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 21 results. Next

A193683 Alternating row sums of Sheffer triangle A143495 (3-restricted Stirling2 numbers).

Original entry on oeis.org

1, 2, 3, 1, -14, -59, -99, 288, 2885, 10365, 1700, -226313, -1535203, -4258630, 17243695, 284513877, 1688253890, 2750940953, -51540956455, -624352447488, -3470378651847, -496964048927, 204678286709292, 2311290490508227, 12611758414937801
Offset: 0

Views

Author

Wolfdieter Lang, Oct 06 2011

Keywords

Comments

In order to have a lower triangular Sheffer matrix for A143495 one uses row and column offsets 0 (not 3).

Examples

			Row no. 3 of A143495 with [0,0] offset is [27,37,12,1], hence a(3)=27-37+12-1=1.
		

References

Crossrefs

Cf. A143495, A074051 (2-restricted Stirling2 case), A193684, A196835, A293037, A346738.

Programs

  • Mathematica
    With[{nn=30},CoefficientList[Series[Exp[3x+1-Exp[x]],{x,0,nn}], x] Range[0,nn]!] (* Harvey P. Dale, Jan 10 2013 *)

Formula

E.g.f.: exp(-exp(x)+3*x+1).
G.f.: (1 - 2/E(0))/x where E(k) = 1 + 1/(1 - 2*x/(1 - 2*(k+1)*x/E(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: 1/U(0) where U(k) = 1 - x*(k+2) + x^2*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 11 2012
G.f.: (1 - G(0) )/(x+1) where G(k) = 1 - 1/(1-k*x-3*x)/(1-x/(x+1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1/Q(0), where Q(k) = 1 - 3*x + x/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
G.f.: G(0)/(1-2*x), where G(k) = 1 - x^2*(2*k+1)/(x^2*(2*k+1) + (1-x*(2*k+2))*(1-x*(2*k+3))/(1 - x^2*(2*k+2)/(x^2*(2*k+2) + (1-x*(2*k+3))*(1-x*(2*k+4))/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Feb 06 2014
a(n) = exp(1) * Sum_{k>=0} (-1)^k * (k + 3)^n / k!. - Ilya Gutkovskiy, Dec 20 2019
a(0) = 1; a(n) = 3 * a(n-1) - Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Seiichi Manyama, Aug 02 2021

A001550 a(n) = 1^n + 2^n + 3^n.

Original entry on oeis.org

3, 6, 14, 36, 98, 276, 794, 2316, 6818, 20196, 60074, 179196, 535538, 1602516, 4799354, 14381676, 43112258, 129271236, 387682634, 1162785756, 3487832978, 10462450356, 31385253914, 94151567436, 282446313698, 847322163876
Offset: 0

Views

Author

Keywords

Comments

a(n)*(-1)^n, n>=0, gives the z-sequence for the Sheffer triangle A049458 ((signed) 3-restricted Stirling1 numbers), which is the inverse triangle of A143495 with offset [0,0] (3-restricted Stirling2 numbers). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The a-sequence for each (signed) r-restricted Stirling1 Sheffer triangle is A027641/A027642 (Bernoulli numbers). - Wolfdieter Lang, Oct 10 2011

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • 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

  • Haskell
    a001550 n = sum $ map (^ n) [1..3]  -- Reinhard Zumkeller, Mar 01 2012
    
  • Magma
    [1^n + 2^n + 3^n : n in [0..30]]; // Wesley Ivan Hurt, Jun 25 2020
    
  • Maple
    A001550:=-(3-12*z+11*z^2)/(z-1)/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    Table[1^n + 2^n + 3^n, {n, 0, 30}]
    CoefficientList[Series[(3-12x+11x^2)/(1-6x+11x^2-6x^3),{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6},{3,6,14},31] (* Harvey P. Dale, Apr 30 2011 *)
    Total[Range[3]^#]&/@Range[0,30] (* Harvey P. Dale, Sep 23 2019 *)
  • PARI
    a(n)=1+2^n+3^n \\ Charles R Greathouse IV, Jun 10 2011
    
  • Python
    def A001550(n): return 3**n+(1<Chai Wah Wu, Nov 01 2024

Formula

From Michael Somos: (Start)
G.f.: (3 -12*x +11*x^2)/(1 -6*x +11*x^2 -6*x^3).
a(n) = 5*a(n-1) - 6*a(n-2) + 2. (End)
E.g.f.: exp(x) + exp(2*x) + exp(3*x). - Mohammad K. Azarian, Dec 26 2008
a(0)=3, a(1)=6, a(2)=14, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3). - Harvey P. Dale, Apr 30 2011
a(n) = A007689(n) + 1. - Reinhard Zumkeller, Mar 01 2012
From Kai Wang, May 18 2020: (Start)
a(n) = 3*A000392(n+3) - 12*A000392(n+2) + 11*A000392(n+1).
A000392(n) = (3*a(n+1) - 12*a(n) + 10*a(n-1))/2. (End)

Extensions

Additional terms from Michael Somos
Attribute "conjectured" removed from Simon Plouffe's g.f. by R. J. Mathar, Mar 11 2009

A005061 a(n) = 4^n - 3^n.

Original entry on oeis.org

0, 1, 7, 37, 175, 781, 3367, 14197, 58975, 242461, 989527, 4017157, 16245775, 65514541, 263652487, 1059392917, 4251920575, 17050729021, 68332056247, 273715645477, 1096024843375, 4387586157901, 17560804984807, 70274600998837, 281192547174175, 1125052618233181
Offset: 0

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Author

Keywords

Comments

Number of 2 X n binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 21 2002
Number of binary vectors (x_1, x_2, ..., x_{2n}) such that in at least one of the disjoint pairs (x_1, x_2), (x_3, x_4), ..., (x_{2n-1}, x_{2n}) both x_{2i-1} and x_{2i} are both 1. Equivalently, number of solutions (x_1, ..., x_n) to the equation x_1*x_2 + x_3*x_4 + x_5*x_6 + ... +x_{2n-1}*x_{2n} = 1 in base-2 lunar arithmetic. - N. J. A. Sloane, Apr 23 2011
a(n)/4^n is the probability that two randomly selected (with replacement) subsets of [n] will have at least one element in common if the probability of selection is equal for all subsets. - Geoffrey Critzer, May 09 2009
This sequence is also the second column of the Sheffer triangle A143495 (3-restricted Stirling2 numbers). (See the e.g.f. given below.) - Wolfdieter Lang, Oct 08 2011
Also, the number of numbers with at most n digits whose largest digit equals 3. See A255463 for the first differences (i.e., ...with exactly n digits...). - M. F. Hasler, May 03 2015
If 2^k | n then a(2^k) | a(n). - Bernard Schott, Oct 08 2020
a(n) is the number of ordered n-tuples with elements from {0,1,2,3} in which any of these elements, say 0, appears at least once. For example, a(2)=7 since 01,10,02,20,03,30,00 are the ordered 2-tuples that contain 0. - Enrique Navarrete, Apr 05 2021
a(n) is the number of n-digit numbers whose smallest decimal digit is 6. - Stefano Spezia, Nov 15 2023

Examples

			G.f. = x + 7*x^2 + 37*x^3 + 175*x^4 + 781*x^5 + 3367*x^6 + 14197*x^7 + ...
		

Crossrefs

Cf. A001047, A002250, A005060, A005062, A143495, A255463 (first differences), A359576.
Array column A047969(n-1, 3), or triangle's subdiagonal A047969(n+2, n-1), for n >= 1.

Programs

Formula

a(n) = 4*a(n-1) + 3^(n-1) for n>=1. - Xavier Acloque, Oct 20 2003
Binomial transform of A001047. - Ross La Haye, Sep 17 2005
From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-4*x)-1/(1-3*x).
E.g.f.: exp(4*x)-exp(3*x). (End)
a(n) = 2^n * Sum_{i=0...n} binomial(n,i)*(2^i-1)/2^i. - Geoffrey Critzer, May 09 2009
a(n) = 7*a(n-1) - 12*a(n-2) for n>=2. - Bruno Berselli, Jan 25 2011
From Joe Slater, Jan 15 2017: (Start)
a(n) = 3*a(n-1) + 4^(n-1) for n>=0.
a(n+1) = Sum_{k=0..n} 4^(n-k) * 3^k. (End)
a(n) = -a(-n) * 12^n for all n in Z. - Michael Somos, Jan 22 2017

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

Views

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

A005494 3-Bell numbers: E.g.f.: exp(3*z + exp(z) - 1).

Original entry on oeis.org

1, 4, 17, 77, 372, 1915, 10481, 60814, 372939, 2409837, 16360786, 116393205, 865549453, 6713065156, 54190360453, 454442481041, 3952241526188, 35590085232519, 331362825860749, 3185554606447814, 31581598272055879, 322516283206446897, 3389017736055752178
Offset: 0

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Author

Keywords

Comments

For further information, references, programs, etc. for r-Bell numbers see A005493. - N. J. A. Sloane, Nov 27 2013
From expansion of falling factorials (binomial transform of A005493).
Row sums of Sheffer triangle (exp(3*x), exp(x)-1). - Wolfdieter Lang, Sep 29 2011

Examples

			G.f. = 1 + 4*x + 17*x^2 + 77*x^3 + 372*x^4 + 1915*x^5 + 10481*x^6 + 60814*x^7 + ...
		

References

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

Crossrefs

A row or column of the array A108087.

Programs

  • Magma
    A005494:= func< n | (&+[Binomial(n,j)*3^(n-j)*Bell(j): j in [0..n]]) >;
    [A005494(n): n in [0..30]]; // G. C. Greubel, Dec 01 2022
    
  • Maple
    seq(add(3^(n-i)*combinat:-bell(i)*binomial(n,i),i=0..n), n=0..50); # Robert Israel, Dec 16 2014
    # second Maple program:
    b:= proc(n, m) option remember; `if`(n=0,
          m^2, m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n+1, 0)-b(n, 0):
    seq(a(n), n=0..23);  # Alois P. Heinz, Aug 03 2025
  • Mathematica
    Range[0, 40]! CoefficientList[Series[Exp[3 x + Exp[x] - 1], {x, 0, 40}], x] (* Vincenzo Librandi, Mar 04 2014 *)
  • SageMath
    def A005494(n): return sum( 3^(n-j)*bell_number(j)*binomial(n,j) for j in range(n+1))
    [A005494(n) for n in range(31)] # G. C. Greubel, Dec 01 2022

Formula

a(n) = Sum_{i=0..n} 3^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n). - Fred Lunnon, Aug 04 2007
a(n) = exp(-1)*Sum_{k>=0} ((k+3)^n)/k!. - Gerald McGarvey, Jun 03 2004. May be rewritten as a(n) = Sum_{k>=3} (k^n*(k-1)*(k-2)/k!)/exp(1), which is a Dobinski-type relation for this sequence. - Karol A. Penson, Aug 18 2006
Define f_1(x), f_2(x), ... such that f_1(x) = x^2*e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n-1) = e^(-1)*f_n(1). - Milan Janjic, May 30 2008
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i <= j), and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = (-1)^(n)charpoly(A,-3). - Milan Janjic, Jul 08 2010
a(n) = Sum_{k=3..n+3} A143495(n+3,k), n >= 0. - Wolfdieter Lang, Sep 29 2011
G.f.: 1/U(0) where U(k)= 1 - x*(k+4) - x^2*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 11 2012
G.f.: Sum_{k>0} x^(k-1) / ((1 - 3*x) * (1 - 4*x) * ... * (1 - (k+2)*x)). - Michael Somos, Feb 26 2014
G.f.: Sum_{k>0} k * x^(k-1) / ((1 - 2*x) * (1 - 3*x) * ... * (1 - (k+1)*x)). - Michael Somos, Feb 26 2014
a(n) ~ exp(n/LambertW(n) - n - 1) * n^(n + 3) / LambertW(n)^(n + 7/2). - Vaclav Kotesovec, Jun 10 2020
a(0) = 1; a(n) = 3 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Ilya Gutkovskiy, Jul 02 2020
a(n) = Sum_{k=0..n} 4^k*A124323(n, k). - Mélika Tebni, Jun 10 2022

A143496 4-Stirling numbers of the second kind.

Original entry on oeis.org

1, 4, 1, 16, 9, 1, 64, 61, 15, 1, 256, 369, 151, 22, 1, 1024, 2101, 1275, 305, 30, 1, 4096, 11529, 9751, 3410, 545, 39, 1, 16384, 61741, 70035, 33621, 7770, 896, 49, 1, 65536, 325089, 481951, 305382, 95781, 15834, 1386, 60, 1, 262144, 1690981, 3216795
Offset: 4

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 4 of the r-Stirling numbers of the second kind. The 4-Stirling numbers of the second kind count the ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2, 3 and 4 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 4-Stirling numbers of the first kind is A143493. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 4-Lah numbers refer to A143499.
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,4), n >= k >= 4, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->4, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(4*x),exp(x)-1) with e.g.f. of column no. m >= 0: exp(4*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|.....4.....5.....6.....7.....8.....9
========================================
4..|.....1
5..|.....4.....1
6..|....16.....9.....1
7..|....64....61....15.....1
8..|...256...369...151....22.....1
9..|..1024..2101..1275...305....30.....1
...
T(6,5) = 9. The set {1,2,3,4,5,6} can be partitioned into five subsets such that 1, 2, 3 and 4 belong to different subsets in 9 ways: {{1,5}{2}{3}{4}{6}}, {{1,6}{2}{3}{4}{5}}, {{2,5}{1}{3}{4}{6}}, {{2,6}{1}{3}{4}{5}}, {{3,5}{1}{2}{4}{6}}, {{3,6}{1}{2}{4}{5}}, {{4,5}{1}{2}{3}{6}}, {{4,6}{1}{2}{3}{5}} and {{5,6}{1}{2}{3}{4}}.
		

Crossrefs

Cf. A003468 (column 7), A005060 (column 5), A008277, A016103 (column 6), A045379 (row sums), A049459 (matrix inverse), A143493, A143494, A143495, A143499.

Programs

  • Maple
    with combinat: T := (n, k) -> 1/(k-4)!*add ((-1)^(k-i)*binomial(k-4,i)*(i+4)^(n-4),i = 0..k-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - 6*StirlingS2[n-1, k] + 11*StirlingS2[n-2, k] - 6*StirlingS2[n-3, k]; Flatten[ Table[ t[n, k], {n, 4, 13}, {k, 4, n}]] (* Jean-François Alcover, Dec 02 2011 *)

Formula

T(n+4,k+4) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+4)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 6*Stirling2(n-1,k) + 11*Stirling2(n-2,k) - 6*Stirling2(n-3,k) for n,k >= 4.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 4 with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4. Special cases: T(n,4) = 4^(n-4); T(n,5) = 5^(n-4) - 4^(n-4).
E.g.f. (k+4)-th column (with offset 4): (1/k!)*exp(4*x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n>=k} T(n,k)*x^n = x^k/((1-4*x)*(1-5*x)*...*(1-k*x)).
E.g.f.: exp(4*t + x*(exp(t)-1)) = Sum_{n = 0..infinity} Sum_(k = 0..n) T(n+4,k+4)*x^k*t^n/n! = Sum_{n = 0..infinity} B_n(4;x)*t^n/n! = 1 + (4+x)*t/1! + (16+9*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(4;x) := Sum_{k = 0..n} T(n+4,k+4)*x^k, may be called the 4-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(4;x) = exp(-x)*Sum_{i = 0..infinity} (i+4)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+4,k+4)*x^k = Sum_{i = 0..infinity} (i+4)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+4)^(n-4). For example, 16 + 9*x + x*(x-1) = (x+4)^2; 64 + 61*x + 15*x*(x-1) + x*(x-1)*(x-2) = (x+4)^3.
This array is the matrix product P^3 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The inverse array is A049459, the signed 4-Stirling numbers of the first kind.
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^(-4)*E^n*x^4 = Sum_{k = 0..n} T(n+4,k+4)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k=4..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_4(x) = x^4. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 4-Eulerian numbers E_4(n,j) := A144698(n,j): T(n,k) = 4!/k!*Sum_{j = n-k..n-4} E_4(n,j)*binomial(j,n-k) for n >= k >= 4.
(End)

A193685 5-Stirling numbers of the second kind.

Original entry on oeis.org

1, 5, 1, 25, 11, 1, 125, 91, 18, 1, 625, 671, 217, 26, 1, 3125, 4651, 2190, 425, 35, 1, 15625, 31031, 19981, 5590, 740, 45, 1, 78125, 201811, 170898, 64701, 12250, 1190, 56, 1, 390625, 1288991, 1398097, 688506, 174951, 24150, 1806, 68, 1, 1953125, 8124571, 11075670, 6906145, 2263065, 416451, 44016, 2622, 81, 1, 9765625, 50700551, 85654261, 66324830, 27273730, 6427575, 900627, 75480, 3675, 95, 1
Offset: 0

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Author

Wolfdieter Lang, Oct 06 2011

Keywords

Comments

This is the lower triangular Sheffer matrix (exp(5*x),exp(x)-1). For Sheffer matrices see the W. Lang link under A006232 with references, and the rules for the conversion to the umbral notation of S. Roman's book.
The general case is Sheffer (exp(r*x),exp(x)-1), r=0,1,..., corresponding to r-Stirling2 numbers with row and column offsets 0. See the Broder link for r-Stirling2 numbers with offset [r,r].
a(n,m), n >= m >= 0, gives the number of partitions of the set {1.2....,n+5} into m+5 nonempty distinct subsets such that 1,2,3,4 and 5 belong to distinct subsets.
a(n,m) appears in the following normal ordering of Bose operators a and a* satisfying the Lie algebra [a,a*]=1: (a*a)^n (a*)^5 = Sum_{m=0..n} a(n,m)*(a*)^(5+m)*a^m, n >= 0. See the Mikhailov papers, where a(n,m) = S(n+5,m+5,5).
With a->D=d/dx and a*->x we also have
(xD)^n x^5 = Sum_{m=0..n} a(n,m)*x^(5+m)*D^m, n >= 0.

Examples

			n\m  0    1    2   3  4  5 ...
0    1
1    5    1
2   25   11    1
3  125   91   18   1
4  625  671  217  26  1
5 3125 4651 2190 425 35  1
...
5-restricted S2: a(1,0)=5 from 1,6|2|3|4|5, 2,6|1|3|4|5,
3,6|1|2|4|5, 4,6|1|2|3|5 and 5,6|1|2|3|4.
Recurrence: a(4,2) = (5+2)*a(3,2)+ a(3,1) = 7*18 + 91 = 217.
Normal ordering (n=1): (xD)^1 x^5 = Sum_{m=0..1} a(1,m)*x^(5+m)*D^m = 5*x^5 + 1*x^6*D.
a(2,1) = Sum_{j=0..1} S1(5,5-j)*S2(7-j,6) = 1*21 - 10*1 = 11.
		

Crossrefs

Cf. A196834 (row sums), A196835 (alternating row sums).
Columns: A000351 (m=0), A005062 (m=1), A019757 (m=2), A028165 (m=3), ...

Programs

  • Mathematica
    a[n_, m_] := Sum[ StirlingS1[5, 5-j]*StirlingS2[n+5-j, m+5], {j, 0, Min[5, n-m]}]; Flatten[ Table[ a[n, m], {n, 0, 10}, {m, 0, n}] ] (* Jean-François Alcover, Dec 02 2011, after Wolfdieter Lang *)

Formula

E.g.f. of row polynomials s(n,x):=Sum_{m=0..n} a(n,m)*x^m: exp(5*z + x(exp(z)-1)).
E.g.f. of column no. m (with leading zeros):
exp(5*x)*((exp(x)-1)^m)/m!, m >= 0 (Sheffer).
O.g.f. of column no. m (without leading zeros):
1/Product_{j=0..m} (1-(5+j)*x), m >= 0. (Compute the first derivative of the column e.g.f. and compare its Laplace transform with the partial fraction decomposition of the o.g.f. x^(m-1)/Product_{j=0..m} (1-(5+j)*x). This works for every r-restricted Stirling2 triangle.)
Recurrence: a(n,m) = (5+m)*a(n-1,m) + a(n-1,m-1), a(0,0)=1, a(n,m)=0 if n < m, a(n,-1)=0.
a(n,m) = Sum_{j=0..min(5,n-m)} S1(5,5-j)*S2(n+5-j,m+5), n >= m >= 0, with S1 and S2 the Stirling1 and Stirling2 numbers A008275 and A048993, respectively (see the Mikailov papers).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k>=0} k*(4+k)^(n-1)*x^(k-1)/k!. - Peter Bala, Jun 23 2014

A049458 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -3, 1, 12, -7, 1, -60, 47, -12, 1, 360, -342, 119, -18, 1, -2520, 2754, -1175, 245, -25, 1, 20160, -24552, 12154, -3135, 445, -33, 1, -181440, 241128, -133938, 40369, -7140, 742, -42, 1, 1814400, -2592720, 1580508, -537628
Offset: 0

Views

Author

Keywords

Comments

a(n,m)= ^3P_n^m in the notation of the given reference with a(0,0) := 1. The monic row polynomials s(n,x) := sum(a(n,m)*x^m,m=0..n) which are s(n,x)= product(x-(3+k),k=0..n-1), n >= 1 and s(0,x)=1 satisfy s(n,x+y) = sum(binomial(n,k)*s(k,x)*S1(n-k,y),k=0..n), with the Stirling1 polynomials S1(n,x)=sum(A008275(n,m)*x^m, m=1..n) and S1(0,x)=1.
In the umbral calculus (see the S. Roman reference given in A048854) the s(n,x) polynomials are called Sheffer polynomials for (exp(3*t),exp(t)-1).
See A143492 for the unsigned version of this array and A143495 for the inverse. - Peter Bala, Aug 25 2008

Examples

			1;
-3, 1;
12, -7, 1;
-60, 47, -12, 1;
360, -342, 119, -18, 1;
s(2,x) = 12-7*x+x^2. S1(2,x) = -x+x^2 (Stirling1 polynomial).
		

References

  • Mitrinovic, D. S.; Mitrinovic, R. S.; Tableaux d'une classe de nombres relies aux nombres de Stirling. Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 1962, 77 pp.

Crossrefs

Unsigned column sequences are: A001710-A001714. Row sums (signed triangle): (n+1)!*(-1)^n. Row sums (unsigned triangle): A001715(n+3).
A143492, A143495. - Peter Bala, Aug 25 2008

Programs

  • Haskell
    a049458 n k = a049458_tabl !! n !! k
    a049458_row n = a049458_tabl !! n
    a049458_tabl = map fst $ iterate (\(row, i) ->
       (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 3)
    -- Reinhard Zumkeller, Mar 11 2014
  • Maple
    A049458_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x+3, n)), x, k), k=0..n): seq(print(A049458_row(n)),n=0..8); # Peter Luschny, May 16 2013
  • Mathematica
    t[n_, k_] := (-1)^(n - k)*Coefficient[ Pochhammer[x + 3, n], x, k]; Table[t[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 17 2013, after Peter Luschny *)

Formula

a(n, m)= a(n-1, m-1) - (n+2)*a(n-1, m), n >= m >= 0; a(n, m) := 0, n
Triangle (signed) = [ -3, -1, -4, -2, -5, -3, -6, -4, -7, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; triangle (unsigned) = [3, 1, 4, 2, 5, 3, 6, 4, 7, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938 (unsigned version in A143492).
E.g.f.: (1+y)^(x-3). - Vladeta Jovovic, May 17 2004
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = f(n,i,3), for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008

Extensions

Second formula corrected by Philippe Deléham, Nov 09 2008

A232472 2-Fubini numbers.

Original entry on oeis.org

2, 10, 62, 466, 4142, 42610, 498542, 6541426, 95160302, 1520385010, 26468935022, 498766780786, 10114484622062, 219641848007410, 5085371491003502, 125055112347154546, 3255163896227709422, 89416052656071565810, 2584886208925055791982, 78447137202259689678706, 2493719594804686310662382
Offset: 2

Author

N. J. A. Sloane, Nov 27 2013

Keywords

Examples

			G.f.: 2*x^2 + 10*x^3 + 62*x^4 + 466*x^5 + 4142*x^6 + 42610*x^7 + 498542*x^8 + ...
		

Programs

  • Magma
    r:=2; r_Fubini:=func;
    [r_Fubini(n, r): n in [r..22]]; // Bruno Berselli, Mar 30 2016
  • Maple
    # r-Stirling numbers of second kind (e.g., A008277, A143494, A143495):
    T := (n,k,r) -> (1/(k-r)!)*add ((-1)^(k+i+r)*binomial(k-r,i)*(i+r)^(n-r),i = 0..k-r):
    # r-Bell numbers (e.g. A000110, A005493, A005494):
    B := (n,r) -> add(T(n,k,r),k=r..n);
    SB := r -> [seq(B(n,r),n=r..30)];
    SB(2);
    # r-Fubini numbers (e.g., A000670, A232472, A232473, A232474):
    F := (n,r) -> add((k)!*T(n,k,r),k=r..n);
    SF := r -> [seq(F(n,r),n=r..30)];
    SF(2);
  • Mathematica
    Rest[max=20; t=Sum[n^(n - 1) x^n / n!, {n, 1, max}]; 2 Range[0, max]!CoefficientList[Series[D[1 / (1 - y (Exp[x] - 1)), y] /.y->1, {x, 0, max}], x]] (* Vincenzo Librandi Jan 03 2016 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Table[Fubini[n, 2], {n, 2, 22}] (* Jean-François Alcover, Mar 30 2016 *)

Formula

Let A(x) be the g.f. A232472, B(x) the g.f. A000670, then A(x) = (1-x)*B(x) - 1. - Sergei N. Gladkovskii, Nov 29 2013
a(n) = Sum_{k>=2} T_k*k^(n-2)/2^k where T_k is the (k-1)-st triangular number (i.e., T_k = k*(k-1)/2). - Derek Orr, Jan 01 2016
a(n) = 2*A069321(n-1). - Vincenzo Librandi, Jan 03 2016, corrected by Vaclav Kotesovec, Jul 01 2018
a(n) ~ n! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Jul 01 2018
From Peter Bala, Dec 08 2020: (Start)
a(n+2) = Sum_{k = 0..n} (k+2)!/k!*( Sum{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+2)^n ).
a(n+2) = Sum_{k = 0..n} 2^(n-k)*binomial(n,k)*( Sum_{i = 0..k} Stirling2(k,i)*(i+2)! ).
a(n) = 2*A069321(n-1) = A000670(n) - A000670(n-1).
a(n+1)= (1/2)*Sum_{k = 0..n} binomial(n,k)*A000670(k+1) for n >= 1.
E.g.f. with offset 0: 2*exp(2*z)/(2 - exp(z))^3 = 2 + 10*z + 62*z^2/2! + 466*z^3/3! + .... (End)

A106516 A Pascal-like triangle based on 3^n.

Original entry on oeis.org

1, 3, 1, 9, 4, 1, 27, 13, 5, 1, 81, 40, 18, 6, 1, 243, 121, 58, 24, 7, 1, 729, 364, 179, 82, 31, 8, 1, 2187, 1093, 543, 261, 113, 39, 9, 1, 6561, 3280, 1636, 804, 374, 152, 48, 10, 1, 19683, 9841, 4916, 2440, 1178, 526, 200, 58, 11, 1, 59049, 29524, 14757, 7356, 3618, 1704, 726, 258, 69, 12, 1
Offset: 0

Author

Paul Barry, May 05 2005

Keywords

Comments

Row sums are A027649. Antidiagonal sums are A106517.
From Wolfdieter Lang, Jan 09 2015: (Start)
Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.
The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)

Examples

			The triangle T(n,k) begins:
n\k     0     1     2    3    4    5   6   7  8  9 10 ...
0:      1
1:      3     1
2:      9     4     1
3:     27    13     5    1
4:     81    40    18    6    1
5:    243   121    58   24    7    1
6:    729   364   179   82   31    8   1
7:   2187  1093   543  261  113   39   9   1
8:   6561  3280  1636  804  374  152  48  10  1
9:  19683  9841  4916 2440 1178  526 200  58 11  1
10: 59049 29524 14757 7356 3618 1704 726 258 69 12  1
... reformatted and extended. - _Wolfdieter Lang_, Jan 06 2015
----------------------------------------------------------
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/ 1        \/1           \/1        \       /1         \
| 3  1     ||0  1        ||0 1      |      | 3  1      |
| 9  4 1   ||0  3  1     ||0 0 1    |... = | 9  7  1   |
|27 13 5 1 ||0  9  4 1   ||0 0 3 1  |      |27 37 12 1 |
|...       ||0 27 13 5 1 ||0 0 9 4 1|      |...        |
|...       ||...         ||...      |      |...        |
= A143495. - _Peter Bala_, Dec 23 2014
		

Crossrefs

Columns 1, 2, 3, 4, 5: A003462, A000340, A052150, A097786, A097787.

Programs

  • Mathematica
    a106516[n_] := Block[{a, k},
    a[x_] := Flatten@ Last@ Reap[For[k = -1, k < x, Sow[Binomial[x, k] +
    2 Sum[3^(i - 1)*Binomial[x - i, k], {i, 1, x}]], k++]]; Flatten@Array[a, n, 0]]; a106516[11] (* Michael De Vlieger, Dec 23 2014 *)

Formula

Riordan array (1/(1-3x), x/(1-x)); Number triangle T(n, 0)=A000244(n), T(n, k)=T(n-1, k-1)+T(n-1, k); T(n, k)=sum{j=0..n, binomial(n, k+j)2^j}.
From Peter Bala, Jul 16 2013: (Start)
T(n,k) = binomial(n,k) + 2*sum {i = 1..n} 3^(i-1)*binomial(n-i,k).
O.g.f.: (1 - t)/( (1 - 3*t)*(1 - (1 + x)*t) ) = 1 + (3 + x)*t + (9 + 4*x + x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x - 2)*( x*(x + 1)^n - 2*3^n ). (End)
Closed-form formula for arbitrary left and right borders of Pascal-like triangle see A228196. - Boris Putievskiy, Aug 19 2013
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 3*T(n-2,k-1), T(0,0)=1, T(1,0)=3, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 26 2013
From Peter Bala, Dec 23 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(27 + 13*x + 5*x^2/2! + x^3/3!) = 27 + 40*x + 58*x^2/2! + 82*x^3/3! + 113*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143495 (but with a different offset). See the Example section. Cf. A055248. (End)
n-th row polynomial R(n, x) = (2*3^n - x*(1 + x)^n)/(2 - x). - Peter Bala, Mar 05 2025
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