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 19 results. Next

A233734 Central terms of triangles A019538 and A090582.

Original entry on oeis.org

1, 6, 150, 8400, 834120, 129230640, 28805736960, 8734434508800, 3457819037312640, 1732015476199008000, 1070842073499515116800, 800968643959240044288000, 712900933001021056900608000, 744602794912654938776355840000, 901893717412811100821094451200000
Offset: 1

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Author

Reinhard Zumkeller, Dec 15 2013

Keywords

Comments

a(n) = A019538(2*n-1,n) = A090582(2*n-1,n).

Programs

  • Haskell
    a233734 n = a019538 (2 * n - 1) n

Formula

a(n) = A129506(n) * n!.

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
Offset: 1

Views

Author

Keywords

Comments

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

Examples

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016
		

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. 835.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
  • C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
  • A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
  • E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
  • J. Stirling, The Differential Method, London, 1749; see p. 7.

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
See also A331155.
Cf. A000110 (row sums), A102661 (partial row sums).

Programs

  • Haskell
    a008277 n k = a008277_tabl !! (n-1) !! (k-1)
    a008277_row n = a008277_tabl !! (n-1)
    a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012
    
  • J
    n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016
    
  • Magma
    [[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
  • Maple
    seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
    stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);
  • Mathematica
    Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 12;
    B = BellMatrix[1&, rows];
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
    a[n_, n_] := 1; a[n_, 1] := 1;
    a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
    Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
    With[{m = 11},
     Flatten@MapIndexed[Take[#, #2[[1]]] &,
       Transpose@
        Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
    Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)
  • Maxima
    create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013
    
  • PARI
    Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012
    
  • Sage
    stirling_number2 # Danny Rorabaugh, Oct 11 2015
    

Formula

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A008292 Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 1

Views

Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010
Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004
T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
Subtriangle for k>=1 and n>=1 of triangle A123125. - Philippe Deléham, Oct 22 2006
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.).
G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246.
(End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008
Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008
For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008
The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009
From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011
The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011
Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011
Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014
Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015
The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016
Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of 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} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 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
For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017
For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018
Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4       5       6      7     8    9 10 ...
1:  1
2:  1    1
3:  1    4     1
4:  1   11    11      1
5:  1   26    66     26       1
6:  1   57   302    302      57       1
7:  1  120  1191   2416    1191     120      1
8:  1  247  4293  15619   15619    4293    247     1
9:  1  502 14608  88234  156190   88234  14608   502    1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - _Michael Somos_, Mar 17 2011
Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - _Vladimir Shevelev_, Jul 01 2011
Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
.   1o (1+t)         1o t         1o t
.   |                / \          / \
.   |               /   \        /   \
.   2o (1+t)      2o     3o    3o    2o
.   |
.   |
.   3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
		

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
  • Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
  • Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
  • K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.

Crossrefs

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Sum([0..k],j->(-1)^j*(k-j)^n*Binomial(n+1,j))))); # Muniru A Asiru, Jun 29 2018
    
  • Haskell
    import Data.List (genericLength)
    a008292 n k = a008292_tabl !! (n-1) !! (k-1)
    a008292_row n = a008292_tabl !! (n-1)
    a008292_tabl = iterate f [1] where
       f xs = zipWith (+)
         (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
         where ks = [1 .. 1 + genericLength xs]
    -- Reinhard Zumkeller, May 07 2013
    
  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n,k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
  • Maple
    A008292 := proc(n,k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1,k)+(n-k+1)*procname(n-1,k-1) ; end if; end proc:
  • Mathematica
    t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
    Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)
    Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)
    Table[Tally[
       Count[#, x_ /; x > 0] & /@ (Differences /@
          Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {A(n,c)=c^(n+c-1)+sum(i=1,c-1,(-1)^i/i!*(c-i)^(n+c-1)*prod(j=1,i,n+c+1-j))}
    
  • Python
    from sympy import binomial
    def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
    for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017
    
  • R
    T <- function(n, k) {
      S <- numeric()
      for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
      return(sum(S))
    }
    for (n in 1:10){
      for (k in 1:n) print(T(n,k))
    } # Indranil Ghosh, Nov 08 2017
    
  • Sage
    [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019
    

Formula

T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
Row sums = n! = A000142(n) unless n=0. - Michael Somos, Mar 17 2011
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011
For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002
From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002
Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Deléham's operator defined in A084938.
Sum_{k=1..n} T(n, k)*2^k = A000629(n). - Philippe Deléham, Jun 05 2004
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008
G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010
If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1.
The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0.
(End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011
T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011
E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - Geoffrey Critzer, Nov 10 2012
From Peter Bala, Mar 12 2013: (Start)
Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i).
(End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013
From Tom Copeland, Sep 18 2014: (Start)
A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271).
E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015
Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015
For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015
From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016
Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024

Extensions

Thanks to Michael Somos for additional comments.
Further comments from Christian G. Bower, May 12 2000

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Views

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Keywords

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021

A008279 Triangle T(n,k) = n!/(n-k)! (0 <= k <= n) read by rows, giving number of permutations of n things k at a time.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 6, 1, 4, 12, 24, 24, 1, 5, 20, 60, 120, 120, 1, 6, 30, 120, 360, 720, 720, 1, 7, 42, 210, 840, 2520, 5040, 5040, 1, 8, 56, 336, 1680, 6720, 20160, 40320, 40320, 1, 9, 72, 504, 3024, 15120, 60480, 181440, 362880, 362880
Offset: 0

Views

Author

Keywords

Comments

Also called permutation coefficients.
Also falling factorials triangle A068424 with column a(n,0)=1 and row a(0,1)=1 otherwise a(0,k)=0, added. - Wolfdieter Lang, Nov 07 2003
The higher-order exponential integrals E(x,m,n) are defined in A163931; for information about the asymptotic expansion of E(x,m=1,n) see A130534. The asymptotic expansions for n = 1, 2, 3, 4, ..., lead to the right hand columns of the triangle given above. - Johannes W. Meijer, Oct 16 2009
The number of injective functions from a set of size k to a set of size n. - Dennis P. Walsh, Feb 10 2011
The number of functions f from {1,2,...,k} to {1,2,...,n} that satisfy f(x) >= x for all x in {1,2,...,k}. - Dennis P. Walsh, Apr 20 2011
T(n,k) = A181511(n,k) for k=1..n-1. - Reinhard Zumkeller, Nov 18 2012
The e.g.f.s enumerating the faces of the permutohedra / permutahedra, Perm(s,t;x) = [e^(sx)-1]/[s-t(e^(sx)-1)], (cf. A090582 and A019538) and the stellahedra / stellohedra, St(s,t;x) = [s e^((s+t)x)]/[s-t(e^(sx)-1)], (cf. A248727) given in Toric Topology satisfy exp[u*d/dt] St(s,t;x) = St(s,u+t;x) = [e^(ux)/(1-u*Perm(s,t;x))]*St(s,t;x), where e^(ux)/(1-uy) is a bivariate e.g.f. for the row polynomials of this entry and A094587. Equivalently, d/dt St = (x+Perm)*St and d/dt Perm = Perm^2, or d/dt log(St) = x + Perm and d/dt log(Perm) = Perm. - Tom Copeland, Nov 14 2016
T(n, k)/n! are the coefficients of the n-th exponential Taylor polynomial, or truncated exponentials, which was proved to be irreducible by Schur. See Coleman link. - Michel Marcus, Feb 24 2020
Given a generic choice of k+2 residues, T(n, k) is the number of meromorphic differentials on the Riemann sphere having a zero of order n and these prescribed residues at its k+2 poles. - Quentin Gendron, Jan 16 2025

Examples

			Triangle begins:
  1;
  1,  1;
  1,  2,  2;
  1,  3,  6,   6;
  1,  4, 12,  24,   24;
  1,  5, 20,  60,  120,   120;
  1,  6, 30, 120,  360,   720,    720;
  1,  7, 42, 210,  840,  2520,   5040,   5040;
  1,  8, 56, 336, 1680,  6720,  20160,  40320,   40320;
  1,  9, 72, 504, 3024, 15120,  60480, 181440,  362880,  362880;
  1, 10, 90, 720, 5040, 30240, 151200, 604800, 1814400, 3628800, 3628800;
  ...
For example, T(4,2)=12 since there are 12 injective functions f:{1,2}->{1,2,3,4}. There are 4 choices for f(1) and then, since f is injective, 3 remaining choices for f(2), giving us 12 ways to construct an injective function. - _Dennis P. Walsh_, Feb 10 2011
For example, T(5,3)=60 since there are 60 functions f:{1,2,3}->{1,2,3,4,5} with f(x) >= x. There are 5 choices for f(1), 4 choices for f(2), and 3 choices for f(3), giving us 60 ways to construct such a function. - _Dennis P. Walsh_, Apr 30 2011
		

References

  • CRC Standard Mathematical Tables and Formulae, 30th ed., 1996, p. 176; 31st ed., p. 215, Section 3.3.11.1.
  • Maple V Reference Manual, p. 490, numbperm(n,k).

Crossrefs

Row sums give A000522.
T(n,0)=A000012, T(n,1)=A000027, T(n+1,2)=A002378, T(n,3)=A007531, T(n,4)=A052762, and T(n,n)=A000142.

Programs

  • Haskell
    a008279 n k = a008279_tabl !! n !! k
    a008279_row n = a008279_tabl !! n
    a008279_tabl = iterate f [1] where
       f xs = zipWith (+) ([0] ++ zipWith (*) xs [1..]) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013, Nov 18 2012
    
  • Magma
    /* As triangle */ [[Factorial(n)/Factorial(n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 11 2015
    
  • Maple
    with(combstruct): for n from 0 to 10 do seq(count(Permutation(n),size=m), m = 0 .. n) od; # Zerinvary Lajos, Dec 16 2007
    seq(seq(n!/(n-k)!,k=0..n),n=0..10); # Dennis P. Walsh, Apr 20 2011
    seq(print(seq(pochhammer(n-k+1,k),k=0..n)),n=0..6); # Peter Luschny, Mar 26 2015
  • Mathematica
    Table[CoefficientList[Series[(1 + x)^m, {x, 0, 20}], x]* Table[n!, {n, 0, m}], {m, 0, 10}] // Grid (* Geoffrey Critzer, Mar 16 2010 *)
    Table[ Pochhammer[n - k + 1, k], {n, 0, 9}, {k, 0, n}] // Flatten (* or *)
    Table[ FactorialPower[n, k], {n, 0, 9}, {k, 0, n}] // Flatten  (* Jean-François Alcover, Jul 18 2013, updated Jan 28 2016 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, n!/(n-k)!)}; /* Michael Somos, Nov 14 2002 */
    
  • PARI
    {T(n, k) = my(A, p); if( k<0 || k>n, 0, if( n==0, 1, A = matrix(n, n, i, j, x + (i==j)); polcoeff( sum(i=1, n!, if( p = numtoperm(n, i), prod(j=1, n, A[j, p[j]]))), k)))}; /* Michael Somos, Mar 05 2004 */
    
  • Python
    from math import factorial, isqrt, comb
    def A008279(n): return factorial(a:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))//factorial(a-n+comb(a+1,2)) # Chai Wah Wu, Nov 13 2024
  • Sage
    for n in range(8): [falling_factorial(n,k) for k in (0..n)] # Peter Luschny, Mar 26 2015
    

Formula

E.g.f.: Sum T(n,k) x^n/n! y^k = exp(x)/(1-x*y). - Vladeta Jovovic, Aug 19 2002
Equals A007318 * A136572. - Gary W. Adamson, Jan 07 2008
T(n, k) = n*T(n-1, k-1) = k*T(n-1, k-1)+T(n-1, k) = n*T(n-1, k)/(n-k) = (n-k+1)*T(n, k-1). - Henry Bottomley, Mar 29 2001
T(n, k) = n!/(n-k)! if n >= k >= 0, otherwise 0.
G.f. for k-th column k!*x^k/(1-x)^(k+1), k >= 0.
E.g.f. for n-th row (1+x)^n, n >= 0.
Sum T(n, k)x^k = permanent of n X n matrix a_ij = (x+1 if i=j, x otherwise). - Michael Somos, Mar 05 2004
Ramanujan psi_1(k, x) polynomials evaluated at n+1. - Ralf Stephan, Apr 16 2004
E.g.f.: Sum T(n,k) x^n/n! y^k/k! = e^{x+xy}. - Franklin T. Adams-Watters, Feb 07 2006
The triangle is the binomial transform of an infinite matrix with (1, 1, 2, 6, 24, ...) in the main diagonal and the rest zeros. - Gary W. Adamson, Nov 20 2006
G.f.: 1/(1-x-xy/(1-xy/(1-x-2xy/(1-2xy/(1-x-3xy/(1-3xy/(1-x-4xy/(1-4xy/(1-... (continued fraction). - Paul Barry, Feb 11 2009
T(n,k) = Sum_{j=0..k} binomial(k,j)*T(x,j)*T(y,k-j) for x+y = n. - Dennis P. Walsh, Feb 10 2011
From Dennis P. Walsh, Apr 20 2011: (Start)
E.g.f (with k fixed): x^k*exp(x).
G.f. (with k fixed): k!*x^k/(1-x)^(k+1). (End)
For n >= 2 and m >= 2, Sum_{k=0..m-2} S2(n, k+2)*T(m-2, k) = Sum_{p=0..n-2} m^p. S2(n,k) are the Stirling numbers of the second kind A008277. - Tony Foster III, Jul 23 2019

A123125 Triangle of Eulerian numbers T(n,k), 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 11, 11, 1, 0, 1, 26, 66, 26, 1, 0, 1, 57, 302, 302, 57, 1, 0, 1, 120, 1191, 2416, 1191, 120, 1, 0, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 0, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 0, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 0

Views

Author

Philippe Deléham, Sep 30 2006

Keywords

Comments

The beginning of this sequence does not quite agree with the usual version, which is A173018. - N. J. A. Sloane, Nov 21 2010
Each row of A123125 is the reverse of the corresponding row in A173018. - Michael Somos, Mar 17 2011
A008292 (subtriangle for k>=1 and n>=1) is the main entry for these numbers.
Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,...] where DELTA is the operator defined in A084938.
Row sums are the factorials. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
If the initial zero column is deleted, the result is A008292. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
This result gives an alternative method of calculating the Eulerian numbers by an Umbral Calculus expansion from Comtet. - Roger L. Bagula, Nov 21 2009
This function seems to be equivalent to the PolyLog expansion. - Roger L. Bagula, Nov 21 2009
A raising operator formed from the e.g.f. of this entry is the generator of a sequence of polynomials p(n,x;t) defined in A046802 that specialize to those for A119879 as p(n,x;-1), A007318 as p(n,x;0), A073107 as p(n,x;1), and A046802 as p(n,0;t). See Copeland link for more associations. - Tom Copeland, Oct 20 2015
The Eulerian numbers in this setup count the permutation trees of power n and width k (see the Luschny link). For the associated combinatorial statistic over permutations see the Sage program below and the example section. - Peter Luschny, Dec 09 2015 [See Elder et al. link. Peter Luschny, Jul 13 2022]
From Wolfdieter Lang, Apr 03 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are the numerator polynomials of the o.g.f. G(n, x) of n-powers {m^n}_{m>=0} (with 0^0 = 1): G(n, x) = R(n, x)/(1-x)^(n+1). See the Aug 14 2008 formula, where f(x,n) = R(n, x). The e.g.f. of R(n, t) is given in Copeland's Oct 14 2015 formula below.
The first nine column sequences are A000007, A000012, A000295, A000460, A000498, A000505, A000514, A001243, A001244. (End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of this entry, A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Let b(n) = (1/(n+1))*Sum_{k=0..n-1} (-1)^(n-k+1)*T(n, k+1) / binomial(n, k+1). Then b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1} F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See also Rzadkowski and Urlinska, example 1.) - Peter Luschny, Feb 15 2021
Patrick J. Burchell (see link) describes the following method: To get the k-th row of the triangle write the nonnegative integers with a fixed exponent k as a sequence, 0^k, 1^k, 2^k, ..., and then apply the first differences to them k + 1 times. - Peter Luschny, Apr 02 2023

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    1
  3:  0 1    4     1
  4:  0 1   11    11      1
  5:  0 1   26    66     26       1
  6:  0 1   57   302    302      57       1
  7:  0 1  120  1191   2416    1191     120      1
  8:  0 1  247  4293  15619   15619    4293    247     1
  9:  0 1  502 14608  88234  156190   88234  14608   502    1
 10:  0 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
...  Reformatted. - _Wolfdieter Lang_, Feb 14 2015
------------------------------------------------------------------
The width statistic over permutations, n=4.
  [1, 2, 3, 4] => 3; [1, 2, 4, 3] => 2; [1, 3, 2, 4] => 2; [1, 3, 4, 2] => 2;
  [1, 4, 2, 3] => 2; [1, 4, 3, 2] => 1; [2, 1, 3, 4] => 3; [2, 1, 4, 3] => 2;
  [2, 3, 1, 4] => 2; [2, 3, 4, 1] => 3; [2, 4, 1, 3] => 2; [2, 4, 3, 1] => 2;
  [3, 1, 2, 4] => 3; [3, 1, 4, 2] => 3; [3, 2, 1, 4] => 2; [3, 2, 4, 1] => 3;
  [3, 4, 1, 2] => 3; [3, 4, 2, 1] => 2; [4, 1, 2, 3] => 4; [4, 1, 3, 2] => 3;
  [4, 2, 1, 3] => 3; [4, 2, 3, 1] => 3; [4, 3, 1, 2] => 3; [4, 3, 2, 1] => 2;
Gives row(4) = [0, 1, 11, 11, 1]. - _Peter Luschny_, Dec 09 2015
------------------------------------------------------------------
From _Wolfdieter Lang_, Apr 03 2017: (Start)
Recurrence: T(5, 3) = (6-3)*T(4, 2) + 3*T(4, 3) = 3*11 + 3*11 = 66.
O.g.f. column k=2: (x/(1 - 2*x))*E_x*(x/(1-x)) = (x/(1-x))^2/(1-2*x).
E.g.f. column k=2: A(2, x) = x*A(1, x) + x*E(1, x) = x*1 + x*(exp(x)-1) = x*exp(x), hence E(2, x) = (1 + int(x*exp(-x),x ))*exp(2*x) = exp(x)*(exp(x) - (1+x)). See A000295. (End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, Holland, 1978, page 245. [Roger L. Bagula, Nov 21 2009]
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed.; Addison-Wesley, 1994, p. 268, Row reversed table 268. - Wolfdieter Lang, Apr 03 2017
  • Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008

Crossrefs

See A008292 (subtriangle for k>=1 and n>=1), which is the main entry for these numbers. Another version has the zeros at the ends of the rows, as in Concrete Mathematics: see A173018.
T(2n,n) gives A180056.

Programs

  • Haskell
    a123125 n k = a123125_tabl !! n !! k
    a123125_row n = a123125_tabl !! n
    a123125_tabl = [1] : zipWith (:) [0, 0 ..] a008292_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    gf := 1/(1 - t*exp(x)): ser := series(gf, x, 12):
    cx := n -> (-1)^(n + 1)*factor(n!*coeff(ser, x, n)*(t - 1)^(n + 1)):
    seq(print(seq(coeff(cx(n), t, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 11 2021
    A123125 := proc(n, k) option remember; if k = n then 1 elif k <= 0 or k > n then 0 else k*procname(n-1, k) + (n-k+1)*procname(n-1, k-1) fi end:
    seq(print(seq(A123125(n, k), k=0..n)), n=0..10); # Peter Luschny, Mar 28 2021
    # Alternative (Patrick J. Burchell):
    t := a -> Statistics:-Difference([0, a]): Trow := k -> (t@@(k+1))([seq(n^k, n = 0..k)]):
    seq(print(Trow(n)), n = 0..6); # Peter Luschny, Apr 02 2023
  • Mathematica
    f[x_, n_] := f[x, n] = (1 - x)^(n + 1)*Sum[k^n*x^k, {k, 0, Infinity}];
    Table[CoefficientList[f[x, n], x], {n,0,9}] // Flatten (* Roger L. Bagula, Aug 14 2008 *)
    t[n_ /; n >= 0, 0] = 1; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = (n-k) t[n-1, k-1] + (k+1) t[n-1, k]; T[n_, k_] := t[n, n-k];
    Table[T[n, k], {n,0,10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019 *)
    A123125[n_, k_] := Sum[(-1)^j*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}];
    Table[A123125[n, k], {n, 0, 9}, {k, 0, n}] // TableForm  (* Peter Luschny, Aug 12 2022 *)
  • Python
    from math import isqrt, comb
    def A123125(n):
        a = (m:=isqrt(k:=n+1<<1))+(k>m*(m+1))
        b = comb(a+1,2)-n
        return sum(-(b-j)**(a-1)*comb(a,j) if j&1 else (b-j)**(a-1)*comb(a,j) for j in range(b)) # Chai Wah Wu, Nov 13 2024
  • Sage
    def statistic_eulerian(pi):
        if not pi: return 0
        h, i, branch, next = 0, len(pi), [0], pi[0]
        while True:
            while next < branch[len(branch)-1]:
                del(branch[len(branch)-1])
            current = 0
            h += 1
            while next > current:
                i -= 1
                if i == 0: return h
                branch.append(next)
                current, next = next, pi[i]
    def A123125_row(n):
        L = [0]*(n+1)
        for p in Permutations(n):
            L[statistic_eulerian(p)] += 1
        return L
    [A123125_row(n) for n in range(7)] # Peter Luschny, Dec 09 2015
    

Formula

Sum_{k=0..n} T(n,k) = n! = A000142(n).
Sum_{k=0..n} 2^k*T(n,k) = A000629(n).
Sum_{k=0..n} 3^k*T(n,k) = abs(A009362(n+1)).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A000670(n).
Sum_{k=0..n} T(n,k)*3^(n-k) = A122704(n). - Philippe Deléham, Nov 07 2007
G.f.: f(x,n) = (1 - x)^(n + 1)*Sum_{k>=0} k^n*x^k. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008. f is not the g.f. of the triangle, it is the polynomial of row n. See an Apr 03 2017 comment above - Wolfdieter Lang, Apr 03 2017
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000629(n), A123227(n), A201355(n), A201368(n) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Dec 01 2011
E.g.f. (1-t)/(1-t*exp((1-t)x)). A123125 * A007318 = A130850 = unsigned A075263, related to reversed A028246. A007318 * A123125 = A046802. Evaluating the row polynomials at -1, giving the alternating-sign row sum, generates A009006. - Tom Copeland, Oct 14 2015
From Wolfdieter Lang, Apr 03 2017: (Start)
T(n, k) = A173018(n, n-k), 0 <= k <= n. Row reversed Euler's triangle. See Graham et al., p. 268.
Recurrence (from A173018): T(n, 0) = 1 if n=0 else 0; T(n, k) = 0 if n < k and T(n, k) = (n+1-k)*T(n-1, k-1) + k*T(n-1, k) else.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n-j, k-j)*S2(n, j)*j!, 0 <= k <= n, else 0. For S2(n, k)*k! see A131689.
The recurrence for the o.g.f. of the sequence of column k is
G(k, x) = (x/(1 - k*x))*(E_x - (k-2))*G(k-1, x), with the Euler operator E_x = x*d_x, for k >= 1, with G(0, x) = 1. (Proof from the recurrence of T(n, k)).
The e.g.f of the sequence of column k is found from E(k, x) = (1 + int(A(k, x),x)*exp(-k*x))*exp(k*x), k >= 1, with the recurrence
A(k, x) = x*A(k-1, x) +(1 + (1-k)*(1-x))*E(k-1, x) for k >= 1, with A(0,x)= 0. (Proof from the recurrence of T(n, k)). (End)
T(n, k) = Sum_{j=0..n-k} (-1)^j*(n-j-k+1)^n*binomial(n + 1, j). - Peter Luschny, Aug 12 2022
G.f.: Sum_{m >= 0} x^m/(1/(1-x)-m*t). - Mamuka Jibladze, Mar 12 2025

A028246 Triangular array a(n,k) = (1/k)*Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 7, 12, 6, 1, 15, 50, 60, 24, 1, 31, 180, 390, 360, 120, 1, 63, 602, 2100, 3360, 2520, 720, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880
Offset: 1

Views

Author

N. J. A. Sloane, Doug McKenzie (mckfam4(AT)aol.com)

Keywords

Comments

Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g., if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11 2001
From Gary W. Adamson, Aug 09 2008: (Start)
Julius Worpitzky's 1883 algorithm generates Bernoulli numbers.
By way of example [Wikipedia]:
B0 = 1;
B1 = 1/1 - 1/2;
B2 = 1/1 - 3/2 + 2/3;
B3 = 1/1 - 7/2 + 12/3 - 6/4;
B4 = 1/1 - 15/2 + 50/3 - 60/4 + 24/5;
B5 = 1/1 - 31/2 + 180/3 - 390/4 + 360/5 - 120/6;
B6 = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7;
...
Note that in this algorithm, odd n's for the Bernoulli numbers sum to 0, not 1, and the sum for B1 = 1/2 = (1/1 - 1/2). B3 = 0 = (1 - 7/2 + 13/3 - 6/4) = 0. The summation for B4 = -1/30. (End)
Pursuant to Worpitzky's algorithm and given M = A028246 as an infinite lower triangular matrix, M * [1/1, -1/2, 1/3, ...] (i.e., the Harmonic series with alternate signs) = the Bernoulli numbers starting [1/1, 1/2, 1/6, ...]. - Gary W. Adamson, Mar 22 2012
From Tom Copeland, Oct 23 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1 x + (2 + t)*x^2/2! + (6 + 6t + t^2)*x^3/3! + ... gives row polynomials for A090582, the f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1 + t)*x^2 / 2! + (1 + 4t + t^2)*x^3 / 3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ... gives row polynomials for the present triangle. (End)
The Worpitzky triangle seems to be an apt name for this triangle. - Johannes W. Meijer, Jun 18 2009
If Pascal's triangle is written as a lower triangular matrix and multiplied by A028246 written as an upper triangular matrix, the product is a matrix where the (i,j)-th term is (i+1)^j. For example,
1,0,0,0 1,1,1, 1 1,1, 1, 1
1,1,0,0 * 0,1,3, 7 = 1,2, 4, 8
1,2,1,0 0,0,2,12 1,3, 9,27
1,3,3,1 0,0,0, 6 1,4,16,64
So, numbering all three matrices' rows and columns starting at 0, the (i,j) term of the product is (i+1)^j. - Jack A. Cohen (ProfCohen(AT)comcast.net), Aug 03 2010
The Fi1 and Fi2 triangle sums are both given by sequence A000670. For the definition of these triangle sums see A180662. The mirror image of the Worpitzky triangle is A130850. - Johannes W. Meijer, Apr 20 2011
Let S_n(m) = 1^m + 2^m + ... + n^m. Then, for n >= 0, we have the following representation of S_n(m) as a linear combination of the binomial coefficients:
S_n(m) = Sum_{i=1..n+1} a(i+n*(n+1)/2)*C(m,i). E.g., S_2(m) = a(4)*C(m,1) + a(5)*C(m,2) + a(6)*C(m,3) = C(m,1) + 3*C(m,2) + 2*C(m,3). - Vladimir Shevelev, Dec 21 2011
Given the set X = [1..n] and 1 <= k <= n, then a(n,k) is the number of sets T of size k of subset S of X such that S is either empty or else contains 1 and another element of X and such that any two elemements of T are either comparable or disjoint. - Michael Somos, Apr 20 2013
Working with the row and column indexing starting at -1, a(n,k) gives the number of k-dimensional faces in the first barycentric subdivision of the standard n-dimensional simplex (apply Brenti and Welker, Lemma 2.1). For example, the barycentric subdivision of the 2-simplex (a triangle) has 1 empty face, 7 vertices, 12 edges and 6 triangular faces giving row 4 of this triangle as (1,7,12,6). Cf. A053440. - Peter Bala, Jul 14 2014
See A074909 and above g.f.s for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
An e.g.f. G(x,t) = exp[P(.,t)x] = 1/t - 1/[t+(1-t)(1-e^(-xt^2))] = (1-t) * x + (-2t + 3t^2 - t^3) * x^2/2! + (6t^2 - 12t^3 + 7t^4 - t^5) * x^3/3! + ... for the shifted, reverse, signed polynomials with the first element nulled, is generated by the infinitesimal generator g(u,t)d/du = [(1-u*t)(1-(1+u)t)]d/du, i.e., exp[x * g(u,t)d/du] u eval. at u=0 generates the polynomials. See A019538 and the G. Rzadkowski link below for connections to the Bernoulli and Eulerian numbers, a Ricatti differential equation, and a soliton solution to the KdV equation. The inverse in x of this e.g.f. is Ginv(x,t) = (-1/t^2)*log{[1-t(1+x)]/[(1-t)(1-tx)]} = [1/(1-t)]x + [(2t-t^2)/(1-t)^2]x^2/2 + [(3t^2-3t^3+t^4)/(1-t)^3]x^3/3 + [(4t^3-6t^4+4t^5-t^6)/(1-t)^4]x^4/4 + ... . The numerators are signed, shifted A135278 (reversed A074909), and the rational functions are the columns of A074909. Also, dG(x,t)/dx = g(G(x,t),t) (cf. A145271). (Analytic G(x,t) added, and Ginv corrected and expanded on Dec 28 2015.) - Tom Copeland, Nov 21 2014
The operator R = x + (1 + t) + t e^{-D} / [1 + t(1-e^(-D))] = x + (1+t) + t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... contains an e.g.f. of the reverse row polynomials of the present triangle, i.e., A123125 * A007318 (with row and column offset 1 and 1). Umbrally, R^n 1 = q_n(x;t) = (q.(0;t)+x)^n, with q_m(0;t) = (t+1)^(m+1) - t^(m+1), the row polynomials of A074909, and D = d/dx. In other words, R generates the Appell polynomials associated with the base sequence A074909. For example, R 1 = q_1(x;t) = (q.(0;t)+x) = q_1(0;t) + q__0(0;t)x = (1+2t) + x, and R^2 1 = q_2(x;t) = (q.(0;t)+x)^2 = q_2(0:t) + 2q_1(0;t)x + q_0(0;t)x^2 = 1+3t+3t^2 + 2(1+2t)x + x^2. Evaluating the polynomials at x=0 regenerates the base sequence. With a simple sign change in R, R generates the Appell polynomials associated with A248727. - Tom Copeland, Jan 23 2015
For a natural refinement of this array, see A263634. - Tom Copeland, Nov 06 2015
From Wolfdieter Lang, Mar 13 2017: (Start)
The e.g.f. E(n, x) for {S(n, m)}{m>=0} with S(n, m) = Sum{k=1..m} k^n, n >= 0, (with undefined sum put to 0) is exp(x)*R(n+1, x) with the exponential row polynomials R(n, x) = Sum_{k=1..n} a(n, k)*x^k/k!. E.g., e.g.f. for n = 2, A000330: exp(x)*(1*x/1!+3*x^2/2!+2*x^3/3!).
The o.g.f. G(n, x) for {S(n, m)}{m >=0} is then found by Laplace transform to be G(n, 1/p) = p*Sum{k=1..n} a(n+1, k)/(p-1)^(2+k).
Hence G(n, x) = x/(1 - x)^(n+2)*Sum_{k=1..n} A008292(n,k)*x^(k-1).
E.g., n=2: G(2, 1/p) = p*(1/(p-1)^2 + 3/(p-1)^3 + 2/(p-1)^4) = p^2*(1+p)/(p-1)^4; hence G(2, x) = x*(1+x)/(1-x)^4.
This works also backwards: from the o.g.f. to the e.g.f. of {S(n, m)}_{m>=0}. (End)
a(n,k) is the number of k-tuples of pairwise disjoint and nonempty subsets of a set of size n. - Dorian Guyot, May 21 2019
From Rajesh Kumar Mohapatra, Mar 16 2020: (Start)
a(n-1,k) is the number of chains of length k in a partially ordered set formed from subsets of an n-element set ordered by inclusion such that the first term of the chains is either the empty set or an n-element set.
Also, a(n-1,k) is the number of distinct k-level rooted fuzzy subsets of an n-set ordered by set inclusion. (End)
The relations on p. 34 of Hasan (also p. 17 of Franco and Hasan) agree with the relation between A019538 and this entry given in the formula section. - Tom Copeland, May 14 2020
T(n,k) is the size of the Green's L-classes in the D-classes of rank (k-1) in the semigroup of partial transformations on an (n-1)-set. - Geoffrey Critzer, Jan 09 2023
T(n,k) is the number of strongly connected binary relations on [n] that have period k (A367948) and index 1. See Theorem 5.4.25(6) in Ki Hang Kim reference. - Geoffrey Critzer, Dec 07 2023

Examples

			The triangle a(n, k) starts:
n\k 1   2    3     4      5      6      7      8     9
1:  1
2:  1   1
3:  1   3    2
4:  1   7   12     6
5:  1  15   50    60     24
6:  1  31  180   390    360    120
7:  1  63  602  2100   3360   2520    720
8:  1 127 1932 10206  25200  31920  20160   5040
9:  1 255 6050 46620 166824 317520 332640 181440 40320
... [Reformatted by _Wolfdieter Lang_, Mar 26 2015]
-----------------------------------------------------
Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.
From _Vladimir Shevelev_, Dec 22 2011: (Start)
Also, for power sums, we have
S_0(n) = C(n,1);
S_1(n) = C(n,1) +    C(n,2);
S_2(n) = C(n,1) +  3*C(n,2) +  2*C(n,3);
S_3(n) = C(n,1) +  7*C(n,2) + 12*C(n,3) +  6*C(n,4);
S_4(n) = C(n,1) + 15*C(n,2) + 50*C(n,3) + 60*C(n,4) + 24*C(n,5); etc.
(End)
For X = [1,2,3], the sets T are {{}}, {{},{1,2}}, {{},{1,3}}, {{},{1,2,3}}, {{},{1,2},{1,2,3}}, {{},{1,3},{1,2,3}} and so a(3,1)=1, a(3,2)=3, a(3,3)=2. - _Michael Somos_, Apr 20 2013
		

References

  • Ki Hang Kim, Boolean Matrix Theory and Applications, Marcel Dekker, New York and Basel (1982).

Crossrefs

Dropping the column of 1's gives A053440.
Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.
Row sums give A000629(n-1) for n >= 1.
Cf. A027642, A002445. - Gary W. Adamson, Aug 09 2008
Appears in A161739 (RSEG2 triangle), A161742 and A161743. - Johannes W. Meijer, Jun 18 2009
Binomial transform is A038719. Cf. A131689.
Cf. A119879.
From Rajesh Kumar Mohapatra, Mar 29 2020: (Start)
A000007(n-1) (column k=1), A000225(n-1) (column k=2), A028243(n-1) (column k=3), A028244(n-1) (column k=4), A028245(n-1) (column k=5), for n > 0.
Diagonal gives A000142(n-1), for n >=1.
Next-to-last diagonal gives A001710,
Third, fourth, fifth, sixth, seventh external diagonal respectively give A005460, A005461, A005462, A005463, A005464. (End)

Programs

  • GAP
    Flat(List([1..10], n-> List([1..n], k-> Stirling2(n,k)* Factorial(k-1) ))); # G. C. Greubel, May 30 2019
    
  • Magma
    [[StirlingSecond(n,k)*Factorial(k-1): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019
    
  • Maple
    a := (n,k) -> add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k)/k;
    seq(print(seq(a(n,k),k=1..n)),n=1..10);
    T := (n,k) -> add(eulerian1(n,j)*binomial(n-j,n-k), j=0..n):
    seq(print(seq(T(n,k),k=0..n)),n=0..9); # Peter Luschny, Jul 12 2013
  • Mathematica
    a[n_, k_] = Sum[(-1)^(k-i) Binomial[k,i]*i^n, {i,0,k}]/k; Flatten[Table[a[n, k], {n, 10}, {k, n}]] (* Jean-François Alcover, May 02 2011 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, n! * polcoeff( (x / log(1 + x + x^2 * O(x^n) ))^(n+1), n-k))}; /* Michael Somos, Oct 02 2002 */
    
  • PARI
    {T(n,k) = stirling(n,k,2)*(k-1)!}; \\ G. C. Greubel, May 31 2019
    
  • Python
    # Assuming offset (n, k) = (0, 0).
    def T(n, k):
        if k >  n: return 0
        if k == 0: return 1
        return k*T(n - 1, k - 1) + (k + 1)*T(n - 1, k)
    for n in range(9):
        print([T(n, k) for k in range(n + 1)])  # Peter Luschny, Apr 26 2022
  • Sage
    def A163626_row(n) :
        x = polygen(ZZ,'x')
        A = []
        for m in range(0, n, 1) :
            A.append((-x)^m)
            for j in range(m, 0, -1):
                A[j - 1] = j * (A[j - 1] - A[j])
        return list(A[0])
    for i in (1..7) : print(A163626_row(i))  # Peter Luschny, Jan 25 2012
    
  • Sage
    [[stirling_number2(n,k)*factorial(k-1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, May 30 2019
    

Formula

E.g.f.: -log(1-y*(exp(x)-1)). - Vladeta Jovovic, Sep 28 2003
a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.
Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938, but the notation is different.
Sum of terms in n-th row = A000629(n) - Gary W. Adamson, May 30 2005
The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t) = t(t+1)(d/dt)P(n, t) for n >= 1 (see the Riskin and Beckwith reference). - Emeric Deutsch, Aug 09 2005
From Gottfried Helms, Jul 12 2006: (Start)
Delta-matrix as can be read from H. Hasse's proof of a connection between the zeta-function and Bernoulli numbers (see link below).
Let P = lower triangular matrix with entries P[row,col] = binomial(row,col).
Let J = unit matrix with alternating signs J[r,r]=(-1)^r.
Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers.
Let V = Vandermonde matrix with V[r,c] = (r+1)^c.
V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0).
Then Delta = P*J * V and B' = N(-1)' * Delta, where B is the column matrix of Bernoulli numbers and ' means transpose, or for the single k-th Bernoulli number B_k with the appropriate column of Delta,
B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k).
Using a single column instead of V and assuming infinite dimension, H. Hasse showed that in x = N(-1) * P*J * N(s), where s can be any complex number and s*zeta(1-s) = x.
His theorem reads: s*zeta(1-s) = Sum_{n>=0..inf} (n+1)^-1*delta(n,s), where delta(n,s) = Sum_{j=0..n} (-1)^j * binomial(n,j) * (j+1)^s.
(End)
a(n,k) = k*a(n-1,k) + (k-1)*a(n-1,k-1) with a(n,1) = 1 and a(n,n) = (n-1)!. - Johannes W. Meijer, Jun 18 2009
Rephrasing the Meijer recurrence above: Let M be the (n+1)X(n+1) bidiagonal matrix with M(r,r) = M(r,r+1) = r, r >= 1, in the two diagonals and the rest zeros. The row a(n+1,.) of the triangle is row 1 of M^n. - Gary W. Adamson, Jun 24 2011
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f.. A(x,t) = G[(t+1)x,-1/(t+1)]-1 (from 2008 comment) = -1 + 1/[1-(1+t)(1-e^(-x))] = (1+t)x + (1+3t+2t^2)x^2/2! + ..., the comp. inverse in x is
B(x,t)= -log(t/(1+t)+1/((1+t)(1+x))) = (1/(1+t))x - ((1+2t)/(1+t)^2)x^2/2 + ((1+3t+3t^2)/(1+t)^3)x^3/3 + .... The numerators are the row polynomials of A074909, and the rational functions are (omitting the initial constants) signed columns of the re-indexed Pascal triangle A007318.
Let h(x,t)= 1/(dB/dx) = (1+x)(1+t(1+x)), then the row polynomial P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(1,t)=1+t. (Series added Dec 29 2015.)(End)
Let denote the Eulerian numbers A173018(n,k), then T(n,k) = Sum_{j=0..n} *binomial(n-j,n-k). - Peter Luschny, Jul 12 2013
Matrix product A007318 * A131689. The n-th row polynomial R(n,x) = Sum_{k >= 1} k^(n-1)*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf A038719. R(n,-1/2) = (-1)^(n-1)*(2^n - 1)*Bernoulli(n)/n. - Peter Bala, Jul 14 2014
a(n,k) = A141618(n,k) / C(n,k-1). - Tom Copeland, Oct 25 2014
For the row polynomials, A028246(n,x) = A019538(n-1,x) * (1+x). - Tom Copeland, Dec 28 2015
n-th row polynomial R(n,x) = (1+x) o (1+x) o ... o (1+x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E11 in the Bala link. - Peter Bala, Jan 12 2018
From Dorian Guyot, May 21 2019: (Start)
Sum_{i=0..k} binomial(k,i) * a(n,i) = (k+1)^n.
Sum_{k=0..n} a(n,k) = 2*A000670(n).
(End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry, A028246, are given by x^n * A_n(1 + 1/x;0). Other specializations of A_n(x;y) give A046802, A090582, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020
The row generating polynomials R(n,x) = Sum_{i=1..n} a(n,i) * x^i satisfy the recurrence equation R(n+1,x) = R(n,x) + Sum_{k=0..n-1} binomial(n-1,k) * R(k+1,x) * R(n-k,x) for n >= 1 with initial value R(1,x) = x. - Werner Schulte, Jun 17 2021

Extensions

Definition corrected by Li Guo, Dec 16 2006
Typo in link corrected by Johannes W. Meijer, Oct 17 2009
Error in title corrected by Johannes W. Meijer, Sep 24 2010
Edited by M. F. Hasler, Oct 29 2014

A074909 Running sum of Pascal's triangle (A007318), or beheaded Pascal's triangle read by beheaded rows.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Wouter Meeussen, Oct 01 2002

Keywords

Comments

This sequence counts the "almost triangular" partitions of n. A partition is triangular if it is of the form 0+1+2+...+k. Examples: 3=0+1+2, 6=0+1+2+3. An "almost triangular" partition is a triangular partition with at most 1 added to each of the parts. Examples: 7 = 1+1+2+3 = 0+2+2+3 = 0+1+3+3 = 0+1+2+4. Thus a(7)=4. 8 = 1+2+2+3 = 1+1+3+3 = 1+1+2+4 = 0+2+3+3 = 0+2+2+4 = 0+1+3+4 so a(8)=6. - Moshe Shmuel Newman, Dec 19 2002
The "almost triangular" partitions are the ones cycled by the operation of "Bulgarian solitaire", as defined by Martin Gardner.
Start with A007318 - I (I = Identity matrix), then delete right border of zeros. - Gary W. Adamson, Jun 15 2007
Also the number of increasing acyclic functions from {1..n-k+1} to {1..n+2}. A function f is acyclic if for every subset B of the domain the image of B under f does not equal B. For example, T(3,1)=4 since there are exactly 4 increasing acyclic functions from {1,2,3} to {1,2,3,4,5}: f1={(1,2),(2,3),(3,4)}, f2={(1,2),(2,3),(3,5)}, f3={(1,2),(2,4),(3,5)} and f4={(1,3),(2,4),(4,5)}. - Dennis P. Walsh, Mar 14 2008
Second Bernoulli polynomials are (from A164555 instead of A027641) B2(n,x) = 1; 1/2, 1; 1/6, 1, 1; 0, 1/2, 3/2, 1; -1/30, 0, 1, 2, 1; 0, -1/6, 0, 5/3, 5/2, 1; ... . Then (B2(n,x)/A002260) = 1; 1/2, 1/2; 1/6, 1/2, 1/3; 0, 1/4, 1/2, 1/4; -1/30, 0, 1/3, 1/2, 1/5; 0, -1/12, 0, 5/12, 1/2, 1/6; ... . See (from Faulhaber 1631) Jacob Bernoulli Summae Potestatum (sum of powers) in A159688. Inverse polynomials are 1; -1, 2; 1, -3, 3; -1, 4, -6, 4; ... = A074909 with negative even diagonals. Reflected A053382/A053383 = reflected B(n,x) = RB(n,x) = 1; -1/2, 1; 1/6, -1, 1; 0, 1/2, -3/2, 1; ... . A074909 is inverse of RB(n,x)/A002260 = 1; -1/2, 1/2; 1/6, -1/2, 1/3; 0, 1/4, -1/2, 1/4; ... . - Paul Curtz, Jun 21 2010
A054143 is the fission of the polynomial sequence (p(n,x)) given by p(n,x) = x^n + x^(n-1) + ... + x + 1 by the polynomial sequence ((x+1)^n). See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Reversal of A135278. - Philippe Deléham, Feb 11 2012
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From A238363, the operator equation d/d(:xD:)f(xD)={exp[d/d(xD)]-1}f(xD) = f(xD+1)-f(xD) follows. Choosing f(x) = x^n and using :xD:^n/n! = binomial(xD,n) and (xD)^n = Bell(n,:xD:), the Bell polynomials of A008277, it follows that the lower triangular matrix [padded A074909]
A) = [St2]*[dP]*[St1] = A048993*A132440*[padded A008275]
B) = [St2]*[dP]*[St2]^(-1)
C) = [St1]^(-1)*[dP]*[St1],
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 whereas [padded A074909]=A007318-I with I=identity matrix. - Tom Copeland, Apr 25 2014
T(n,k) generated by m-gon expansions in the case of odd m with "vertex to side" version or even m with "vertex to vertes" version. Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as a triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to side" version or (ii) m is even and expansion is "vertex to vertex" version. m*Sum_{i=1..k} T(n,k) gives the total label value at the n-th iteration. See also A247976. Vertex to vertex: A061777, A247618, A247619, A247620. Vertex to side: A101946, A247903, A247904, A247905. - Kival Ngaokrajang Sep 28 2014
From Tom Copeland, Nov 12 2014: (Start)
With P(n,x) = [(x+1)^(n+1)-x^(n+1)], the row polynomials of this entry, Up(n,x) = P(n,x)/(n+1) form an Appell sequence of polynomials that are the umbral compositional inverses of the Bernoulli polynomials B(n,x), i.e., B[n,Up(.,x)] = x^n = Up[n,B(.,x)] under umbral substitution, e.g., B(.,x)^n = B(n,x).
The e.g.f. for the Bernoulli polynomials is [t/(e^t - 1)] e^(x*t), and for Up(n,x) it's exp[Up(.,x)t] = [(e^t - 1)/t] e^(x*t).
Another g.f. is G(t,x) = log[(1-x*t)/(1-(1+x)*t)] = log[1 + t /(1 + -(1+x)t)] = t/(1-t*Up(.,x)) = Up(0,x)*t + Up(1,x)*t^2 + Up(2,x)*t^3 + ... = t + (1+2x)/2 t^2 + (1+3x+3x^2)/3 t^3 + (1+4x+6x^2+4x^3)/4 t^4 + ... = -log(1-t*P(.,x)), expressed umbrally.
The inverse, Ginv(t,x), in t of the g.f. may be found in A008292 from Copeland's list of formulas (Sep 2014) with a=(1+x) and b=x. This relates these two sets of polynomials to algebraic geometry, e.g., elliptic curves, trigonometric expansions, Chebyshev polynomials, and the combinatorics of permutahedra and their duals.
Ginv(t,x) = [e^((1+x)t) - e^(xt)] / [(1+x) * e^((1+x)t) - x * e^(xt)] = [e^(t/2) - e^(-t/2)] / [(1+x)e^(t/2) - x*e^(-t/2)] = (e^t - 1) / [1 + (1+x) (e^t - 1)] = t - (1 + 2 x) t^2/2! + (1 + 6 x + 6 x^2) t^3/3! - (1 + 14 x + 36 x^2 + 24 x^3) t^4/4! + ... = -exp[-Perm(.,x)t], where Perm(n,x) are the reverse face polynomials, or reverse f-vectors, for the permutahedra, i.e., the face polynomials for the duals of the permutahedra. Cf. A090582, A019538, A049019, A133314, A135278.
With L(t,x) = t/(1+t*x) with inverse L(t,-x) in t, and Cinv(t) = e^t - 1 with inverse C(t) = log(1 + t). Then Ginv(t,x) = L[Cinv(t),(1+x)] and G(t,x) = C[L[t,-(1+x)]]. Note L is the special linear fractional (Mobius) transformation.
Connections among the combinatorics of the permutahedra, simplices (cf. A135278), and the associahedra can be made through the Lagrange inversion formula (LIF) of A133437 applied to G(t,x) (cf. A111785 and the Schroeder paths A126216 also), and similarly for the LIF A134685 applied to Ginv(t,x) involving the simplicial Whitehouse complex, phylogenetic trees, and other structures. (See also the LIFs A145271 and A133932). (End)
R = x - exp[-[B(n+1)/(n+1)]D] = x - exp[zeta(-n)D] is the raising operator for this normalized sequence UP(n,x) = P(n,x) / (n+1), that is, R UP(n,x) = UP(n+1,x), where D = d/dx, zeta(-n) is the value of the Riemann zeta function evaluated at -n, and B(n) is the n-th Bernoulli number, or constant B(n,0) of the Bernoulli polynomials. The raising operator for the Bernoulli polynomials is then x + exp[-[B(n+1)/(n+1)]D]. [Note added Nov 25 2014: exp[zeta(-n)D] is abbreviation of exp(a.D) with (a.)^n = a_n = zeta(-n)]. - Tom Copeland, Nov 17 2014
The diagonals T(n, n-m), for n >= m, give the m-th iterated partial sum of the positive integers; that is A000027(n+1), A000217(n), A000292(n-1), A000332(n+1), A000389(n+1), A000579(n+1), A000580(n+1), A000581(n+1), A000582(n+1), ... . - Wolfdieter Lang, May 21 2015
The transpose gives the numerical coefficients of the Maurer-Cartan form matrix for the general linear group GL(n,1) (cf. Olver, but note that the formula at the bottom of p. 6 has an error--the 12 should be a 15). - Tom Copeland, Nov 05 2015
The left invariant Maurer-Cartan form polynomial on p. 7 of the Olver paper for the group GL^n(1) is essentially a binomial convolution of the row polynomials of this entry with those of A133314, or equivalently the row polynomials generated by the product of the e.g.f. of this entry with that of A133314, with some reindexing. - Tom Copeland, Jul 03 2018
From Tom Copeland, Jul 10 2018: (Start)
The first column of the inverse matrix is the sequence of Bernoulli numbers, which follows from the umbral definition of the Bernoulli polynomials (B.(0) + x)^n = B_n(x) evaluated at x = 1 and the relation B_n(0) = B_n(1) for n > 1 and -B_1(0) = 1/2 = B_1(1), so the Bernoulli numbers can be calculated using Cramer's rule acting on this entry's matrix and, therefore, from the ratios of volumes of parallelepipeds determined by the columns of this entry's square submatrices. - Tom Copeland, Jul 10 2018
Umbrally composing the row polynomials with B_n(x), the Bernoulli polynomials, gives (B.(x)+1)^(n+1) - (B.(x))^(n+1) = d[x^(n+1)]/dx = (n+1)*x^n, so multiplying this entry as a lower triangular matrix (LTM) by the LTM of the coefficients of the Bernoulli polynomials gives the diagonal matrix of the natural numbers. Then the inverse matrix of this entry has the elements B_(n,k)/(k+1), where B_(n,k) is the coefficient of x^k for B_n(x), and the e.g.f. (1/x) (e^(xt)-1)/(e^t-1). (End)

Examples

			T(4,2) = 0+0+1+3+6 = 10 = binomial(5, 2).
Triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7   8   9 10 11
0:  1
1:  1  2
2:  1  3  3
3:  1  4  6   4
4:  1  5 10  10   5
5:  1  6 15  20  15   6
6:  1  7 21  35  35  21   7
7:  1  8 28  56  70  56  28   8
8:  1  9 36  84 126 126  84  36  9
9:  1 10 45 120 210 252 210 120 45   10
10: 1 11 55 165 330 462 462 330 165  55 11
11: 1 12 66 220 495 792 924 792 495 220 66 12
... Reformatted. - _Wolfdieter Lang_, Nov 04 2014
.
Can be seen as the square array A(n, k) = binomial(n + k + 1, n) read by descending antidiagonals. A(n, k) is the number of monotone nondecreasing functions f: {1,2,..,k} -> {1,2,..,n}. - _Peter Luschny_, Aug 25 2019
[0]  1,  1,   1,   1,    1,    1,     1,     1,     1, ... A000012
[1]  2,  3,   4,   5,    6,    7,     8,     9,    10, ... A000027
[2]  3,  6,  10,  15,   21,   28,    36,    45,    55, ... A000217
[3]  4, 10,  20,  35,   56,   84,   120,   165,   220, ... A000292
[4]  5, 15,  35,  70,  126,  210,   330,   495,   715, ... A000332
[5]  6, 21,  56, 126,  252,  462,   792,  1287,  2002, ... A000389
[6]  7, 28,  84, 210,  462,  924,  1716,  3003,  5005, ... A000579
[7]  8, 36, 120, 330,  792, 1716,  3432,  6435, 11440, ... A000580
[8]  9, 45, 165, 495, 1287, 3003,  6435, 12870, 24310, ... A000581
[9] 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, ... A000582
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n+1,k)))); # Muniru A Asiru, Jul 10 2018
    
  • Haskell
    a074909 n k = a074909_tabl !! n !! k
    a074909_row n = a074909_tabl !! n
    a074909_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Feb 25 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+1,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2018
    
  • Maple
    A074909 := proc(n,k)
        if k > n or k < 0 then
            0;
        else
            binomial(n+1,k) ;
        end if;
    end proc: # Zerinvary Lajos, Nov 09 2006
  • Mathematica
    Flatten[Join[{1}, Table[Sum[Binomial[k, m], {k, 0, n}], {n, 0, 12}, {m, 0, n}] ]] (* or *) Flatten[Join[{1}, Table[Binomial[n, m], {n, 12}, {m, n}]]]
  • PARI
    print1(1);for(n=1,10,for(k=1,n,print1(", "binomial(n,k)))) \\ Charles R Greathouse IV, Mar 26 2013
    
  • Python
    from math import comb, isqrt
    def A074909(n): return comb(r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),n-comb(r,2)) # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = Sum_{i=0..n} C(i, n-k) = C(n+1, k).
Row n has g.f. (1+x)^(n+1)-x^(n+1).
E.g.f.: ((1+x)*e^t - x) e^(x*t). The row polynomials p_n(x) satisfy dp_n(x)/dx = (n+1)*p_(n-1)(x). - Tom Copeland, Jul 10 2018
T(n, k) = T(n-1, k-1) + T(n-1, k) for k: 0Reinhard Zumkeller, Apr 18 2005
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. for column k (with leading zeros): x^(k-1)*(1/(1-x)^(k+1)-1), k >= 0. - Wolfdieter Lang, Nov 04 2014
Up(n, x+y) = (Up(.,x)+ y)^n = Sum_{k=0..n} binomial(n,k) Up(k,x)*y^(n-k), where Up(n,x) = ((x+1)^(n+1)-x^(n+1)) / (n+1) = P(n,x)/(n+1) with P(n,x) the n-th row polynomial of this entry. dUp(n,x)/dx = n * Up(n-1,x) and dP(n,x)/dx = (n+1)*P(n-1,x). - Tom Copeland, Nov 14 2014
The o.g.f. GF(x,t) = x / ((1-t*x)*(1-(1+t)x)) = x + (1+2t)*x^2 + (1+3t+3t^2)*x^3 + ... has the inverse GFinv(x,t) = (1+(1+2t)x-sqrt(1+(1+2t)*2x+x^2))/(2t(1+t)x) in x about 0, which generates the row polynomials (mod row signs) of A033282. The reciprocal of the o.g.f., i.e., x/GF(x,t), gives the free cumulants (1, -(1+2t) , t(1+t) , 0, 0, ...) associated with the moments defined by GFinv, and, in fact, these free cumulants generate these moments through the noncrossing partitions of A134264. The associated e.g.f. and relations to Grassmannians are described in A248727, whose polynomials are the basis for an Appell sequence of polynomials that are umbral compositional inverses of the Appell sequence formed from this entry's polynomials (distinct from the one described in the comments above, without the normalizing reciprocal). - Tom Copeland, Jan 07 2015
T(n, k) = (1/k!) * Sum_{i=0..k} Stirling1(k,i)*(n+1)^i, for 0<=k<=n. - Ridouane Oudra, Oct 23 2022

Extensions

I added an initial 1 at the suggestion of Paul Barry, which makes the triangle a little nicer but may mean that some of the formulas will now need adjusting. - N. J. A. Sloane, Feb 11 2003
Formula section edited, checked and corrected by Wolfdieter Lang, Nov 04 2014

A046802 T(n, k) = Sum_{j=k..n} binomial(n, j)*E1(j, j-k), where E1 are the Eulerian numbers A173018. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 33, 15, 1, 1, 31, 131, 131, 31, 1, 1, 63, 473, 883, 473, 63, 1, 1, 127, 1611, 5111, 5111, 1611, 127, 1, 1, 255, 5281, 26799, 44929, 26799, 5281, 255, 1, 1, 511, 16867, 131275, 344551, 344551, 131275, 16867, 511, 1, 1, 1023, 52905
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of positroid cells of the totally nonnegative Grassmannian G+(k,n) (cf. Postnikov/Williams). It is the triangle of the h-vectors of the stellahedra. - Tom Copeland, Oct 10 2014
See A248727 for a simple transformation of the row polynomials of this entry that produces the umbral compositional inverses of the polynomials of A074909, related to the face polynomials of the simplices. - Tom Copeland, Jan 21 2015
From Tom Copeland, Jan 24 2015: (Start)
The reciprocal of this entry's e.g.f. is [t e^(-xt) - e^(-x)] / (t-1) = 1 - (1+t) x + (1+t+t^2) x^2/2! - (1+t+t^2+t^3) x^3/3! + ... = e^(q.(0;t)x), giving the base sequence (q.(0;t))^n = q_n(0;t) = (-1)^n [1-t^(n+1)] / (1-t) for the umbral compositional inverses (q.(0;t)+z)^n = q_n(z;t) of the Appell polynomials associated with this entry, p_n(z;t) below, i.e., q_n(p.(z;t)) = z^n = p_n(q.(z;t)), in umbral notation. The relations in A133314 then apply between the two sets of base polynomials. (Inserted missing index in a formula - Mar 03 2016.)
The associated o.g.f. for the umbral inverses is Og(x) = x / (1-x q.(0:t)) = x / [(1+x)(1+tx)] = x / [1+(1+t)x+tx^2]. Applying A134264 to h(x) = x / Og(x) = 1 + (1+t) x + t x^2 leads to an o.g.f. for the Narayana polynomials A001263 as the comp. inverse Oginv(x) = [1-(1+t)x-sqrt[1-2(1+t)x+((t-1)x)^2]] / (2xt). Note that Og(x) gives the signed h-polynomials of the simplices and that Oginv(x) gives the h-polynomials of the simplicial duals of the Stasheff polynomials, or type A associahedra. Contrast this with A248727 = A046802 * A007318, which has o.g.f.s related to the corresponding f-polynomials. (End)
The Appell polynomials p_n(x;t) in the formulas below specialize to the Swiss-knife polynomials of A119879 for t = -1, so the Springer numbers A001586 are given by 2^n p_n(1/2;-1). - Tom Copeland, Oct 14 2015
The row polynomials are the h-polynomials associated to the stellahedra, whose f-polynomials are the row polynomials of A248727. Cf. page 60 of the Buchstaber and Panov link. - Tom Copeland, Nov 08 2016
The row polynomials are the h-polynomials of the stellohedra, which enumerate partial permutations according to descents. Cf. Section 10.4 of the Postnikov-Reiner-Williams reference. - Lauren Williams, Jul 05 2022
From p. 60 of the Buchstaber and Panov link, S = P * C / T where S, P, C, and T are the bivariate e.g.f.s of the h vectors of the stellahedra, permutahedra, hypercubes, and (n-1)-simplices, respectively. - Tom Copeland, Jan 09 2017
The number of Le-diagrams of type (k, n) this means the diagram uses the bounding box size k x (n-k), equivalently the number of Grassmann necklaces of type (k, n) and also the number of decorated permutations with k anti-exceedances. - Thomas Scheuerle, Dec 29 2024

Examples

			The triangle T(n, k) begins:
n\k 0   1     2      3      4      5      6     7
0:  1
1:  1   1
2:  1   3     1
3:  1   7     7      1
4:  1  15    33     15      1
5:  1  31   131    131     31      1
6:  1  63   473    883    473     63      1
7:  1 127  1611   5111   5111   1611    127     1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, Holland, 1974, page 245 [From Roger L. Bagula, Nov 21 2009]
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.

Crossrefs

Programs

  • Maple
    T := (n, k) -> add(binomial(n, r)*combinat:-eulerian1(r, r-k), r = k .. n):
    for n from 0 to 8 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jun 27 2018
  • Mathematica
    t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n-1)-1;
    t[n_, k_] = Sum[((i-k+1)^i*(k-i)^(n-i-1) - (i-k+2)^i*(k-i-1)^(n-i-1))*Binomial[n-1, i], {i, 0, k-1}];
    T[n_, k_] := t[n+1, k+1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Jan 22 2015, after Tom Copeland *)
    T[ n_, k_] := Coefficient[n! SeriesCoefficient[(1-x) Exp[t] / (1 - x Exp[(1-x) t]), {t, 0, n}] // Simplify, x, k];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Michael Somos, Jan 22 2015 *)

Formula

E.g.f.: (y-1)*exp(x*y)/(y-exp((y-1)*x)). - Vladeta Jovovic, Sep 20 2003
p(t,x) = (1 - x)*exp(t)/(1 - x*exp(t*(1 - x))). - Roger L. Bagula, Nov 21 2009
With offset=0, T(n,0)=1 otherwise T(n,k) = sum_{i=0..k-1} C(n,i)((i-k)^i*(k-i+1)^(n-i) - (i-k+1)^i*(k-i)^(n-i)) (cf. Williams). - Tom Copeland, Oct 10 2014
With offset 0, T = A007318 * A123125. Second column is A000225; 3rd, appears to be A066810. - Tom Copeland, Jan 23 2015
A raising operator (with D = d/dx) associated with this entry's row polynomials is R = x + t + (1-t) / [1-t e^{(1-t)D}] = x + t + 1 + t D + (t+t^2) D^2/2! + (t+4t^2+t^3) D^3/3! + ... , containing the e.g.f. for the Eulerian polynomials of A123125. Then R^n 1 = (p.(0;t)+x)^n = p_n(x;t) are the Appell polynomials with this entry's row polynomials p_n(0;t) as the base sequence. Examples of this formalism are given in A028246 and A248727. - Tom Copeland, Jan 24 2015
With offset 0, T = A007318*(padded A090582)*(inverse of A097805) = A007318*(padded A090582)*(padded A130595) = A007318*A123125 = A007318*(padded A090582)*A007318*A097808*A130595, where padded matrices are of the form of padded A007318, which is A097805. Inverses of padded matrices are just the padded versions of inverses of the unpadded matrices. This relates the face vectors, or f-vectors, and h-vectors of the permutahedra / permutohedra to those of the stellahedra / stellohedra. - Tom Copeland, Nov 13 2016
Umbrally, the row polynomials (offset 0) are r_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A123125. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = (1-x)/(1-x*exp((1-x)y)), the e.g.f. of A123125, so OP(x,d/dy) y^n evaluated at y = 1 is r_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A248727, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry (A046802, the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
From Peter Luschny, Apr 30 2021: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A122045(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A007047(n).
Sum_{k=0..n} T(n, n-k) = A000522(n).
Sum_{k=0..n} T(n-k, k) = Sum_{k=0..n} (n - k)^k = A026898(n-1) for n >= 1.
Sum_{k=0..n} k*T(n, k) = A036919(n) = floor(n*n!*e/2).
(End)

Extensions

More terms from Vladeta Jovovic, Sep 20 2003
First formula corrected by Wolfdieter Lang, Feb 14 2015
Offset set to 0 and edited by Peter Luschny, Apr 30 2021

A119879 Exponential Riordan array (sech(x),x).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1
Offset: 0

Views

Author

Paul Barry, May 26 2006

Keywords

Comments

Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020

Examples

			Triangle begins:
     1;
     0,    1;
    -1,    0,     1;
     0,   -3,     0,   1;
     5,    0,    -6,   0,   1;
     0,   25,     0, -10,   0,   1;
   -61,    0,    75,   0, -15,   0,   1;
     0, -427,     0, 175,   0, -21,   0,  1;
  1385,    0, -1708,   0, 350,   0, -28,  0,  1;
		

Crossrefs

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.

Programs

  • Maple
    T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009
  • Mathematica
    T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, after Peter Luschny *)
  • PARI
    {T(n,k) = binomial(n,k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
    for(n=0,5, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 25 2019
  • Sage
    @CachedFunction
    def A119879_poly(n, x) :
        return 1 if n == 0  else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
    def A119879_row(n) :
        R = PolynomialRing(ZZ, 'x')
        return R(A119879_poly(n,x)).coeffs()  # Peter Luschny, Jul 16 2012
    # Alternatively:
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015
    

Formula

Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
E.g.f.: exp(x*z)/cosh(x). - Peter Luschny, Aug 01 2012
Sum_{k=0..n} T(n,k)*x^k = A122045(n), A155585(n), A119880(n), A119881(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 27 2013
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024
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