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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.

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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

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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

A008290 Triangle T(n,k) of rencontres numbers (number of permutations of n elements with k fixed points).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 9, 8, 6, 0, 1, 44, 45, 20, 10, 0, 1, 265, 264, 135, 40, 15, 0, 1, 1854, 1855, 924, 315, 70, 21, 0, 1, 14833, 14832, 7420, 2464, 630, 112, 28, 0, 1, 133496, 133497, 66744, 22260, 5544, 1134, 168, 36, 0, 1, 1334961, 1334960, 667485, 222480, 55650, 11088, 1890, 240, 45, 0, 1
Offset: 0

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Author

Keywords

Comments

This is a binomial convolution triangle (Sheffer triangle) of the Appell type: (exp(-x)/(1-x),x), i.e., the e.g.f. of column k is (exp(-x)/(1-x))*(x^k/k!). See the e.g.f. given by V. Jovovic below. - Wolfdieter Lang, Jan 21 2008
The formula T(n,k) = binomial(n,k)*A000166(n-k), with the derangements numbers (subfactorials) A000166 (see also the Charalambides reference) shows the Appell type of this triangle. - Wolfdieter Lang, Jan 21 2008
T(n,k) is the number of permutations of {1,2,...,n} having k pairs of consecutive right-to-left minima (0 is considered a right-to-left minimum for each permutation). Example: T(4,2)=6 because we have 1243, 1423, 4123, 1324, 3124 and 2134; for example, 1324 has right-to-left minima in positions 0-1,3-4 and 2134 has right-to-left minima in positions 0,2-3-4, the consecutive ones being joined by "-". - Emeric Deutsch, Mar 29 2008
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for the sequence aC(0,1). - Tom Copeland, Sep 10 2008
A refinement of this triangle is given by A036039. - Tom Copeland, Nov 06 2012
This triangle equals (A211229(2*n,2*k)) n,k >= 0. - Peter Bala, Dec 17 2014

Examples

			exp((y-1)*x)/(1-x) = 1 + y*x + (1/2!)*(1+y^2)*x^2 + (1/3!)*(2 + 3*y + y^3)*x^3 + (1/4!)*(9 + 8*y + 6*y^2 + y^4)*x^4 + (1/5!)*(44 + 45*y + 20*y^2 + 10*y^3 + y^5)*x^5 + ...
Triangle begins:
       1
       0      1
       1      0     1
       2      3     0     1
       9      8     6     0    1
      44     45    20    10    0    1
     265    264   135    40   15    0   1
    1854   1855   924   315   70   21   0  1
   14833  14832  7420  2464  630  112  28  0 1
  133496 133497 66744 22260 5544 1134 168 36 0 1
...
From _Peter Bala_, Feb 13 2017: (Start)
The infinitesimal generator has integer entries given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2 and begins
   0
   0  0
   1  0  0
   2  3  0  0
   6  8  6  0 0
  24 30 20 10 0 0
...
It is essentially A238363 (unsigned and omitting the main diagonal), A211603 (with different offset) and appears to be A092271, again without the main diagonal. (End)
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 173, Table 5.2 (without row n=0 and column k=0).
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 194.
  • Arnold Kaufmann, Introduction à la combinatorique en vue des applications, Dunod, Paris, 1968. See p. 92.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.

Crossrefs

Mirror of triangle A098825.
Cf. A080955.
Cf. A000012, A000142 (row sums), A000354.
Cf. A170942. Sub-triangle of A211229.
T(2n,n) gives A281262.

Programs

  • Haskell
    a008290 n k = a008290_tabl !! n !! k
    a008290_row n = a008290_tabl !! n
    a008290_tabl = map reverse a098825_tabl
    -- Reinhard Zumkeller, Dec 16 2013
  • Maple
    T:= proc(n,k) T(n, k):= `if`(k=0, `if`(n<2, 1-n, (n-1)*
          (T(n-1, 0)+T(n-2, 0))), binomial(n, k)*T(n-k, 0))
        end:
    seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Mar 15 2013
  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := Round[n!/E] /; n >= 1 size = 8; Table[Binomial[n, k]a[n - k], {n, 0, size}, {k, 0, n}] // TableForm (* Harlan J. Brothers, Mar 19 2007 *)
    T[n_, k_] := Subfactorial[n-k]*Binomial[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2017 *)
    T[n_, k_] := If[n<1, Boole[n==0 && k==0], T[n, k] = T[n-1, k-1] + T[n-1, k]*(n-1-k) + T[n-1, k+1]*(k+1)]; (* Michael Somos, Sep 13 2024 *)
    T[0, 0]:=1; T[n_, 0]:=T[n, 0]=n  T[n-1, 0]+(-1)^n; T[n_, k_]:=T[n, k]=n/k T[n-1, k-1];
    Flatten@Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Oliver Seipel, Nov 26 2024 *)
  • PARI
    {T(n, k) = if(k<0 || k>n, 0, n!/k! * sum(i=0, n-k, (-1)^i/i!))}; /* Michael Somos, Apr 26 2000 */
    

Formula

T(n, k) = T(n-1, k)*n + binomial(n, k)*(-1)^(n-k) = T(n, k-1)/k + binomial(n, k)*(-1)^(n-k)/(n-k+1) = T(n-1, k-1)*n/k = T(n-k, 0)*binomial(n, k) = A000166(n-k)*binomial(n,k) [with T(0, 0) = 1]; so T(n, n) = 1, T(n, n-1) = 0, T(n, n-2) = n*(n-1)/2 for n >= 0.
Sum_{k=0..n} T(n, k) = Sum_{k=0..n} k * T(n, k) = n! for all n > 0, n, k integers. - Wouter Meeussen, May 29 2001
From Vladeta Jovovic, Aug 12 2002: (Start)
O.g.f. for k-th column: (1/k!)*Sum_{i>=k} i!*x^i/(1+x)^(i+1).
O.g.f. for k-th row: k!*Sum_{i=0..k} (-1)^i/i!*(1-x)^i. (End)
E.g.f.: exp((y-1)*x)/(1-x). - Vladeta Jovovic, Aug 18 2002
E.g.f. for number of permutations with exactly k fixed points is x^k/(k!*exp(x)*(1-x)). - Vladeta Jovovic, Aug 25 2002
Sum_{k=0..n} T(n, k)*x^k is the permanent of the n X n matrix with x's on the diagonal and 1's elsewhere; for x = 0, 1, 2, 3, 4, 5, 6 see A000166, A000142, A000522, A010842, A053486, A053487, A080954. - Philippe Deléham, Dec 12 2003; for x = 1+i see A009551 and A009102. - John M. Campbell, Oct 11 2011
T(n, k) = Sum_{j=0..n} A008290(n, j)*k^(n-j) is the permanent of the n X n matrix with 1's on the diagonal and k's elsewhere; for k = 0, 1, 2 see A000012, A000142, A000354. - Philippe Deléham, Dec 13 2003
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*binomial(j,k)*n!/j!. - Paul Barry, May 25 2006
T(n,k) = (n!/k!)*Sum_{j=0..n-k} ((-1)^j)/j!, 0 <= k <= n. From the Appell type of the triangle and the subfactorial formula.
T(n,0) = n*Sum_{j=0..n-1} (j/(j+1))*T(n-1,j), T(0,0)=1. From the z-sequence of this Sheffer triangle z(j)=j/(j+1) with e.g.f. (1-exp(x)*(1-x))/x. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
T(n,k) = (n/k)*T(n-1,k-1) for k >= 1. See above. From the a-sequence of this Sheffer triangle a(0)=1, a(n)=0, n >= 1 with e.g.f. 1. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
From Henk P. J. van Wijk, Oct 29 2012: (Start)
T(n,k) = T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k=0 and
T(n,k) = T(n-1,k-1) + T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k>=1.
(End)
T(n,k) = A098825(n,n-k). - Reinhard Zumkeller, Dec 16 2013
Sum_{k=0..n} k^2 * T(n, k) = 2*n! if n > 1. - Michael Somos, Jun 06 2017
From Tom Copeland, Jul 26 2017: (Start)
The lowering and raising operators of this Appell sequence of polynomials P(n,x) are L = d/dx and R = x + d/dL log[exp(-L)/(1-L)] = x-1 + 1/(1-L) = x + L + L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L)^(-1) exp(-L) x^n = (1+L+L^2+...)(x-1)^n = n! Sum_{k=0..n} (x-1)^k / k!.
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
For more on the infinitesimal generator, noted by Bala below, see A238385. (End)
Sum_{k=0..n} k^m * T(n,k) = A000110(m)*n! if n >= m. - Zhujun Zhang, May 24 2019
Sum_{k=0..n} (k+1) * T(n,k) = A098558(n). - Alois P. Heinz, Mar 11 2022
From Alois P. Heinz, May 20 2023: (Start)
Sum_{k=0..n} (-1)^k * T(n,k) = A000023(n).
Sum_{k=0..n} (-1)^k * k * T(n,k) = A335111(n). (End)
T(n,k) = A145224(n,k)+A145225(n,k), refined by even and odd perms. - R. J. Mathar, Jul 06 2023

Extensions

Comments and more terms from Michael Somos, Apr 26 2000 and Christian G. Bower, Apr 26 2000

A000311 Schroeder's fourth problem; also series-reduced rooted trees with n labeled leaves; also number of total partitions of n.

Original entry on oeis.org

0, 1, 1, 4, 26, 236, 2752, 39208, 660032, 12818912, 282137824, 6939897856, 188666182784, 5617349020544, 181790703209728, 6353726042486272, 238513970965257728, 9571020586419012608, 408837905660444010496, 18522305410364986906624
Offset: 0

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Comments

a(n) is the number of labeled series-reduced rooted trees with n leaves (root has degree 0 or >= 2); a(n-1) = number of labeled series-reduced trees with n leaves. Also number of series-parallel networks with n labeled edges, divided by 2.
A total partition of n is essentially what is meant by the first part of the previous line: take the numbers 12...n, and partition them into at least two blocks. Partition each block with at least 2 elements into at least two blocks. Repeat until only blocks of size 1 remain. (See the reference to Stanley, Vol. 2.) - N. J. A. Sloane, Aug 03 2016
Polynomials with coefficients in triangle A008517, evaluated at 2. - Ralf Stephan, Dec 13 2004
Row sums of unsigned A134685. - Tom Copeland, Oct 11 2008
Row sums of A134991, which contains an e.g.f. for this sequence and its compositional inverse. - Tom Copeland, Jan 24 2018
From Gus Wiseman, Dec 28 2019: (Start)
Also the number of singleton-reduced phylogenetic trees with n labels. A phylogenetic tree is a series-reduced rooted tree whose leaves are (usually disjoint) nonempty sets. It is singleton-reduced if no non-leaf node covers only singleton branches. For example, the a(4) = 26 trees are:
{1,2,3,4} {{1},{2},{3,4}} {{1},{2,3,4}}
{{1},{2,3},{4}} {{1,2},{3,4}}
{{1,2},{3},{4}} {{1,2,3},{4}}
{{1},{2,4},{3}} {{1,2,4},{3}}
{{1,3},{2},{4}} {{1,3},{2,4}}
{{1,4},{2},{3}} {{1,3,4},{2}}
{{1,4},{2,3}}
{{{1},{2,3}},{4}}
{{{1,2},{3}},{4}}
{{1},{{2},{3,4}}}
{{1},{{2,3},{4}}}
{{{1},{2,4}},{3}}
{{{1,2},{4}},{3}}
{{1},{{2,4},{3}}}
{{{1,3},{2}},{4}}
{{{1},{3,4}},{2}}
{{{1,3},{4}},{2}}
{{{1,4},{2}},{3}}
{{{1,4},{3}},{2}}
(End)

Examples

			E.g.f.: A(x) = x + x^2/2! + 4*x^3/3! + 26*x^4/4! + 236*x^5/5! + 2752*x^6/6! + ...
where exp(A(x)) = 1 - x + 2*A(x), and thus
Series_Reversion(A(x)) = x - x^2/2! - x^3/3! - x^4/4! - x^5/5! - x^6/6! + ...
O.g.f.: G(x) = x + x^2 + 4*x^3 + 26*x^4 + 236*x^5 + 2752*x^6 + 39208*x^7 + ...
where
G(x) = x/2 + x/(2*(2-x)) + x/(2*(2-x)*(2-2*x)) + x/(2*(2-x)*(2-2*x)*(2-3*x)) + x/(2*(2-x)*(2-2*x)*(2-3*x)*(2-4*x)) + x/(2*(2-x)*(2-2*x)*(2-3*x)*(2-4*x)*(2-5*x)) + ...
From _Gus Wiseman_, Dec 28 2019: (Start)
A rooted tree is series-reduced if it has no unary branchings, so every non-leaf node covers at least two other nodes. The a(4) = 26 series-reduced rooted trees with 4 labeled leaves are the following. Each bracket (...) corresponds to a non-leaf node.
  (1234)  ((12)34)  ((123)4)
          (1(23)4)  (1(234))
          (12(34))  ((124)3)
          (1(24)3)  ((134)2)
          ((13)24)  (((12)3)4)
          ((14)23)  ((1(23))4)
                    ((12)(34))
                    (1((23)4))
                    (1(2(34)))
                    (((12)4)3)
                    ((1(24))3)
                    (1((24)3))
                    (((13)2)4)
                    ((13)(24))
                    (((13)4)2)
                    ((1(34))2)
                    (((14)2)3)
                    ((14)(23))
                    (((14)3)2)
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 224.
  • J. Felsenstein, Inferring phyogenies, Sinauer Associates, 2004; see p. 25ff.
  • L. R. Foulds and R. W. Robinson, Enumeration of phylogenetic trees without points of degree two. Ars Combin. 17 (1984), A, 169-183. Math. Rev. 85f:05045
  • T. S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 197.
  • E. Schroeder, Vier combinatorische Probleme, Z. f. Math. Phys., 15 (1870), 361-376.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see "total partitions", Example 5.2.5, Equation (5.27), and also Fig. 5-3 on page 14. See also the Notes on page 66.

Crossrefs

Row sums of A064060 and A134991.
The unlabeled version is A000669.
Unlabeled phylogenetic trees are A141268.
The node-counting version is A060356, with unlabeled version A001678.
Phylogenetic trees with n labels are A005804.
Chains of set partitions are A005121, with maximal version A002846.
Inequivalent leaf-colorings of series-reduced rooted trees are A318231.
For n >= 2, A000311(n) = A006351(n)/2 = A005640(n)/2^(n+1).
Cf. A000110, A000669 = unlabeled hierarchies, A119649.

Programs

  • Maple
    M:=499; a:=array(0..500); a[0]:=0; a[1]:=1; a[2]:=1; for n from 0 to 2 do lprint(n,a[n]); od: for n from 2 to M do a[n+1]:=(n+2)*a[n]+2*add(binomial(n,k)*a[k]*a[n-k+1],k=2..n-1); lprint(n+1,a[n+1]); od:
    Order := 50; t1 := solve(series((exp(A)-2*A-1),A)=-x,A); A000311 := n-> n!*coeff(t1,x,n);
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(combinat[multinomial](n, n-i*j, i$j)/j!*
          a(i)^j*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> `if`(n<2, n, b(n, n-1)):
    seq(a(n), n=0..40);  # Alois P. Heinz, Jan 28 2016
    # faster program:
    b:= proc(n, i) option remember;
        `if`(i=0 and n=0, 1, `if`(i<=0 or i>n, 0,
        i*b(n-1, i) + (n+i-1)*b(n-1, i-1))) end:
    a:= n -> `if`(n<2, n, add(b(n-1, i), i=0..n-1)):
    seq(a(n), n=0..40);  # Peter Luschny, Feb 15 2021
  • Mathematica
    nn = 19; CoefficientList[ InverseSeries[ Series[1+2a-E^a, {a, 0, nn}], x], x]*Range[0, nn]! (* Jean-François Alcover, Jul 21 2011 *)
    a[ n_] := If[ n < 1, 0, n! SeriesCoefficient[ InverseSeries[ Series[ 1 + 2 x - Exp[x], {x, 0, n}]], n]]; (* Michael Somos, Jun 04 2012 *)
    a[n_] := (If[n < 2,n,(column = ConstantArray[0, n - 1]; column[[1]] = 1; For[j = 3, j <= n, j++, column = column * Flatten[{Range[j - 2], ConstantArray[0, (n - j) + 1]}] + Drop[Prepend[column, 0], -1] * Flatten[{Range[j - 1, 2*j - 3], ConstantArray[0, n - j]}];]; Sum[column[[i]], {i, n - 1}]  )]); Table[a[n], {n, 0, 20}] (* Peter Regner, Oct 05 2012, after a formula by Felsenstein (1978) *)
    multinomial[n_, k_List] := n!/Times @@ (k!); b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, Sum[multinomial[n, Join[{n-i*j}, Array[i&,j]]]/j!*a[i]^j *b[n-i*j, i-1], {j, 0, n/i}]]]; a[n_] := If[n<2, n, b[n, n-1]]; Table[ a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 07 2016, after Alois P. Heinz *)
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    mtot[m_]:=Prepend[Join@@Table[Tuples[mtot/@p],{p,Select[sps[m],1Gus Wiseman, Dec 28 2019 *)
    (* Lengthy but easy to follow *)
      lead[, n /; n < 3] := 0
      lead[h_, n_] := Module[{p, i},
            p = Position[h, {_}];
            Sum[MapAt[{#, n} &, h, p[[i]]], {i, Length[p]}]
            ]
      follow[h_, n_] := Module[{r, i},
            r = Replace[Position[h, {_}], {a__} -> {a, -1}, 1];
            Sum[Insert[h, n, r[[i]]], {i, Length[r]}]
            ]
      marry[, n /; n < 3] := 0
      marry[h_, n_] := Module[{p, i},
            p = Position[h, _Integer];
            Sum[MapAt[{#, n} &, h, p[[i]]], {i, Length[p]}]
            ]
      extend[a_ + b_, n_] := extend[a, n] + extend[b, n]
      extend[a_, n_] := lead[a, n] + follow[a, n] + marry[a, n]
      hierarchies[1] := hierarchies[1] = extend[hier[{}], 1]
      hierarchies[n_] := hierarchies[n] = extend[hierarchies[n - 1], n] (* Daniel Geisler, Aug 22 2022 *)
  • Maxima
    a(n):=if n=1 then 1 else sum((n+k-1)!*sum(1/(k-j)!*sum((2^i*(-1)^(i)*stirling2(n+j-i-1,j-i))/((n+j-i-1)!*i!),i,0,j),j,1,k),k,1,n-1); /* Vladimir Kruchinin, Jan 28 2012 */
    
  • PARI
    {a(n) = local(A); if( n<0, 0, for( i=1, n, A = Pol(exp(A + x * O(x^i)) - A + x - 1)); n! * polcoeff(A, n))}; /* Michael Somos, Jan 15 2004 */
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = O(x); for( i=1, n, A = intformal( 1 / (1 + x - 2*A))); n! * polcoeff(A, n))}; /* Michael Somos, Oct 25 2014 */
    
  • PARI
    {a(n) = n! * polcoeff(serreverse(1+2*x - exp(x +x^2*O(x^n))), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Oct 27 2014
    
  • PARI
    \p100 \\ set precision
    {A=Vec(sum(n=0, 600, 1.*x/prod(k=0, n, 2 - k*x + O(x^31))))}
    for(n=0, 25, print1(if(n<1,0,round(A[n])),", ")) \\ Paul D. Hanna, Oct 27 2014
    
  • Python
    from functools import lru_cache
    from math import comb
    @lru_cache(maxsize=None)
    def A000311(n): return n if n <= 1 else -(n-1)*A000311(n-1)+comb(n,m:=n+1>>1)*(0 if n&1 else A000311(m)**2) + (sum(comb(n,i)*A000311(i)*A000311(n-i) for i in range(1,m))<<1) # Chai Wah Wu, Nov 10 2022

Formula

E.g.f. A(x) satisfies exp A(x) = 2*A(x) - x + 1.
a(0)=0, a(1)=a(2)=1; for n >= 2, a(n+1) = (n+2)*a(n) + 2*Sum_{k=2..n-1} binomial(n, k)*a(k)*a(n-k+1).
a(1)=1; for n>1, a(n) = -(n-1) * a(n-1) + Sum_{k=1..n-1} binomial(n, k) * a(k) * a(n-k). - Michael Somos, Jun 04 2012
From the umbral operator L in A135494 acting on x^n comes, umbrally, (a(.) + x)^n = (n * x^(n-1) / 2) - (x^n / 2) + Sum_{j>=1} j^(j-1) * (2^(-j) / j!) * exp(-j/2) * (x + j/2)^n giving a(n) = 2^(-n) * Sum_{j>=1} j^(n-1) * ((j/2) * exp(-1/2))^j / j! for n > 1. - Tom Copeland, Feb 11 2008
Let h(x) = 1/(2-exp(x)), an e.g.f. for A000670, then the n-th term of A000311 is given by ((h(x)*d/dx)^n)x evaluated at x=0, i.e., A(x) = exp(x*a(.)) = exp(x*h(u)*d/du) u evaluated at u=0. Also, dA(x)/dx = h(A(x)). - Tom Copeland, Sep 05 2011 (The autonomous differential eqn. here is also on p. 59 of Jones. - Tom Copeland, Dec 16 2019)
A134991 gives (b.+c.)^n = 0^n, for (b_n)=A000311(n+1) and (c_0)=1, (c_1)=-1, and (c_n)=-2* A000311(n) = -A006351(n) otherwise. E.g., umbrally, (b.+c.)^2 = b_2*c_0 + 2 b_1*c_1 + b_0*c_2 =0. - Tom Copeland, Oct 19 2011
a(n) = Sum_{k=1..n-1} (n+k-1)!*Sum_{j=1..k} (1/(k-j)!)*Sum_{i=0..j} 2^i*(-1)^i*Stirling2(n+j-i-1, j-i)/((n+j-i-1)!*i!), n>1, a(0)=0, a(1)=1. - Vladimir Kruchinin, Jan 28 2012
Using L. Comtet's identity and D. Wasserman's explicit formula for the associated Stirling numbers of second kind (A008299) one gets: a(n) = Sum_{m=1..n-1} Sum_{i=0..m} (-1)^i * binomial(n+m-1,i) * Sum_{j=0..m-i} (-1)^j * ((m-i-j)^(n+m-1-i))/(j! * (m-i-j)!). - Peter Regner, Oct 08 2012
G.f.: x/Q(0), where Q(k) = 1 - k*x - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: x*Q(0), where Q(k) = 1 - x*(k+1)/(x*(k+1) - (1-k*x)*(1-x-k*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
a(n) ~ n^(n-1) / (sqrt(2) * exp(n) * (2*log(2)-1)^(n-1/2)). - Vaclav Kotesovec, Jan 05 2014
E.g.f. A(x) satisfies d/dx A(x) = 1 / (1 + x - 2 * A(x)). - Michael Somos, Oct 25 2014
O.g.f.: Sum_{n>=0} x / Product_{k=0..n} (2 - k*x). - Paul D. Hanna, Oct 27 2014
E.g.f.: (x - 1 - 2 LambertW(-exp((x-1)/2) / 2)) / 2. - Vladimir Reshetnikov, Oct 16 2015 (This e.g.f. is given in A135494, the entry alluded to in my 2008 formula, and in A134991 along with its compositional inverse. - Tom Copeland, Jan 24 2018)
a(0) = 0, a(1) = 1; a(n) = n! * [x^n] exp(Sum_{k=1..n-1} a(k)*x^k/k!). - Ilya Gutkovskiy, Oct 17 2017
a(n+1) = Sum_{k=0..n} A269939(n, k) for n >= 1. - Peter Luschny, Feb 15 2021

Extensions

Name edited by Gus Wiseman, Dec 28 2019

A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440, 8064, 3360, 960, 180, 20, 1
Offset: 0

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Comments

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
As an upper right triangle, table rows give number of points, edges, faces, cubes,
4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed. - Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129). - Gerald McGarvey, May 17 2005
Riordan array (1/(1-2x), x/(1-2x)). - Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n-1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist). - Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x. - Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR. - Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0). - Joerg Arndt, Jul 01 2011
T(i,j) is the number of i-permutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n. - N-E. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthal-numbers: A001045. - Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...). - Gary W. Adamson, Feb 07 2010
f-vectors ("face"-vectors) for n-dimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.) - Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(-x*z) governed by the Fokker-Planck equation as given in the Dattoli et al. ref. below. - Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array. - Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found. - Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]->[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(n-k) ways to map the other (n-k) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k). - Dennis P. Walsh, Sep 26 2017
Since this array is the square of the Pascal lower triangular matrix, the row polynomials of this array are obtained as the umbral composition of the row polynomials P_n(x) of the Pascal matrix with themselves. E.g., P_3(P.(x)) = 1 P_3(x) + 3 P_2(x) + 3 P_1(x) + 1 = (x^3 + 3 x^2 + 3 x + 1) + 3 (x^2 + 2 x + 1) + 3 (x + 1) + 1 = x^3 + 6 x^2 + 12 x + 8. - Tom Copeland, Nov 12 2018
T(n,k) is the number of 2-compositions of n+1 with some zeros allowed that have k zeros; see the Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
Also the convolution triangle of A000079. - Peter Luschny, Oct 09 2022

Examples

			Triangle begins with T(0,0):
   1;
   2,  1;
   4,  4,  1;
   8, 12,  6,  1;
  16, 32, 24,  8,  1;
  32, 80, 80, 40, 10,  1;
  ... -  corrected by _Clark Kimberling_, Aug 05 2011
Seen as an array read by descending antidiagonals:
[0] 1, 2,  4,   8,    16,    32,    64,     128,     256, ...     [A000079]
[1] 1, 4,  12,  32,   80,    192,   448,    1024,    2304, ...    [A001787]
[2] 1, 6,  24,  80,   240,   672,   1792,   4608,    11520, ...   [A001788]
[3] 1, 8,  40,  160,  560,   1792,  5376,   15360,   42240, ...   [A001789]
[4] 1, 10, 60,  280,  1120,  4032,  13440,  42240,   126720, ...  [A003472]
[5] 1, 12, 84,  448,  2016,  8064,  29568,  101376,  329472, ...  [A054849]
[6] 1, 14, 112, 672,  3360,  14784, 59136,  219648,  768768, ...  [A002409]
[7] 1, 16, 144, 960,  5280,  25344, 109824, 439296,  1647360, ... [A054851]
[8] 1, 18, 180, 1320, 7920,  41184, 192192, 823680,  3294720, ... [A140325]
[9] 1, 20, 220, 1760, 11440, 64064, 320320, 1464320, 6223360, ... [A140354]
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
  • H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

Crossrefs

Programs

  • GAP
    Flat(List([0..15], n->List([0..n], k->Binomial(n, k)*2^(n-k)))); # Stefano Spezia, Nov 21 2018
  • Haskell
    a038207 n = a038207_list !! n
    a038207_list = concat $ iterate ([2,1] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Haskell
    a038207' n k = a038207_tabl !! n !! k
    a038207_row n = a038207_tabl !! n
    a038207_tabl = iterate f [1] where
       f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
    -- Reinhard Zumkeller, Feb 27 2013
    
  • Magma
    /* As triangle */ [[(&+[Binomial(n,i)*Binomial(i,k): i in [k..n]]): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Nov 16 2018
    
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*2^(i-j), j = 0 .. i) end do; # yields sequence in triangular form - Emeric Deutsch, Nov 04 2007
    # Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
    PMatrix(10, n -> 2^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x -> 1, {n, 0,10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)
    Table[Binomial[n,k]2^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, May 22 2020 *)
  • PARI
    {T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */
    
  • Sage
    def A038207_triangle(dim):
        M = matrix(ZZ,dim,dim)
        for n in range(dim): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+2*M[n-1,k]
        return M
    A038207_triangle(9)  # Peter Luschny, Sep 20 2012
    

Formula

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (-1)^k*A065109(n,k).
G.f.: 1/(1-2*z-t*z). - Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number. - Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n. - Philippe Deléham, Dec 15 2009
n-th row is obtained by taking pairwise sums of triangle A112857 terms starting from the right. - Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n-1,k-1) + 2*T(n-1,k) for kJon Perry, Oct 11 2012
The e.g.f. for the n-th row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n, -L(.,-x))/n! = 2^n L(n, -x/2)/n!. E.g., L(2,x) = 2 -4*x +x^2, so p(2,x)= (1/2)*L(2, -L(.,-x)) = (1/2)*(2*L(0,-x) + 4*L(1,-x) + L(2,-x)) = (1/2)*(2 + 4*(1+x) + (2+4*x+x^2)) = 4 + 4*x + x^2/2. - Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
exp(x*z/(1-2*z))/(1-2*z)= exp(2*z D_z z) e^(x*z)= exp(2*D_x (x D_x)) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
= Sum_{n>=0} z^n * 2^n Lag_n(-x/2)= exp[z*EF(.,x)], an o.g.f. for the f-vectors (rows) of A038207 where EF(n,x) is an e.g.f. for the n-th f-vector. (Lag_n(x) are the un-normalized Laguerre polynomials.)
Conversely,
exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
= (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
= (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
= exp(z*OF(.,x)), an e.g.f for the f-vectors of A038207 where
OF(n,x)= (2+x)^n is an o.g.f. for the n-th f-vector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n). - Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*|A167374| = |A118801|*|A167374| = |A118801*A167374| = |P*A167374*P^(-1)*A167374| = |P*NpdP*A167374|. Cf. A118801. - Tom Copeland, Nov 17 2016
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + .... - Peter Bala, Mar 05 2017
T(3*k+2,k) = T(3*k+2,k+1), T(2*k+1,k) = 2*T(2*k+1,k+1). - Yuchun Ji, May 26 2020
From Robert A. Russell, Aug 05 2020: (Start)
G.f. for column k: x^k / (1-2*x)^(k+1).
E.g.f. for column k: exp(2*x) * x^k / k!. (End)
Also the array A(n, k) read by descending antidiagonals, where A(n, k) = (-1)^n*Sum_{j= 0..n+k} binomial(n + k, j)*hypergeom([-n, j+1], [1], 1). - Peter Luschny, Nov 09 2021

A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

Original entry on oeis.org

1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600, 2324549427200, 81359229952000, 3091650738176000, 126757680265216000, 5577337931669504000, 262134882788466688000
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

a(n-1), n >= 1, enumerates increasing plane (aka ordered) trees with n vertices (one of them a root labeled 1) where each vertex with outdegree r >= 0 comes in r+1 types (like an (r+1)-ary vertex). See the increasing tree comments under A004747. - Wolfdieter Lang, Oct 12 2007
An example for the case of 3 vertices is shown below. For the enumeration of non-plane trees of this type see A029768. - Peter Bala, Aug 30 2011
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016789. - Reinhard Zumkeller, Sep 20 2013
The Mathar conjecture is true. Generally from the factorial form, the last term is the "extra" product beyond the prior term, from k=n-1 and 3k+2 evaluates to 3*(n-1)+2 = 3n-1, yielding a(n) = a(n-1)*(3n-1) (eqn1). Similarly, a(n) = a(n-2)*(3n-1)*(3(n-2)+2) = a(n-2)*(3n-1)*(3n-4) (eqn2) and a(n) = a(n-3)*(3n-1)*(3n-4)*(3*(n-2)+2) = a(n-3)*(3n-1)*(3n-4)*(3n-7) (eqn3). We equate (eqn2) and (eqn3) to get a(n-2)*(3n-1)*(3n-4) = a(n-3)*(3n-1)*(3n-4)*(3n-7) or a(n-2)+(7-3n)*a(n-3) = 0 (eqn4). From (eqn1) we have a(n)+(1-3n)*a(n-1) = 0 (eqn5). Combining (eqn4) and (eqn5) yields a(n)+(1-3n)*a(n-1)+a(n-2)+(7-3n)*a(n-3) = 0. - Bill McEachen, Jan 01 2016
a(n-1), n>=1, is the dimension of the n-th component of the operad encoding the multilinearization of the following identity in nonassociative algebras: s*(a,a,b)-(s+t)*(a,b,a)+t*(b,a,a)=0, for any given pair of scalars (s,t). Here (a,b,c) is the associator (ab)c-a(bc). This is proved in the referenced article on associator dependent algebras by Bremner and me. - Vladimir Dotsenko, Mar 22 2022

Examples

			a(2) = 10 from the described trees with 3 vertices: there are three trees with a root vertex (label 1) with outdegree r=2 (like the three 3-stars each with one different ray missing) and the four trees with a root (r=1 and label 1) a vertex with (r=1) and a leaf (r=0). Assigning labels 2 and 3 yields 2*3+4=10 such trees.
a(2) = 10. The 10 possible plane increasing trees on 3 vertices, where vertices of outdegree 1 come in 2 colors (denoted a or b) and vertices of outdegree 2 come in 3 colors (a, b or c), are:
.
   1a    1b    1a    1b        1a       1b       1c
   |     |     |     |        / \      / \      / \
   2a    2b    2b    2a      2   3    2   3    2   3
   |     |     |     |
   3     3     3     3         1a       1b       1c
                              / \      / \      / \
                             3   2    3   2    3   2
		

Crossrefs

a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.
Cf. A225470, A290596 (first columns).
Subsequence of A007661.

Programs

  • Haskell
    a008544 n = a008544_list !! n
    a008544_list = scanl (*) 1 a016789_list
    -- Reinhard Zumkeller, Sep 20 2013
    
  • Magma
    [Round((Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6) )/ Sqrt(3)*3^n/4^(n-1)): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
    
  • Magma
    [Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // G. C. Greubel, Mar 31 2019
  • Maple
    a := n -> mul(3*k-1, k = 1..n);
    A008544 := n -> mul(k, k = select(k-> k mod 3 = 2, [$1 .. 3*n])): seq(A008544(n), n = 0 .. 16); # Peter Luschny, Jun 23 2011
  • Mathematica
    k = 3; b[1]=2; b[n_]:= b[n] = b[n-1]+k; a[0]=1; a[1]=2; a[n_]:= a[n] = a[n-1]*b[n]; Table[a[n], {n,0,20}] (* Roger L. Bagula, Sep 17 2008 *)
    Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* Michael De Vlieger, Jan 02 2016 *)
    Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* G. C. Greubel, Mar 31 2019 *)
  • Maxima
    a(n):=((n)!*sum(binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k),k,floor(n/2),n)); /* Vladimir Kruchinin, Sep 28 2013 */
    
  • PARI
    a(n) = prod(k=0,n-1, 3*k+2 );
    
  • PARI
    vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ G. C. Greubel, Mar 31 2019
    
  • Sage
    @CachedFunction
    def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
    [A008544(n) for n in (0..16)]  # Peter Luschny, May 20 2013
    
  • Sage
    [3^n*rising_factorial(2/3, n) for n in (0..20)] # G. C. Greubel, Mar 31 2019
    

Formula

a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).
E.g.f.: (1-3*x)^(-2/3).
a(n) = 2*A034000(n), n >= 1, a(0) = 1.
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(2/3)^-1*n^(1/6)*3^n*e^-n*n^n*{1 - 1/36*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = (Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6))/sqrt(3)*3^n/4^(n-1). - Jeremy L. Martin, Mar 31 2002 (typo fixed by Vincenzo Librandi, Feb 21 2015)
From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)
a(n) = A084939(n)/A000142(n)*A000079(n).
a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)
Let T = A094638 and c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), then A008544 = unsigned [ T * c(-3) ] and the list partition transform A133314 of [1,T * c(-3)] gives [1,T * c(3)] with all odd terms negated, which equals a signed version of A007559; i.e., LPT[(1,signed A008544)] = signed A007559. Also LPT[A007559] = (1,-A008544) and e.g.f. [1,T * c(t)] = (1-x*t)^(-1/t) for t = 3 or -3. Analogous results hold for the double factorial, quadruple factorial and so on. - Tom Copeland, Dec 22 2007
G.f.: 1/(1-2x/(1-3x/(1-5x/(1-6x/(1-8x/(1-9x/(1-11x/(1-12x/(1-...))))))))) (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-1)^n*Sum_{k=0..n} 3^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+2)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k+2)/(x*(3*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
D-finite with recurrence: a(n) = (9*(n-2)*(n-1)+2)*a(n-2) + 4*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 09 2013
a(n) = n!*Sum_{k=floor(n/2)..n} binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k). - Vladimir Kruchinin, Sep 28 2013
Recurrence equation: a(n) = 3*a(n-1) + (3*n - 4)^2*a(n-2) with a(0) = 1 and a(1) = 2. A024396 satisfies the same recurrence (but with different initial conditions). This observation leads to a continued fraction expansion for the constant A193534 due to Euler. - Peter Bala, Feb 20 2015
a(n) = A225470(n, 0), n >= 0. - Wolfdieter Lang, May 29 2017
G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - G. C. Greubel, Mar 31 2019
D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
G.f.: 1/(1-2*x-6*x^2/(1-8*x-30*x^2/(1-14*x-72*x^2/(1-20*x-132*x^2/(1-...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020
G.f.: 1/G(0), where G(k) = 1 - (6*k+2)*x - 3*(k+1)*(3*k+2)*x^2/G(k+1). - Nikolaos Pantelidis, Feb 28 2020
Sum_{n>=0} 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - Amiram Eldar, Mar 01 2022

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344, 302408598355968
Offset: 0

Views

Author

Keywords

Comments

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) = 1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

Examples

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013
		

References

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

Crossrefs

Row sums of A059474. - David W. Wilson, Aug 14 2006
Equals 2 * A003480, n>0.
Row sums of A140071.

Programs

  • Haskell
    a007070 n = a007070_list !! n
    a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
       (tail $ map (* 2) a007070_list) a007070_list)
    -- Reinhard Zumkeller, Jan 16 2012
  • Magma
    Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011
    
  • Magma
    [n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011
    
  • Maple
    A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
    seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015
  • Mathematica
    LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)
  • PARI
    a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)
    
  • PARI
    a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))
    
  • Sage
    [lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024

A008517 Second-order Eulerian triangle T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 8, 6, 1, 22, 58, 24, 1, 52, 328, 444, 120, 1, 114, 1452, 4400, 3708, 720, 1, 240, 5610, 32120, 58140, 33984, 5040, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
Offset: 1

Views

Author

Keywords

Comments

Second-order Eulerian numbers <> = T(n,k+1) count the permutations of the multiset {1,1,2,2,...,n,n} with k ascents with the restriction that for all m, all integers between the two copies of m are less than m. In particular, the two 1s are always next to each other.
When seen as coefficients of polynomials with descending exponents, evaluations are in A000311 (x=2) and A001662 (x=-1).
The row reversed triangle is A112007. There one can find comments on the o.g.f.s for the diagonals of the unsigned Stirling1 triangle |A008275|.
Stirling2(n,n-k) = Sum_{m=0..k-1} T(k,m+1)*binomial(n+k-1+m, 2*k), k>=1. See the Graham et al. reference p. 271 eq. (6.43).
This triangle is the coefficient triangle of the numerator polynomials appearing in the o.g.f. for the k-th diagonal (k >= 1) of the Stirling2 triangle A048993.
The o.g.f. for column k satisfies the recurrence G(k,x) = x*(2*x*(d/dx)G(k-1,x) + (2-k)*G(k-1,x))/(1-k*x), k >= 2, with G(1,x) = 1/(1-x). - Wolfdieter Lang, Oct 14 2005
This triangle is in some sense generated by the differential equation y' = 1 - 2/(1+x+y). (This is the differential equation satisfied by the function defined implicitly as x+y=exp(x-y).) If we take y = a(0) + a(1)x + a(2)x^2 + a(3)x^3 + ... and assume a(0)=c then all the a's may be calculated formally in terms of c and we have a(1) = (c-1)/(c+1) and, for n > 1, a(n) = 2^n/n! (1+c)^(1-2n)( T(n,1)c - T(n,2)c^2 + T(n,3)c^3 - ... + (-1)^(n-1) T(n,n)c^n ). - Moshe Shmuel Newman, Aug 08 2007
From the recurrence relation, the generating function F(x,y) := 1 + Sum_{n>=1, 1<=k<=n} [T(n,k)x^n/n!*y^k] satisfies the partial differential equation F = (1/y-2x)F_x + (y-1)F_y, with (non-elementary) solution F(x,y) = (1-y)/(1-Phi(w)) where w = y*exp(x(y-1)^2-y) and Phi(x) is defined by Phi(x) = x*exp(Phi(x)). By Lagrange inversion (see Wilf's book "generatingfunctionology", page 168, Example 1), Phi(x) = Sum_{n>=1} n^(n-1)*x^n/n!. Thus Phi(x) can alternatively be described as the e.g.f. for rooted labeled trees on n vertices A000169. - David Callan, Jul 25 2008
A method for solving PDEs such as the one above for F(x,y) is described in the Klazar reference (see pages 207-208). In his case, the auxiliary ODE dy/dx = b(x,y)/a(x,y) is exact; in this case it is not exact but has an integrating factor depending on y alone, namely y-1. The e.g.f. for the row sums A001147 is 1/sqrt(1-2*x) and the demonstration that F(x,1) = 1/sqrt(1-2*x) is interesting: two applications of l'Hopital's rule to lim_{y->1}F(x,y) yield F(x,1) = 1/(1-2x)*1/F(x,1). So l'Hopital's rule doesn't directly yield F(x,1) but rather an equation to be solved for F(x,1)!. - David Callan, Jul 25 2008
From Tom Copeland, Oct 12 2008; May 19 2010: (Start)
Let P(0,t)= 0, P(1,t)= 1, P(2,t)= t, P(3,t)= t + 2 t^2, P(4,t)= t + 8 t^2 + 6 t^3, ... be the row polynomials of the present array, then
exp(x*P(.,t)) = ( u + Tree(t*exp(u)) ) / (1-t) = WD(x*(1-t), t/(1-t)) / (1-t)
where u = x*(1-t)^2 - t, Tree(x) is the e.g.f. of A000169 and WD(x,t) is the e.g.f. for A134991, relating the Ward and 2-Eulerian polynomials by a simple transformation.
Note also apparently P(4,t) / (1-t)^3 = Ward Poly(4, t/(1-t)) = essentially an e.g.f. for A093500.
The compositional inverse of f(x,t) = exp(P(.,t)*x) about x=0 is
g(x,t) = ( x - (t/(1-t)^2)*(exp(x*(1-t))-x*(1-t)-1) )
= x - t*x^2/2! - t*(1-t)*x^3/3! - t*(1-t)^2*x^4/4! - t*(1-t)^3*x^5/5! - ... .
Can apply A134685 to these coefficients to generate f(x,t). (End)
Triangle A163936 is similar to the one given above except for an extra right hand column [1, 0, 0, 0, ... ] and that its row order is reversed. - Johannes W. Meijer, Oct 16 2009
From Tom Copeland, Sep 04 2011: (Start)
Let h(x,t) = 1/(1-(t/(1-t))*(exp(x*(1-t))-1)), an e.g.f. in x for row polynomials in t of A008292, then the n-th row polynomial in t of the table A008517 is given by ((h(x,t)*D_x)^(n+1))x with the derivative evaluated at x=0.
Also, df(x,t)/dx = h(f(x,t),t) where f(x,t) is an e.g.f. in x of the row polynomials in t of A008517, i.e., exp(x*P(.,t)) in Copeland's 2008 comment. (End)
The rows are the h-vectors of A134991. - Tom Copeland, Oct 03 2011
Hilbert series of the pre-WDVV ring, thus h-vectors of the Whitehouse simplicial complex (cf. Readdy, Table 1). - Tom Copeland, Sep 20 2014
Arises in Buckholtz's analysis of the error term in the series for exp(nz). - N. J. A. Sloane, Jul 05 2016

Examples

			Triangle begins:
  1;
  1,   2;
  1,   8,   6;
  1,  22,  58,  24;
  1,  52, 328, 444, 120; ...
Row 3: There are three plane increasing 0-1-2-3 trees on 3 vertices. The number of colors are shown to the right of a vertex.
.
    1o (2*t+1)         1o t*(t+2)      1o t*(t+2)
     |                 / \             / \
     |                /   \           /   \
    2o (2*t+1)      2o    3o        3o    2o
     |
     |
    3o
.
The total number of trees is (2*t+1)^2 + t*(t+2) + t*(t+2) = 1 + 8*t + 6*t^2.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270. [with offsets [0,0]: see A201637]

Crossrefs

Columns include A005803, A004301, A006260.
Right-hand columns include A000142, A002538, A002539.
Row sums are A001147.
For a (0,0) based version as used in 'Concrete Mathematics' and by Maple see A201637. For a (0,0) based version which has this triangle as a subtriangle see A340556.

Programs

  • Maple
    with(combinat): A008517 := proc(n, m) local k: add((-1)^(n+k)* binomial(2*n+1, k)* stirling1(2*n-m-k+1, n-m-k+1), k=0..n-m) end: seq(seq(A008517(n, m), m=1..n), n=1..8);
    # Johannes W. Meijer, Oct 16 2009, revised Nov 22 2012
    A008517 := proc(n,k) option remember; `if`(n=1,`if`(k=0,1,0), A008517(n-1,k)* (k+1) + A008517(n-1,k-1)*(2*n-k-1)) end: seq(print(seq(A008517(n,k), k=0..n-1)), n=1..9);
    # Peter Luschny, Apr 20 2011
  • Mathematica
    a[n_, m_] = Sum[(-1)^(n + k)*Binomial[2 n + 1, k]*StirlingS1[2n-m-k+1, n-m-k+1], {k, 0, n-m}]; Flatten[Table[a[n, m], {n, 1, 9}, {m, 1, n}]][[1 ;; 44]] (* Jean-François Alcover, May 18 2011, after Johannes W. Meijer *)
  • PARI
    {T(n, k) = my(z); if( n<1, 0, z = 1 + O(x); for( k=1, n, z = 1 + intformal( z^2 * (z+y-1))); n! * polcoeff( polcoeff(z, n),k))}; /* Michael Somos, Oct 13 2002 */
    
  • PARI
    {T(n,k)=polcoeff((1-x)^(2*n+1)*sum(j=0,2*n+1,j^(n+j)*x^j/j!*exp(-j*x +x*O(x^k))),k)} \\ Paul D. Hanna, Oct 31 2012
    for(n=1,10,for(k=1,n,print1(T(n,k),", "));print(""))
    
  • PARI
    T(n, m) = sum(k=0, n-m, (-1)^(n+k)*binomial(2*n+1, k)*stirling(2*n-m-k+1, n-m-k+1, 1)); \\ Michel Marcus, Dec 07 2021
    
  • Sage
    @CachedFunction
    def A008517(n, k):
        if n==1: return 1 if k==0 else 0
        return A008517(n-1,k)*(k+1)+A008517(n-1,k-1)*(2*n-k-1)
    for n in (1..9): [A008517(n, k) for k in(0..n-1)] # Peter Luschny, Oct 31 2012

Formula

T(n,k) = 0 if n < k, T(1,1) = 1, T(n,-1) = 0, T(n,k) = k*T(n-1,k) + (2*n-k)*T(n-1,k-1).
a(n,m) = Sum_{k=0..n-m} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-m-k+1, n-m-k+1). - Johannes W. Meijer, Oct 16 2009
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 = 1 and for the Eulerian numbers A008292 when k = 0.
Let v = -t*exp((1-t)^2*x-t) and let B(x,t) = -(1+1/t*LambertW(v))/(1+LambertW(v)). From the e.g.f. given by Copeland above we find B(x,t) = compositional inverse with respect to x of G(1,x,t) = Sum_{n>=1} R(n,t)*x^n/n! = x+(1+2*t)*x^2/2!+(1+8*t+6*t^2)*x^3/3!+.... The function B(x,t) satisfies the differential equation dB/dx = (1+B)*(1+t*B)^2 = 1 + (2*t+1)*B + t*(t+2)*B^2 + t^2*B^3.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees where each vertex has outdegree <= 3, the vertices of outdegree 1 come in 2*t+1 colors, the vertices of outdegree 2 come in t*(t+2) colors and the vertices of outdegree 3 come in t^2 colors. An example is given below. Cf. A008292. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x,t) = (1+x)*(1+t*x)^2 and let D be the operator f(x,t)*d/dx. Then R(n+1,t) = D^n(f(x,t)) evaluated at x = 0. (End)
From Tom Copeland, Oct 03 2011: (Start)
a(n,k) = Sum_{i=0..k} (-1)^(k-i) binomial(n-i,k-i) A134991(n,i), offsets 0.
P(n+1,t) = (1-t)^(2n+1) Sum_{k>=1} k^(n+k) [t*exp(-t)]^k / k! for n>0; consequently, Sum_{k>=1} (-1)^k k^(n+k) x^k/k!= [1+LW(x)]^(-(2n+1))P[n+1,-LW(x)] where LW(x) is the Lambert W-Function and P(n,t), for n > 0, are the row polynomials as given in Copeland's 2008 comment. (End)
The e.g.f. A(x,t) = -v * { Sum_{j>=1} D(j-1,u) (-z)^j / j! } where u=x*(1-t)^2-t, v=(1+u)/(1-t), z=(t+u)/[(1+u)^2] and D(j-1,u) are the polynomials of A042977. dA(x,t)/dx=(1-t)/[1+u-(1-t)A(x,t)]=(1-t)/{1+LW[-t exp(u)]}, (Copeland's e.g.f. in 2008 comment). - Tom Copeland, Oct 06 2011
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=-t, and (a_n)=-P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 = 0. - Tom Copeland, Oct 08 2011
The compositional inverse (with respect to x) of y = y(t;x) = (x-t*(exp(x)-1)) is 1/(1-t)*y + t/(1-t)^3*y^2/2! + (t+2*t^2)/(1-t)^5*y^3/3! + (t+8*t^2+6*t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed in the Comments section, the rational functions in t are the generating functions for the diagonals of the triangle of Stirling numbers of the second kind (A048993). See the Bala link for a proof. Cf. A112007 and A134991. - Peter Bala, Dec 04 2011
O.g.f. of row n: (1-x)^(2*n+1) * Sum_{k>=0} k^(n+k) * exp(-k*x) * x^k/k!. - Paul D. Hanna, Oct 31 2012
T(n, k) = n!*[x^n][t^k](egf) where egf = (1-t)/(1 + LambertW(-exp(t^2*x - 2*t*x - t + x)*t)) and after expansion W(-exp(-t)t) is substituted by (-t). - Shamil Shakirov, Feb 17 2025

A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700, 1026576, 362880
Offset: 1

Views

Author

André F. Labossière, May 17 2004

Keywords

Comments

Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008
Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019
T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006
Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006
Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007
From Tom Copeland, Dec 15 2007: (Start)
Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).
Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.
Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.
Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.
Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.
Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.
Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.
The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).
The above results hold for t any real or complex number. (End)
Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*Stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*Stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008
T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008
From Tom Copeland, Nov 02 2010: (Start)
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.
E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].
(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)
With offset 0, 0 <= k <= n: T(n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011
The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011
T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011
Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011
Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014
From Tom Copeland, Dec 09 2016: (Start)
The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.
Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.
(End)
From Tom Copeland, Dec 16 2019: (Start)
The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.
Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.
The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)

Examples

			Triangle starts:
  1;
  1,  1;
  1,  3,  2;
  1,  6, 11,  6;
  1, 10, 35, 50, 24;
...
		

References

  • M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 18, equations 18:4:2 - 18:4:8 at page 151.
  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

Crossrefs

A008276 gives the (signed) Stirling numbers of the first kind.
Cf. A000108, A014137, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A000217 (2nd column), A000914 (3rd column), A001303 (4th column), A000915 (5th column), A053567 (6th column), A000142 (row sums).
Triangle sums (see the comments): A124380 (Kn11), A001710 (Fi1, Fi2). - Johannes W. Meijer, Apr 20 2011

Programs

  • GAP
    Flat(List([1..10], n-> List([1..n], k-> Stirling1(n,n-k+1) ))); # G. C. Greubel, Dec 29 2019
  • Haskell
    a094638 n k = a094638_tabl !! (n-1) !! (k-1)
    a094638_row n = a094638_tabl !! (n-1)
    a094638_tabl = map reverse a130534_tabl
    -- Reinhard Zumkeller, Aug 01 2014
    
  • Magma
    [(-1)^(k+1)*StirlingFirst(n,n-k+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019
    
  • Maple
    T:=(n,k)->abs(Stirling1(n,n+1-k)): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form. # Emeric Deutsch, Aug 14 2006
  • Mathematica
    Table[CoefficientList[Series[Product[1 + i x, {i,n}], {x,0,20}], x], {n,0,6}] (* Geoffrey Critzer, Feb 04 2011 *)
    Table[Abs@StirlingS1[n, n-k+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)
  • Maxima
    create_list(abs(stirling1(n+1,n-k+1)),n,0,10,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    {T(n,k)=if(n<1 || k>n,0,(n-1)!*polcoeff(polcoeff(x*y/(1 - x*y+x*O(x^n))^(1 + 1/y),n,x),k,y))} /* Paul D. Hanna, Jul 21 2011 */
    
  • Sage
    [[stirling_number1(n, n-k+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019
    

Formula

With P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0. (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t=m-1. See Bergeron et al. in "Varieties of Increasing Trees". - Tom Copeland, Dec 09 2007
First comment and formula above rephrased as o.g.f. for row n: Product_{i=0...n} (1+i*x). - Geoffrey Critzer, Feb 04 2011
n-th row polynomials with alternate signs are the characteristic polynomials of the (n-1)x(n-1) matrices with 1's in the superdiagonal, (1,2,3,...) in the main diagonal, and the rest zeros. For example, the characteristic polynomial of [1,1,0; 0,2,1; 0,0,3] is x^3 - 6*x^2 + 11*x - 6. - Gary W. Adamson, Jun 28 2011
E.g.f.: A(x,y) = x*y/(1 - x*y)^(1 + 1/y) = Sum_{n>=1, k=1..n} T(n,k)*x^n*y^k/(n-1)!. - Paul D. Hanna, Jul 21 2011
With F(x,t) = (1-t*x)^(-1/t) - 1 an e.g.f. for the row polynomials P(n,t) of A094638 with P(0,t)=0, G(x,t)= [1-(1+x)^(-t)]/t is the comp. inverse in x. Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+x)^(t+1),
P(n,t) = [(H(x,t)*d/dx)^n] x evaluated at x=0; i.e.,
F(x,t) = exp[x*P(.,t)] = exp[x*H(u,t)*d/du] u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 20 2011
T(n,k) = |A008276(n,k)|. - R. J. Mathar, May 19 2016
The row polynomials of this entry are the reversed row polynomials of A143491 multiplied by (1+x). E.g., (1+x)(1 + 5x + 6x^2) = (1 + 6x + 11x^2 + 6x^3). - Tom Copeland, Dec 11 2016
Regarding the row e.g.f.s in Copeland's 2007 formulas, e.g.f.s for A001710, A001715, and A001720 give the compositional inverses of the e.g.f. here for t = 2, 3, and 4 respectively. - Tom Copeland, Dec 28 2019

Extensions

Edited by Emeric Deutsch, Aug 14 2006

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

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1
		

Crossrefs

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
    a094587_row n = a094587_tabl !! n
    a094587_tabl = map fst $ iterate f ([1], 1)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
    # Alternative: Note that if you leave out 'abs' you get A021009.
    T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
    for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011
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