2(n,k) = A000142(k)*A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
(1,1,1) (1,2,2) (1,2,3)
(2,1,2) (1,3,2)
(2,2,1) (2,1,3)
(1,1,2) (2,3,1)
(1,2,1) (3,1,2)
(2,1,1) (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

Examples

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)
		

Crossrefs

Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) + T(n-1, k))
    end
    for n in 0:7
        println([T(n, k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
    
  • Maple
    A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
    # Alternatively:
    A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017
  • Mathematica
    t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
    T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    
  • SageMath
    @cached_function
    def F(n): # Fubini polynomial
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

Formula

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018

A134055 a(n) = Sum_{k=1..n} C(n-1,k-1) * S2(n,k) for n>0, a(0)=1, where S2(n,k) = A048993(n,k) are Stirling numbers of the 2nd kind.

Original entry on oeis.org

1, 1, 2, 8, 41, 252, 1782, 14121, 123244, 1169832, 11960978, 130742196, 1518514076, 18645970943, 241030821566, 3268214127548, 46338504902485, 685145875623056, 10538790233183702, 168282662416550040, 2784205185437851772, 47646587512911994120
Offset: 0

Views

Author

Paul D. Hanna, Oct 08 2007

Keywords

Examples

			O.g.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 41*x^4 + 252*x^5 + 1782*x^6 + 14121*x^7 +...
where
A(x) = 1 + x/(1-x)*exp(-x/(1-x)) + 2^2*x^2/(1-2*x)^2*exp(-2*x/(1-2*x))/2! + 3^3*x^3/(1-3*x)^3*exp(-3*x/(1-3*x))/3! + 4^4*x^4/(1-4*x)^4*exp(-4*x/(1-4*x))/4! +...
simplifies to a power series in x with integer coefficients.
Illustrate the definition of the terms by:
a(4) = 1*1 + 3*7 + 3*6 + 1*1 = 41;
a(5) = 1*1 + 4*15 + 6*25 + 4*10 + 1*1 = 252;
a(6) = 1*1 + 5*31 + 10*90 + 10*65 + 5*15 + 1*1 = 1782.
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local b; b:=
          proc(h, m) option remember; `if`(h=0,
            binomial(n-1, m-1), m*b(h-1, m)+b(h-1, m+1) )
          end; b(n, 0)
        end:
    seq(a(n), n=0..22);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n-1,k-1] * StirlingS2[n,k],{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Aug 11 2014 *)
  • PARI
    a(n)=if(n==0,1,sum(k=1, n, binomial(n-1, k-1)*polcoeff(1/prod(i=0, k, 1-i*x +x*O(x^(n-k))), n-k)))
    
  • PARI
    a(n)=polcoeff(sum(k=0,n+1,(k*x)^k/(1-k*x)^k*exp(-k*x/(1-k*x+x*O(x^n)))/k!),n)
    for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 04 2012

Formula

O.g.f.: Sum_{n>=0} (n*x)^n/(1-n*x)^n * exp(-n*x/(1-n*x)) / n!. - Paul D. Hanna, Nov 04 2012
From Alois P. Heinz, Jun 24 2023: (Start)
a(n) mod 2 = A037011(n) for n >= 1.
a(n) mod 2 = 1 <=> n in { A048297 } or n = 0. (End)

Extensions

An initial '1' was added and definition changed slightly by Paul D. Hanna, Nov 04 2012

A130191 Square of the Stirling2 matrix A048993.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 5, 6, 1, 0, 15, 32, 12, 1, 0, 52, 175, 110, 20, 1, 0, 203, 1012, 945, 280, 30, 1, 0, 877, 6230, 8092, 3465, 595, 42, 1, 0, 4140, 40819, 70756, 40992, 10010, 1120, 56, 1, 0, 21147, 283944, 638423, 479976, 156072, 24570, 1932, 72, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jun 01 2007

Keywords

Comments

Without row n=0 and column k=0 this is triangle A039810.
This is an associated Sheffer matrix with e.g.f. of the m-th column ((exp(f(x))-1)^m)/m! with f(x)=:exp(x)-1.
The triangle is also called the exponential Riordan array [1, exp(exp(x)-1)]. - Peter Luschny, Apr 19 2015
Also the Bell transform of shifted Bell numbers A000110(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle starts:
  1;
  0,   1;
  0,   2,    1;
  0,   5,    6,    1;
  0,  15,   32,   12,    1;
  0,  52,  175,  110,   20,   1;
  0, 203, 1012,  945,  280,  30,  1;
  0, 877, 6230, 8092, 3465, 595, 42, 1;
		

Crossrefs

Columns k=0..3 give A000007, A000110 (for n > 0), A000558, A000559.
Row sums: A000258.
Alternating row sums: A130410.
T(2n,n) gives A321712.
Cf. A039810 (another version), A048993.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> combinat:-bell(n+1), 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 10;
    M = BellMatrix[BellB[# + 1]&, rows];
    Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
    a[n_, m_]:= Sum[StirlingS2[n, k]*StirlingS2[k, m], {k,m,n}]; Table[a[n, m], {n, 0, 100}, {m, 0, n}]//Flatten (* G. C. Greubel, Jul 10 2018 *)
  • PARI
    for(n=0, 9, for(k=0, n, print1(sum(j=k, n, stirling(n, j, 2)*stirling(j, k, 2)), ", "))) \\ G. C. Greubel, Jul 10 2018
  • Sage
    # uses[riordan_array from A256893]
    riordan_array(1, exp(exp(x) - 1), 8, exp=true) # Peter Luschny, Apr 19 2015
    

Formula

a(n,k) = Sum_{j=k..n} S2(n,j) * S2(j,k), n>=k>=0.
E.g.f. row polynomials with argument x: exp(x*f(f(z))).
E.g.f. column k: ((exp(exp(x) - 1) - 1)^k)/k!.

A171367 Antidiagonal sums of triangle of Stirling numbers of 2nd kind A048993.

Original entry on oeis.org

1, 0, 1, 1, 2, 4, 9, 22, 58, 164, 495, 1587, 5379, 19195, 71872, 281571, 1151338, 4902687, 21696505, 99598840, 473466698, 2327173489, 11810472444, 61808852380, 333170844940, 1847741027555, 10532499571707, 61649191750137, 370208647200165, 2278936037262610, 14369780182166215
Offset: 0

Views

Author

Paul Barry, Dec 06 2009

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, m) option remember; `if`(n<=m,
         `if`(n=m, 1, 0), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..30);  # Alois P. Heinz, May 16 2023
  • Mathematica
    Table[Sum[StirlingS2[n-k, k], {k, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 18 2016 *)
  • Maxima
    makelist(sum(stirling2(n-k,k),k,0,n),n,0,60); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    a(n) = sum(k=0, n, stirling(n-k, k,2)); /* Joerg Arndt, Jan 16 2013 */

Formula

G.f.: 1/(1-x^2/(1-x/(1-x^2/(1-2x/(1-x^2/1-3x/(1-x^2/(1-4x/(1-x^2/(1-5x/(1-... (continued fraction).
G.f.: (G(0) - 1)/(x-1)/x where G(k) = 1 - x/(1-k*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: T(0)/(1-x^2), where T(k) = 1-x^3*(k+1)/(x^3*(k+1)-(1-x*(x+k))*(1-x*(x+1+k))/T(k+1) ); (continued fraction, after P. Flajolet, p. 140). - Sergei N. Gladkovskii, Oct 30 2013
G.f. (alternating signs): Sum_{k>=0} S(x,k)*x^k, where S(x,k)*exp(-x) is the inverse Mellin transform of Gamma(s)*s^k. - Benedict W. J. Irwin, Oct 14 2016

A112494 Sixth diagonal of the Stirling2 triangle A048993 and sixth column of triangle A008278.

Original entry on oeis.org

1, 63, 966, 7770, 42525, 179487, 627396, 1899612, 5135130, 12662650, 28936908, 62022324, 125854638, 243577530, 452329200, 809944464, 1404142047, 2364885369, 3880739170, 6220194750, 9759104355, 15015551265, 22693687380, 33738295500, 49402080000, 71327958156
Offset: 6

Views

Author

Wolfdieter Lang, Oct 14 2005

Keywords

Crossrefs

Cf. A001298 (fifth diagonal, resp. column).

Programs

  • Mathematica
    Table[StirlingS2[n, n-5], {n, 6, 100}] (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
  • PARI
    for(n=6,50, print1(stirling(n,n-5,2), ", ")) \\ G. C. Greubel, Oct 22 2017
    
  • PARI
    Vec(x^6*(1 + 52*x + 328*x^2 + 444*x^3 + 120*x^4) / (1 - x)^11 + O(x^40)) \\ Colin Barker, Nov 04 2017
  • Sage
    [stirling_number2(n,n-5) for n in range(6, 30)] # Zerinvary Lajos, May 16 2009
    

Formula

a(n) = Stirling2(n, n-5) with Stirling2(n, m)=A048993(n, m). a(n) = A008278(n+5, 6).
a(n) = sum(A008517(5, m+1)*binomial(n+5-m, 2*5), m=0..4) from the o.g.f. See p. 257 eq. (6.43) of the R. L. Graham et al. book quoted in A008517.
O.g.f.: x*sum(A008517(5, m+1)*x^m, m=0..4)/(1-x)^11 with the fifth row [1, 52, 328, 444, 120] of the second-order Eulerian triangle A008517.
E.g.f. with offset n=-4: exp(x)*sum(A112493(5, m)*(x^(m+5))/(m+5)!, m=0..5) with the k=5 row [1, 57, 546, 1750, 2205, 945] of triangle A112493.
a(n) = sum(A112493(5, m)*binomial(n+4, 5+m), m=0..5) from the e.g.f. (coefficients from A112493(5, m) are [1, 57, 546, 1750, 2205, 945]).
With an offset of 1 the o.g.f. is D^5(x/(1-x)), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
G.f.: x^6*(1 + 52*x + 328*x^2 + 444*x^3 + 120*x^4) / (1 - x)^11. - Colin Barker, Nov 04 2017

A321331 Triangle read by rows: T(n, k) = (k+1)*S2(n+1, k+1), for n >= k >= 0, and S2 = A048993 (Stirling2).

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 14, 18, 4, 1, 30, 75, 40, 5, 1, 62, 270, 260, 75, 6, 1, 126, 903, 1400, 700, 126, 7, 1, 254, 2898, 6804, 5250, 1596, 196, 8, 1, 510, 9075, 31080, 34755, 15876, 3234, 288, 9, 1, 1022, 27990, 136420, 212625, 136962, 41160, 6000, 405, 10, 1, 2046, 85503, 583000, 1233650, 1076922, 447909, 95040, 10395, 550, 11
Offset: 0

Views

Author

Wolfdieter Lang, Dec 03 2018

Keywords

Comments

This lower triangular matrix T is the inverse of the triangular matrix with elements Narumi[-1](n,m)/(m+1) = S1(n+1, m+1)/(n+1), with the Narumi triangle for parameter a = -1, and S1 = A048994 (Stirling1), i.e., Sum_{k=m..n} T(n, k) * S1(k+1, m+1)/(k+1) = delta_{n,m} (Kronecker symbol).
This triangle arises from the inverse of the rational Sheffer matrix Narumi[-1] = (log(1+x)/x, log(1+x) (such special Sheffer matrices (g(x), x*g(x)) define elements of the Narumi subgroup). The inverse matrix is (Narumi[-1])^(-1) = ((exp(x) - 1)/x, exp(x) - 1).
In order to have an integer matrix one takes T(n, k) := (n+1)*(Narumi[-1])^(-1)(n, k) = (k+1)*S2(n+1, k+1). The connection to S2 = A048993 results from the general relation between each Narumi-type matrix N = (g(x), x*g(x)) and its associated Sheffer matrix J = (1, x*g(x)) (this is of the Jabotinsky-type), i.e., N(n, m) = (m+1)*J(n+1, m+1)/(n+1), or with the row polynomials Npol(n, x) = (1/(n+1))*(d/dx)Jpol(n+1, x).
The signed triangle (-1)^(n-k)*A028421(n, k) (with upper diagonals filled with zeros) gives the integer matrix Nscaled with elements (n+1)*Narumi[-1](n,k). This inverse of Nscaled has the rational elements (Narumi[-1])^(-1)(k, m)/(m+1) = (1/(k+1))*S2(k+1, m+1).
The a- and z- sequence for the Sheffer matrix (Narumi[-1])^(-1) (see A006232 for a link on these sequences) have e.g.f.s Ea(x) = x/log(1 + x) and Ez(x) = 1/log(1 + x) - 1/x, hence a(n) = A006232(n)/A006233(n) and z(n) = A006232(n+1)/A075178(n), for n >= 0. This leads to the recurrence for T(n, k) given in the formula section.
The Boas-Buck-type column recurrence (see the link, also for references) uses the sequence with o.g.f. GBB(y) = exp(y)/(exp(y) - 1) - 1/y, with BB(n) = (-1)^(n+1)*A060054(n+1 ) / A227830(n+1), for n >= 1. For the recurrence see the formula section.
The Meixner-type identity (see the Meixner link) for the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, derived from the one for the Narumi[-1]^(-1) row polynomials is Sum_{k=1..n} (-1)^{k+1}*(1/k)*(d/dx)^k R(n, x)/(n+1) = R(n-1, x), for n >= 1, and R(0, x) = 1. Here d/dx is a differentiation operator.
The Roman-type recurrence for the row polynomials (see the reference, Corollary 3.7.2. p. 50) becomes, with the z-sequence from above: R(n, x) = ((n+1)/n)*{(x + 1/2)*1 + (x - z(1))*d/dx - Sum_{k=2..n-1} (1/k!)*z(k)*(d/dx)^k}*R(n-1, x), for n >= 1, and R(0, x) = 1.
The triangle is the exponential Riordan square (cf. A321620) of exp(x)-1 with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
  n\k  0    1     2      3       4       5      6     7     8   9 10 ...
  ----------------------------------------------------------------------
  0:   1
  1:   1    2
  2:   1    6     3
  3:   1   14    18      4
  4:   1   30    75     40       5
  5:   1   62   270    260      75       6
  6:   1  126   903   1400     700     126      7
  7:   1  254  2898   6804    5250    1596    196     8
  8:   1  510  9075  31080   34755   15876   3234   288     9
  9:   1 1022 27990 136420  212625  136962  41160  6000   405  10
  10:  1 2046 85503 583000 1233650 1076922 447909 95040 10395 550 11
  ...
Recurrence (from Stirling2): T(4, 2) = 3*(T(3, 2) + T(3, 1)/2) = 3*(18 + 14/2) = 75.
Recurrence (from a- and z-sequence): T(4, 0) = 5*((1/2)*T(3, 0) - (1/12)*T(3, 1) + (1/12)*T(3, 2) - (19/120)*T(3, 3)) = 5*(1/2 - 14/12 + 18/12 - 4*19/120) = 1; T(4,2) = (5/2)*(1*1*T(3, 1) + 2*(1/2)*T(3, 2) + 3*(-1/6)* T(3, 3)) = (5/2)*(14 + 18 - 2) = 75.
Recurrence for column k=2 (Boas-Buck-type): T(4, 2) = (5!*3/2)*((1/3!)*(1/12)*T(2, 2) + (1/4!)*(1/2)*T(3, 2)) = (5!*3/2)*((1/72)*3 + (1/48)*18) = 75.
Meixner identity for the row polynomials, for n = 3: {d/dx  - (1/2)*(d/dx)^2 + (1/3)*(d/dx)^3)}*R(3, x)/4) = ((14 - 36/2 + 24/3) + (36 - 24/2)*x + 12*x^2)/4 = (1 + 6*x + 3*x^2) = R(2, x).
Roman type recurrence for row polynomials: R(n, 3) = (3/2)*{(x + 1/12)*(1 + 6*x + 3*x^2) + (x - (-1/2))*(6 + 6*x) - (1/2!)*(1/12)*6} = 1 + 14*x + 18*x^2 + 4*x^3.
		

References

  • Steven Roman, The umbral calculus, Academic Press, 1984.

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->(k+1)*Stirling2(n+1,k+1)))); # Muniru A Asiru, Dec 03 2018
    
  • Maple
    T:=(n,k)->(k+1)*Stirling2(n+1,k+1): seq(seq(T(n,k),k=0..n),n=0..10); # Muniru A Asiru, Dec 03 2018
  • Mathematica
    T[n_, k_] := (k+1) * StirlingS2[n+1, k+1];  Table[T[n, k], {n,0,10}, {k, 0, n}] //Flatten (* Amiram Eldar, Dec 03 2018 *)
  • PARI
    T(n, k) = (k+1)*stirling(n+1, k+1, 2) \\ Thomas Scheuerle, Nov 10 2023
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(exp(x) - 1, 10, True) # Peter Luschny, Jan 03 2019
    

Formula

T(n, k) = (k+1)*A048993(n+1, k+1), with A048993 = Stirling2, for n >= k >= 0, and 0 otherwise.
T(n, k) = (n+1)*(Narumi[a=-1])^(-1)(n, k), with the Narumi[a=-1] matrix with entries (-1)^(n-k)*A028421(n, k)/(n+1).
E.g.f. for column k sequence: E(k, x) = (x*d/dx + 1)*EN(k, x), where EN(k, x) = (exp(x) - 1)^(k+1)/(x*k!) is the e.g.f. for the (Narumi[a=-1])^(-1) columns. Hence E(k, x) = exp(x)*(exp(x) - 1)*(k+1)/k!, for k >= 0.
E.g.f. for (ordinary) row polynomials R(n, x): Epol(z, x) = exp(z)*exp(x*(exp(z) - 1))*(1 + x*(exp(z) - 1)).
Recurrence (from Stirling2): T(n, k) = 0 for n < k; T(n, 0) = (k + 1)*T(n-1, k), for n <= 1, T(0, 0) = 1; T(n, k) = (k+1)*(T(n-1, k) + T(n-1, k-1)/k), for n >= 1, k >= 1.
Recurrence (from a- and z-sequence, see above): a = {1, 1/2, -1/6, 1/4, -19/30, 9/4, ...}, z = {1/2, -1/12, 1/12, -19/120, 9/20, -863/504, ...}.
T(n, k) = 0, for n < k; T(n, 0) = (n+1)*Sum_{j=0..n-1} z(j)*T(n-1, j), for n >= 1, with T(0, 0) = 1; T(n, k) = ((n+1)/k)*Sum_{j=0..n-m} binomial(k-1+j, j)*a(j)*T(n-1, k-1+j).
Recurrence for column k, from the Boas-Buck-type sequence BB(n) = (-1)^(n+1)*A060054(n+1)/A227830(n+1), for n >= 0; BB = {1/2, 1/12, 0, -1/720, 0, 1/30240, 0, -1/1209600, ...}: T(n, k) = 0, for n < k; T(n, n) = n+1, for n >= 0; T(n, k) = ((n+1)!*(k+1)/(n-k))*Sum_{j=k..n-1} (1/(j+1)!)*BB(n-(j+1))*T(j, k), for n >= 0 and k = 0, 1, ..., n-1.
T(n, k) = Stirling2(n+2, k+1) - Stirling2(n+1, k). - Peter Luschny, May 26 2020

A002871 a(n) = max_{k=0..n} 2^k*A048993(n,k).

Original entry on oeis.org

1, 2, 4, 12, 48, 200, 1040, 5600, 33600, 222432, 1460928, 11487168, 84713728, 731574272, 6314147840, 55456727040, 548291597568, 5226494727168, 54361802626560, 586042688924160, 6149776714099200, 72895623466265600, 855187250563024896
Offset: 0

Views

Author

Keywords

Comments

Original name: Sorting numbers (see Motzkin article for details).
For n>0, a(n) is also the maximum term in row n of the triangle in A227450. - Danny Rorabaugh, Oct 24 2015

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    a:= n-> max(seq(2^k*Stirling2(n, k), k=0..n)):
    seq(a(n), n=0..30);  # Alois P. Heinz, Mar 26 2013
  • Mathematica
    a[n_] := Max[Table[2^k*StirlingS2[n, k], {k, 0, n}]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 25 2015 *)
  • PARI
    a(n) = vecmax(vector(n+1, k, 2^(k-1)*stirling(n, k-1, 2))); \\ Michel Marcus, Feb 25 2015

Formula

a(n) = max{2^k*Stirling2(n,k), k=0..n}. - Sean A. Irvine, Mar 26 2013

Extensions

More terms from Sean A. Irvine, Mar 26 2013
New name from Danny Rorabaugh, Oct 24 2015

A226156 a(n) = BS(n) * W(n) where BS = Sum_{k=0..n} ((-1)^k*k!/(k+1)) S(n, k) and S(n, k) the Stirling subset numbers A048993(n, k). W(n) = Product_{ p primes <= n+1 such that p divides n+1 or p-1 divides n } = A225481(n).

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 35, 0, -3617, 0, 43867, 0, -1222277, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -84802531453387, 0, 90219075042845, 0, -26315271553053477373, 0, 38089920879940267
Offset: 0

Views

Author

Peter Luschny, May 30 2013

Keywords

Comments

a(n)/A225481(n) is a representation of the Bernoulli numbers. This is case m = 1 of the scaled generalized Bernoulli numbers defined as Sum_{k=0..n} ((-1)^k*k!/(k+1)) S_{m}(n,k) where S_{m}(n,k) are generalized Stirling subset numbers. A225481(n) can be seen as an analog of the Clausen numbers A141056(n). Reduced to lowest terms a(n)/A225481(n) becomes A027641(n)/A027642(n).

Examples

			The numerators of 1/1, -1/2, 1/6, 0/2, -1/30, 0/6, 1/42, 0/2, -1/30, 0/10, 5/66, 0/6, -691/2730, 0/14, 35/30, 0/2, -3617/510, 0/6, 43867/798, ... (the denominators are A225481(n)).
		

Crossrefs

Programs

  • Mathematica
    BS[n_] := Sum[((-1)^k*k!/(k + 1)) StirlingS2[n, k], {k, 0, n}];
    W[n_] := Product[If[Divisible[n + 1, p] || Divisible[n, p - 1], p, 1], {p, Prime /@ Range[PrimePi[n + 1]]}];
    a[n_] := BS[n] W[n];
    Table[a[n], {n, 0, 38}] (* Jean-François Alcover, Jul 08 2019 *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :   # -- The Eulerian numbers --
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + \
               (m*k+1)*EulerianNumber(n-1, k, m)
    @CachedFunction
    def SF_BS(n, m):   # -- The scaled Stirling-Frobenius Bernoulli numbers --
        return add(add(EulerianNumber(n, j, m)*binomial(j, n - k) \
               for j in (0..n))/((-m)^k*(k+1)) for k in (0..n))
    def A226156(n):    # -- The numerators of SF_BS(n, 1) relative to A225481 --
        C = mul(filter(lambda p: ((n+1)%p == 0) or (n%(p-1) == 0), primes(n+2)))
        return C*SF_BS(n, 1)
    [A226156(n) for n in (0..25)]

A245059 a(n) = Sum_{k=1..n} C(n-1,k-1) * S2(n,k) * 2^(n-k) for n>0, a(0)=1, where S2(n,k) = A048993(n,k) are Stirling numbers of the 2nd kind.

Original entry on oeis.org

1, 1, 3, 17, 129, 1177, 12463, 149053, 1975473, 28628865, 449059179, 7562334793, 135837896769, 2588529249737, 52093016105575, 1102851978691749, 24480094135644513, 568066476383361793, 13745454515733689427, 346020796943921077057, 9043636093339718229697, 244954584886648170627641
Offset: 0

Views

Author

Paul D. Hanna, Jul 10 2014

Keywords

Examples

			O.g.f.: A(x) = 1 + x + 3*x^2 + 17*x^3 + 129*x^4 + 1177*x^5 + 12463*x^6 +...
where
A(x) = 1 + x/(1-2*x)*exp(-x/(1-2*x)) + 2^2*x^2/(1-4*x)^2*exp(-2*x/(1-4*x))/2! + 3^3*x^3/(1-6*x)^3*exp(-3*x/(1-6*x))/3! + 4^4*x^4/(1-8*x)^4*exp(-4*x/(1-8*x))/4! +...
simplifies to a power series in x with integer coefficients.
Illustrate the definition of the terms by:
a(2) = 1*1*2 + 1*1 = 3;
a(3) = 1*1*2^2 + 2*3*2 + 1*1 = 17;
a(4) = 1*1*2^3 + 3*7*2^2 + 3*6*2 + 1*1 = 129;
a(5) = 1*1*2^4 + 4*15*2^3 + 6*25*2^2 + 4*10*2 + 1*1 = 1177;
a(6) = 1*1*2^5 + 5*31*2^4 + 10*90*2^3 + 10*65*2^2 + 5*15*2 + 1*1 = 12463; ...
		

Crossrefs

Programs

  • PARI
    {a(n)=if(n==0, 1, sum(k=1, n, binomial(n-1, k-1)*polcoeff(1/prod(i=0, k, 1-i*x +x*O(x^(n-k))), n-k)*2^(n-k)))}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff(sum(k=0, n+1, (k*x)^k/(1-2*k*x)^k*exp(-k*x/(1-2*k*x+x*O(x^n)))/k!), n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

O.g.f.: Sum_{n>=0} (n*x)^n/(1-2*n*x)^n * exp(-n*x/(1-2*n*x)) / n!.

A245060 a(n) = Sum_{k=1..n} C(n-1,k-1) * S2(n,k) * 3^(n-k) for n>0, a(0)=1, where S2(n,k) = A048993(n,k) are Stirling numbers of the 2nd kind.

Original entry on oeis.org

1, 1, 4, 28, 271, 3172, 43174, 666577, 11445214, 215478712, 4401799930, 96757165012, 2273105615356, 56755763435503, 1499039156935948, 41714498328290992, 1218787798107634291, 37275555462806318512, 1190200470204107432854, 39581409916012393962280, 1368112674516484881342244
Offset: 0

Views

Author

Paul D. Hanna, Jul 10 2014

Keywords

Examples

			O.g.f.: A(x) = 1 + x + 4*x^2 + 28*x^3 + 271*x^4 + 3172*x^5 + 43174*x^6 +...
where
A(x) = 1 + x/(1-3*x)*exp(-x/(1-3*x)) + 2^2*x^2/(1-6*x)^2*exp(-2*x/(1-6*x))/2! + 3^3*x^3/(1-9*x)^3*exp(-3*x/(1-9*x))/3! + 4^4*x^4/(1-12*x)^4*exp(-4*x/(1-12*x))/4! +...
simplifies to a power series in x with integer coefficients.
Illustrate the definition of the terms by:
a(2) = 1*1*3 + 1*1 = 4;
a(3) = 1*1*3^2 + 2*3*3 + 1*1 = 28;
a(4) = 1*1*3^3 + 3*7*3^2 + 3*6*3 + 1*1 = 271;
a(5) = 1*1*3^4 + 4*15*3^3 + 6*25*3^2 + 4*10*3 + 1*1 = 3172;
a(6) = 1*1*3^5 + 5*31*3^4 + 10*90*3^3 + 10*65*3^2 + 5*15*3 + 1*1 = 43174; ...
		

Crossrefs

Programs

  • PARI
    {a(n)=if(n==0, 1, sum(k=1, n, binomial(n-1, k-1)*polcoeff(1/prod(i=0, k, 1-i*x +x*O(x^(n-k))), n-k)*3^(n-k)))}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff(sum(k=0, n+1, (k*x)^k/(1-3*k*x)^k*exp(-k*x/(1-3*k*x+x*O(x^n)))/k!), n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

O.g.f.: Sum_{n>=0} (n*x)^n/(1-3*n*x)^n * exp(-n*x/(1-3*n*x)) / n!.
Showing 1-10 of 261 results. Next