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 10 results.

A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

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

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)
		

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
    A143494 = lambda n,k: stirling2r(n, k, 2)
    for n in (2..6):
        [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012

A028025 Expansion of 1/((1-3x)*(1-4x)*(1-5x)*(1-6x)).

Original entry on oeis.org

1, 18, 205, 1890, 15421, 116298, 830845, 5709330, 38119741, 249026778, 1599719485, 10142356770, 63639854461, 396031348458, 2448208592125, 15053605980210, 92160458747581, 562225198873338, 3419937140824765
Offset: 0

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Keywords

Comments

This gives the fourth column of the Sheffer triangle A143495 (3-restricted Stirling2 numbers). See the e.g.f. given below, and comments on the general case under A193685. - Wolfdieter Lang, Oct 08 2011

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/((1-3x)(1-4x)(1-5x)(1-6x)),{x,0,30}],x] (* or *) LinearRecurrence[{18,-119,342,-360},{1,18,205,1890},30] (* Harvey P. Dale, Jan 29 2024 *)
  • PARI
    Vec(1/((1-3*x)*(1-4*x)*(1-5*x)*(1-6*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,3), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = -5^(n+3)/2 + 2*4^(n+2)+ 6^(n+2) - 3^(n+2)/2. - R. J. Mathar, Mar 22 2011
O.g.f.: 1/((1-3*x)*(1-4*x)*(1-5*x)*(1-6*x)).
E.g.f.: (d^3/dx^3)(exp(3*x)*((exp(x)-1)^3)/3!). - Wolfdieter Lang, Oct 08 2011

A016075 Expansion of 1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)).

Original entry on oeis.org

1, 38, 905, 17290, 289821, 4453638, 64331905, 887339330, 11810819141, 152832918238, 1933092302505, 23997027406170, 293289532268461, 3537885908902838, 42204462297434705, 498697803478957810, 5844588402226277781, 68011678300853991438, 786547256602640400505
Offset: 0

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Keywords

Crossrefs

Programs

  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)))); // Vincenzo Librandi, Jun 24 2013
    
  • Magma
    I:=[1, 38, 905, 17290]; [n le 4 select I[n] else 38*Self(n-1)-539*Self(n-2)+3382*Self(n-3)-7920*Self(n-4): n in [1..20]]; // Vincenzo Librandi, Jun 24 2013
    
  • Mathematica
    CoefficientList[Series[1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)), {x,0,20}], x] (* Vincenzo Librandi, Jun 23 2013 *)
  • PARI
    x='x+O('x^30); Vec(1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x))) \\ G. C. Greubel, Feb 07 2018

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,8), (n>=3). - Milan Janjic, Apr 26 2009
a(n) = 38*a(n-1) - 539*a(n-2) + 3382*a(n-3) - 7920*a(n-4), n>=4. - Vincenzo Librandi, Mar 17 2011
a(n) = 21*a(n-1) - 110*a(n-2) + 9^(n+1) - 8^(n+1), n>=2. - Vincenzo Librandi, Mar 17 2011
a(n) = 11^(n+3)/6 -5*10^(n+2) -4*8^(n+2)/3 + 9^(n+3)/2. - R. J. Mathar, Mar 18 2011

A028165 Expansion of 1/((1-5x)*(1-6x)*(1-7x)*(1-8x)).

Original entry on oeis.org

1, 26, 425, 5590, 64701, 688506, 6906145, 66324830, 616252901, 5580303586, 49508360265, 432061044870, 3720287489101, 31681154472266, 267320885100785, 2238337148081710, 18621251375573301, 154069635600426546
Offset: 0

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Keywords

Comments

This is the column m=2 sequence (without leading zeros) of the Sheffer triangle (exp(5*x), exp(x)-1) of the 5-restricted Stirling2 numbers A193685. For a proof see the column o.g.f. formula there. - Wolfdieter Lang, Oct 07 2011

Crossrefs

Programs

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,5), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = 26*a(n-1) - 251*a(n-2) + 1066*a(n-3) - 1680*a(n-4), n >= 4. - Vincenzo Librandi, Mar 19 2011
a(n) = 15*a(n-1) - 56*a(n-2) + 6^(n+1) - 5^(n+1), a(0)=1, a(1)=26. - Vincenzo Librandi, Mar 19 2011
E.g.f.: (d^3/dx^3)(exp(5*x)*((exp(x)-1)^3)/3!). See the Sheffer triangle comment above. - Wolfdieter Lang, Oct 07 2011
a(n) = -125*5^n/6 + 108*6^n - 343*7^n/2 + 256*8^n/3. - R. J. Mathar, Jun 23 2013

A028200 Expansion of 1/((1-6x)*(1-7x)*(1-8x)*(1-9x)).

Original entry on oeis.org

1, 30, 565, 8550, 113701, 1388310, 15958405, 175419750, 1863406501, 19269697590, 195034120645, 1939826329350, 19018419228901, 184245490086870, 1767124523521285, 16805853434269350, 158682246543588901, 1489103597614860150, 13900428943759584325
Offset: 0

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Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[ 1/((1-6x)(1-7x)(1-8x)(1-9x)), {x, 0, 20} ], x]
    LinearRecurrence[{30,-335,1650,-3024},{1,30,565,8550},20] (* Harvey P. Dale, Mar 27 2023 *)
  • PARI
    Vec(1/((1-6*x)*(1-7*x)*(1-8*x)*(1-9*x)) + O(x^30)) \\ Michel Marcus, Feb 12 2017

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,6), (n >= 3). [Milan Janjic, Apr 26 2009]
a(n) = 17*a(n-1) - 72*a(n-2) + 7^(n+1) - 6^(n+1), a(0)=1, a(1)=30. - Vincenzo Librandi, Mar 11 2011
a(n) = (9^(n+3) - 3*8^(n+3) + 3*7^(n+3) - 6^(n+3))/6. [Yahia Kahloune, Jun 12 2013]
a(n) = 30*a(n-1) - 335*a(n-2) + 1650*a(n-3) - 3024*a(n-4). - Matthew House, Feb 11 2017

A016094 Expansion of 1/((1-9*x)*(1-10*x)*(1-11*x)*(1-12*x)).

Original entry on oeis.org

1, 42, 1105, 23310, 431221, 7309722, 116419465, 1769717670, 25948716541, 369730963602, 5147200519825, 70298695224030, 944897655707461, 12530341519244682, 164265473257148185, 2132247784185258390
Offset: 0

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Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/((1-9x)(1-10x)(1-11x)(1-12x)) ,{x,0,20}],x] (* or *) LinearRecurrence[{42,-659,4578,-11880},{1,42,1105,23310},20] (* Harvey P. Dale, Dec 14 2021 *)

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,9), n >= 3. - Milan Janjic, Apr 26 2009
a(n) = 42*a(n-1) - 659*a(n-2) + 4578*a(n-3) - 11880*a(n-4), n >= 4. - Vincenzo Librandi, Mar 18 2011
a(n) = 23*a(n-1) - 132*a(n-2) + 10^(n+1) - 9^(n+1), n >= 2. - Vincenzo Librandi, Mar 18 2011
a(n) = 5*10^(n+2) + 2*12^(n+2) - 11^(n+3)/2 - 3*9^(n+2)/2. - R. J. Mathar, Mar 19 2011

A016109 Expansion of 1/((1-7*x)*(1-8*x)*(1-9*x)*(1-10*x)).

Original entry on oeis.org

1, 34, 725, 12410, 186501, 2571114, 33339685, 413066170, 4941549701, 57504755594, 654463491045, 7314256515930, 80522026412101, 875355238834474, 9415203971344805, 100355146006589690
Offset: 0

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Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/((1-7x)(1-8x)(1-9x)(1-10x)),{x,0,20}],x] (* or *) LinearRecurrence[{34,-431,2414,-5040},{1,34,725,12410},21] (* Harvey P. Dale, Jan 26 2012 *)

Formula

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,7), n >= 3. - Milan Janjic, Apr 26 2009; adapted by R. J. Mathar, Mar 15 2011
a(n) = 19*a(n-1) - 90*a(n-2) + 8^(n+1) - 7^(n+1), n >= 2. - Vincenzo Librandi, Mar 12 2011
a(n) = (10^(n+3) - 3*9^(n+3) + 3*8^(n+3) - 7^(n+3))/6. - Bruno Berselli, Mar 12 2011
a(n) = 34*a(n-1) - 431*a(n-2) + 2414*a(n-3) - 5040*a(n-4); a(0)=1, a(1)=34, a(2)=725, a(3)=12410. - Harvey P. Dale, Jan 26 2012

Extensions

Offset changed to 0 by Vincenzo Librandi, Mar 12 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

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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

A099111 Expansion of 1 / ((1+x)*(1-2*x)*(1-3*x)*(1-4*x)*(1-5*x)).

Original entry on oeis.org

1, 13, 112, 798, 5103, 30471, 173734, 958936, 5170165, 27396369, 143295516, 742128114, 3814432987, 19490910907, 99140278258, 502476527532, 2539579950369, 12806979393285, 64472086878760, 324111808432390
Offset: 0

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Author

Ralf Stephan, Sep 28 2004

Keywords

Crossrefs

First differences are in A004058. Pairwise sums are in A025211.

Programs

  • Magma
    [(1/360)*(6250*5^n - 9216*4^n + 3645*3^n - 320*2^n + (-1)^n): n in [0..20]]; // Vincenzo Librandi, Oct 11 2011
  • Mathematica
    CoefficientList[Series[1/((1+x)(1-2x)(1-3x)(1-4x)(1-5x)),{x,0,20}],x] (* or *) LinearRecurrence[{13,-57,83,34,-120},{1,13,112,798,5103},20] (* Harvey P. Dale, Jan 17 2024 *)

Formula

a(n) = (1/360)*(6250*5^n - 9216*4^n + 3645*3^n - 320*2^n + (-1)^n).

A269952 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j,-n)*S2(j,k), S2 the Stirling set numbers A048993, for n>=0 and 0<=k<=n.

Original entry on oeis.org

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

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Author

Peter Luschny, Apr 10 2016

Keywords

Examples

			1,
0, 1,
0, 2, 1,
0, 4, 5, 1,
0, 8, 19, 9, 1,
0, 16, 65, 55, 14, 1,
0, 32, 211, 285, 125, 20, 1,
0, 64, 665, 1351, 910, 245, 27, 1.
		

Crossrefs

Variant: A143494 (the main entry for this triangle).
A005493 (row sums), A074051 (alt. row sums), A000079 (col. 1), A001047 (col. 2),
A016269 (col. 3), A025211 (col. 4), A000096 (diag. n,n-1), A215862 (diag. n,n-2),
A049444, A136124, A143491 (matrix inverse).

Programs

  • Maple
    A269952 := (n,k) -> Stirling2(n+1, k+1) - Stirling2(n, k+1):
    seq(seq(A269952(n,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[ Table[ Sum[(-1)^(n-j) Binomial[-j,-n] StirlingS2[j,k], {j,0,n}], {n,0,9}, {k,0,n}]]

Formula

T(n, k) = S2(n+1, k+1) - S2(n, k+1).
Showing 1-10 of 10 results.