

A094638


Triangle read by rows: T(n,k) = s(n,n+1k), 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).


61



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
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OFFSET

1,5


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,...,n1. 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 columnconvex 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
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(n1,k1).  Philippe Deléham, Jan 06 2007
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)] = (1xt)^(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+12*k)*x^(n+12*k), I_n(x) = Sum_{k=0..floor(n/2)}(1)^(k+1)*Stirling1(n+1,n2*k)*x^(n2*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
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n1) + ... + T(n,n)y^(n1)(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+n1)!/(x1)!. 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
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)
The e.g.f. given by Copeland in the formula section appears in a combinatorial DysonSchwinger 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+n1)!/(xD1)! = 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 = (kn) L_(n+k), as for the dual Hopf algebra. (End)


REFERENCES

M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109116 (MR1768839).
R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.


LINKS

F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.C. Raoult, Springer 1992, pp. 2448. Added Mar 01 2014


FORMULA

With P(n,t) = Sum_{k=0..n1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1tx)^(1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1tx)^(1/t)  1 ] evaluated at t=x=0. (1tx)^(1/t)  1 is the e.g.f. for a plane mary tree when t=m1. 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
nth row polynomials with alternate signs are the characteristic polynomials of the (n1)x(n1) 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/(n1)!.  Paul D. Hanna, Jul 21 2011
With F(x,t) = (1t*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.
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


EXAMPLE

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


MAPLE

T:=(n, k)>abs(Stirling1(n, n+1k)): 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, nk+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)


PROG

(PARI) {T(n, k)=if(n<1  k>n, 0, (n1)!*polcoeff(polcoeff(x*y/(1  x*y+x*O(x^n))^(1 + 1/y), n, x), k, y))} /* Paul D. Hanna, Jul 21 2011 */
(Maxima) create_list(abs(stirling1(n+1, nk+1)), n, 0, 10, k, 0, n); / * Emanuele Munarini, Jun 01 2012 */
(Haskell)
a094638 n k = a094638_tabl !! (n1) !! (k1)
a094638_row n = a094638_tabl !! (n1)
a094638_tabl = map reverse a130534_tabl
(Magma) [(1)^(k+1)*StirlingFirst(n, nk+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019
(Sage) [[stirling_number1(n, nk+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019
(GAP) Flat(List([1..10], n> List([1..n], k> Stirling1(n, nk+1) ))); # G. C. Greubel, Dec 29 2019


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).


KEYWORD



AUTHOR



EXTENSIONS



STATUS

approved



