A130534 -id:A130534 - cp's OEIS frontend

A164652 Triangle read by rows: Hultman numbers: a(n,k) is the number of permutations of n elements whose cycle graph (as defined by Bafna and Pevzner) contains k cycles for n >= 0 and 1 <= k <= n+1.

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 8, 0, 15, 0, 1, 0, 84, 0, 35, 0, 1, 180, 0, 469, 0, 70, 0, 1, 0, 3044, 0, 1869, 0, 126, 0, 1, 8064, 0, 26060, 0, 5985, 0, 210, 0, 1, 0, 193248, 0, 152900, 0, 16401, 0, 330, 0, 1, 604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1, 0, 19056960, 0, 18128396, 0, 2641925, 0, 88803, 0, 715, 0, 1

Offset: 0

Anthony Labarre, Aug 19 2009

a(n,k) is also the number of ways to express a given (n+1)-cycle as the product of an (n+1)-cycle and a permutation with k cycles (see Doignon and Labarre). a(n,n+1-2k) is the number of permutations of n elements whose block-interchange distance is k (see Christie, Doignon and Labarre).
Named after the Swedish mathematician Axel Hultman. - Amiram Eldar, Jun 11 2021
a(2*n,1) is the number of spanning trees in certain graphs with 2*n+1 vertices and n*(n+1) edges (see Ishikawa, Miezaki, and Tanaka). - Tsuyoshi Miezaki, Feb 08 2023

			Triangle begins:
  n=0:  1;
  n=1:  0, 1;
  n=2:  1, 0, 1;
  n=3:  0, 5, 0, 1;
  n=4:  8, 0, 15, 0, 1;
  n=5:  0, 84, 0, 35, 0, 1;
  n=6:  180, 0, 469, 0, 70, 0, 1;
  n=7:  0, 3044, 0, 1869, 0, 126, 0, 1;
  n=8:  8064, 0, 26060, 0, 5985, 0, 210, 0, 1;
  n=9:  0, 193248, 0, 152900, 0, 16401, 0, 330, 0, 1;
  n=10: 604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1;
  ...
From _Jon E. Schoenfield_, May 20 2023: (Start)
As a right-aligned triangle:
                                                      1; n=0
                                                   0, 1; n=1
                                                1, 0, 1; n=2
                                           0,   5, 0, 1; n=3
                                        8, 0,  15, 0, 1; n=4
                                 0,    84, 0,  35, 0, 1; n=5
                            180, 0,   469, 0,  70, 0, 1; n=6
                      0,   3044, 0,  1869, 0, 126, 0, 1; n=7
                8064, 0,  26060, 0,  5985, 0, 210, 0, 1; n=8
          0,  193248, 0, 152900, 0, 16401, 0, 330, 0, 1; n=9
  604800, 0, 2286636, 0, 696905, 0, 39963, 0, 495, 0, 1; n=10
  ...
(End)

Axel Hultman, Toric permutations, Master's thesis, Department of Mathematics, KTH, Stockholm, Sweden, 1999.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Nikita Alexeev, Anna Pologova, and Max A. Alekseyev, Generalized Hultman Numbers and Cycle Structures of Breakpoint Graphs, Journal of Computational Biology, Vol. 24, No. 2 (2017), pp. 93-105; arXiv preprint, arXiv:1503.05285 [q-bio.GN], 2015-2017.
Nikita Alexeev and Peter Zograf, Hultman numbers, polygon gluings and matrix integrals, arXiv preprint arXiv:1111.3061 [math.PR], 2011.
Nikita Alexeev and Peter Zograf, Random matrix approach to the distribution of genomic distance, Journal of Computational Biology, Vol. 21, No. 8 (2014), pp. 622-631.
Miklos Bona and Ryan Flynn, The Average Number of Block Interchanges Needed to Sort A Permutation and a recent result of Stanley, arXiv:0811.0740 [math.CO], 2008.
Miklos Bona and Ryan Flynn, The Average Number of Block Interchanges Needed to Sort A Permutation and a recent result of Stanley, Inf. Process. Lett., Vol. 109 (2009), pp. 927-931.
David A. Christie, Sorting Permutations by Block-Interchanges, Inf. Process. Lett., Vol. 60, No. 4 (1996), pp. 165-169.
Robert Cori and Gábor Hetyei, On reduced unicellular hypermonopoles, arXiv:2403.19569 [math.CO], 2024. See at p. 4.
Jean-Paul Doignon and Anthony Labarre, On Hultman Numbers, J. Integer Seq., Vol. 10 (2007), Article 07.6.2, 13 pages.
Simona Grusea and Anthony Labarre, The distribution of cycles in breakpoint graphs of signed permutations, arXiv:1104.3353 [cs.DM], 2011-2012.
Reina Ishikawa, Tsuyoshi Miezaki, and Yuuho Tanaka, A new interpretation of Hultman numbers.

Cf. A000142 (row sums), A000217, A060593, A128174, A130534, A132393, A185259, A189507, A260695.

Cf. A185263 (rows reversed without 0's).

a164652 n k = a164652_tabl !! n !! k
a164652_row n = a164652_tabl !! n
a164652_tabl = [0] : tail (zipWith (zipWith (*)) a128174_tabl $
   zipWith (map . flip div) (tail a000217_list) (map init $ tail a130534_tabl))
-- Reinhard Zumkeller, Aug 01 2014

A164652:= (n, k)-> `if`(n-k mod 2 = 1, -Stirling1(n+2, k)/binomial(n+2, 2), 0):
for n from 0 to 7 do seq(A164652(n,k),k=1..n+1) od; # Peter Luschny, Mar 22 2015

T[n_, k_] := If[OddQ[n-k], Abs[StirlingS1[n+2, k]]/Binomial[n+2, 2], 0];
Table[T[n, k], {n, 0, 11}, {k, 1, n+1}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

T(n,k)= my(s=(n-k)%2); (-1)^s*s*stirling(n+2,k,1)/binomial(n+2,2);
concat(vector(12, n, vector(n, k, T(n-1,k)))) \\ Gheorghe Coserea, Jan 23 2018

def A164652(n, k):
    return stirling_number1(n+2,k)/binomial(n+2,2) if is_odd(n-k) else 0
for n in (0..7): print([A164652(n,k) for k in (1..n+1)]) # Peter Luschny, Mar 22 2015

T(n,k) = S1(n+2,k)/C(n+2,2) if n-k is odd, and 0 otherwise. Here S1(n,k) are the unsigned Stirling numbers of the first kind A132393 and C(n,k) is the binomial coefficient (see Bona and Flynn).
For n > 0: T(n,k) = A128174(n+1,k) * A130534(n+1,k-1) / A000217(n+1). - Reinhard Zumkeller, Aug 01 2014
n-th row polynomial R(n,x) = (x/2)*( P(n+1,x) + (-1)^n * P(n+1,-x) ) / binomial(n+2,2), where P(k,x) = (x + 1)*(x + 2)*...*(x + k). - Peter Bala, May 14 2023

T(0,1) set to 1 by Peter Luschny, Mar 24 2015

Edited to match values of k to the range 1 to n+1. - Max Alekseyev, Nov 20 2020

A051431 a(n) = (n+10)!/10!.

Original entry on oeis.org

1, 11, 132, 1716, 24024, 360360, 5765760, 98017920, 1764322560, 33522128640, 670442572800, 14079294028800, 309744468633600, 7124122778572800, 170978946685747200, 4274473667143680000, 111136315345735680000, 3000680514334863360000, 84019054401376174080000

Offset: 0

Wolfdieter Lang

The p=10 member of the p-family of sequences {(n+p-1)!/p!}, n >= 1.

The asymptotic expansion of the higher-order exponential integral E(x,m=1,n=11) ~ exp(-x)/x*(1 - 11/x + 132/x^2 - 1716/x^3 + 24024/x^4 - 360360/x^5 + 5765760/x^6 - ...) leads to the sequence given above. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A000142, A001710, A001715, A001720, A001725, A001730, A049388, A049389, A049398.

Cf. A130534, A163931, A173333, A245334.

a051431 = (flip div 3628800) . a000142 . (+ 10)
-- Reinhard Zumkeller, Aug 31 2014

[Factorial(n+10)/3628800: n in [0..25]]; // Vincenzo Librandi, Jul 20 2011

a[n_] := (n + 10)!/10!; Array[a, 20, 0] (* Amiram Eldar, Jan 15 2023 *)

a(n) = (n+10)!/10!
E.g.f.: 1/(1-x)^11.
a(n) = A173333(n+10,10). - Reinhard Zumkeller, Feb 19 2010
a(n) = A245334(n+10,n) / 11. - Reinhard Zumkeller, Aug 31 2014
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=0} 1/a(n) = 3628800*e - 9864100.
Sum_{n>=0} (-1)^n/a(n) = 3628800/e - 1334960. (End)

A139605 Weights for expansion of iterated derivatives, powers of a Lie derivative, or infinitesimal generator in vector form, (f(x)D_x)^n; coefficients of A-polynomials of Comtet; Scherk partition polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 4, 7, 6, 1, 1, 7, 4, 11, 1, 5, 30, 15, 10, 25, 10, 1, 1, 11, 15, 32, 34, 26, 1, 6, 57, 34, 146, 31, 15, 120, 90, 20, 65, 15, 1, 1, 16, 26, 15, 76, 192, 34, 122, 180, 57, 1, 7, 98, 140, 406, 462, 588, 63, 21, 252, 154, 896, 301

Offset: 1

Tom Copeland, Jun 12 2008

This entry and the references differ slightly among themselves in the order of coefficients for higher order terms. Table on p. 167 of Comtet has at least one index error.
Let F[FI(x)] = FI[F(x)] = x (i.e., F and FI are a compositional inverse pair) about x=0 with F(0)=FI(0)=0. Define f(x) = 1/[dFI(x)/dx], then for w(x) analytic about the origin, exp[t f(x)d/dx] w(x) = w{F[t+FI(x)]} = q(t,x) with q{t,F[s+FI(x)]} = q(t+s,x). See A145271 for w(x)=x and note that A145271 is embedded in A139605. E.g.f. of the binomial Sheffer sequence associated to F(x) is exp[x f(z)d/dz] exp(t*z)= exp{t*F[x+FI(z)]} evaluated at z=0. - Tom Copeland, Oct 19 2011
dq(t,x)/dt - f(x)dq(t,x)/dx = 0, so (1,-f(x)) gives the components of a vector orthogonal to the gradient of q and therefore tangent to the contour of q at (t,x). - Tom Copeland, Oct 26 2011
The formula exp[t f(x)d/dx] w(x)= w{F[t+FI(x)]} above is implicit in the chain rule formulas on pages 10 and 12 of Mathemagical Forests. Another derivation is alluded to in the Dattoli reference in A080635 (repeated below). - Tom Copeland, Nov 28 2011
Let f(x) and g(x) be two infinitely differential functions. Denote f_0 = f(x), f_1 = df_0/dx, f_2 = df_1/dx, and so on. Same with g_0 = g(x). Define the linear operator L(u(x)) = g(x) * du(x)/dx. Denote F_1 = L(f(x)), F_2 = L(F_1), and so on. When n>0, F_n is a linear combination of f_1, ..., f_n where each f_k is multiplied by a homogeneous polynomial (P(n,k)) of degree n in g_0, ..., g_{n-1}. The triangle of the sum of the coefficients of P(n,k) is A130534. - Michael Somos, Mar 23 2014
Triangle with row n length A000070(n+1) and row n consists of the coefficients:  P(n,1), ..., P(n,n). The order of coefficients in P(n,k) is Abramowitz and Stegun order for partitions of n-k with parts g_1, ..., g_{n-k}. - Michael Somos, Mar 23 2014
A130534(n,k) gives the number of rooted trees with (k+1) trunks that are associated with D^(k+1) in the forest of "naturally grown" rooted trees with (n+2) nodes, or vertices, that are associated with R^(n+1) in the example below. Cf. MF link. - Tom Copeland, Mar 23 2014
These partition polynomials appeared in 1823 in a dissertation by Heinrich Scherk. See p. 76 of Blasiak and Flajolet. - Tom Copeland, Jul 14 2021
Schröder made use of iterated derivatives, or iterated infinitesimal generators (IGs), ((1/f') D)^n in his investigations of functional iteration, or iterated functional composition, related to extensions of Newton's method of finding zeros of an equation. He constructs the series, in terms of the IGs, for finv[t+f(z)] evaluated at t = -f(z), giving z_1 = finv(0) although he doesn't present his analysis this way. - Tom Copeland, Jul 19 2021

			Let R = f(x)d/dx = f(x)D and (j,k) = [(d/dx)^j f(x)]^k ; then
R^0  = 1.
R^1  = (0,1)D.
R^2  = (0,1)(1,1)D + (0,2)D^2.
R^3  = [(0,1)(1,2) + (0,2)(2,1)]D + 3 (0,2)(1,1)D^2 + (0,3)D^3.
R^4  = [(0,1)(1,3) + 4 (0,2)(1,1)(2,1) + (0,3)(3,1)]D +
       [7 (0,2)(1,2) + 4 (0,3)(2,1)]D^2 + 6 (0,3)(1,1)D^3 + (0,4)D^4. - _Tom Copeland_, Jun 12 2008
R^5  = [(0,1)(1,4) + 11 (0,2)(1,2)(2,2) + 4 (0,3)(2,2) + (0,4)(4,1) + 7 (0,3)(1,1)(3,1)]D + [15 (0,2)(1,3) + 30 (0,3)(1,1)(2,1) + 5 (0,4)(1,3)]D^2 + [25 (0,3)(1,2) + 10 (0,4)(2,1) + 25 (0,3)(1,2)]D^3  + 10 (0,4)(1,1)D^4 + (0,5)D^5. - _Tom Copeland_, Jul 17 2016
R^6  = [(0,1)(1,5) + 26 (0,2)(1,3)(2,1) + 34 (0,3)(1,1)(2,2) + 32 (0,3)(1,2)(3,1) + 11 (0,4)(1,1)(4,1) + 15 (0,4)(2,1)(3,1) + (0,5)(1,5)]D + [31 (0,2)(1,4) + 146 (0,3)(1,2)(2,1) + 57 (0,4)(1,1)(3,1) + 34 (0,4)(2,2) + 6 (0,5)(4,1)]D^2 + [90 (0,3)(1,3) + 120 (0,4)(1,1)(2,1) + 15 (0,5)(3,1)]D^3 + [65 (0,4)(1,2) + 20 (0,5)(2,1)]D^4 + 15 (0,5)(1,1)D^5 + (0,6)D^6. - _Tom Copeland_, Jul 17 2016
------------
F_1 = (1*g_0) * f_1, F_2 = (1*g_0*g_1) * f_1 + (1*g_0^2) * f_2, F_3 = (1*g_0*g_1^2 + 1*g_0^2*g_2) * f_1 + (3*g_0^2*g_1) * f_2 + (1*g_0^3) * f_3. - _Michael Somos_, Mar 23 2014
P(4,2) = 4*g0^3*g2 + 7*g0^2*g1^2. P(5,2) = 5*g0^4*g3 + 30*g0^3*g1*g2 + 15*g0^2*g1^3. - _Michael Somos_, Mar 23 2014
1
1 , 1
1 1 , 3 , 1
1 4 1 , 4 7 , 6 , 1
1 7 4 11 1, 5 30 15 , 10 25 , 10 , 1
1 11 15 32 34 26 1 , 6 57 34 146 31 , 15 120 90 , 20 65 , 15 , 1

F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-like Structures, (1997), Cambridge University Press, p. 386.
H. Davis, The Theory of Linear Operators, (1936), The Principia Press, p. 13.
T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

P. Blasiak and P. Flajolet, Combinatorial Models of Creation-Annihilation, arXiv:1010.0354 [math.CO], 2011.
Louis Comtet, Une formule explicite pour les puissances successives de l'opérateur de dérivation de Lie, Comptes Rendus Acad. Sc. Paris, Serie A tome 276 (1973), pp. 165 - 168.
Tom Copeland, Mathemagical Forests v2, 2008.
Tom Copeland, Infinigens, the Pascal Triangle, and the Witt and Virasoro Algebras, 2012.
Tom Copeland, Pre-Lie algebras, Cayley's analytic trees, and mathemagical forests, 2018.
G. Datolli, P. L. Ottaviani, A. Torre and L. Vazquez, Evolution operator equations: integration with algebraic and finite differences methods.[...], La Rivista del Nuovo Cimento 20,2 (1997) 1-133. eq. (I.2.18).
D. Grinberg, Commutators, matrices, and an identity of Copeland, arXiv:1908.09179 [math.RA], 2019.
Guo-Niu Han and Shi-mei Ma, Eulerian polynomials and the g-indices of Young tableaux, Proc. Amer. Math. Soc., (2023).
Kentaro Ihara, Derivations and automorphisms on non-commutative power series, Journal of Pure and Applied Algebra, Volume 216, Issue 1, January 2012, Pages 192-201.
T. Mansour and M. Shork, The generalized Touchard polynomials revisited, Journal of Applied Mathematics and Computation, Volume 219, Issue 19, June 2013, Pages 9978-9991.
MathOverflow, Important formulas in combinatorics: The combinatorics underlying iterated derivatives (infinitesimal Lie generators) for compositional inversion and flow maps for vector fields, answer by Tom Copeland to an MO question posed by Gil Kalai, 2015.
MathOverflow, Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation, a MO question posed by Tom Copeland, answered by Darij Grinberg, 2019.
Mathoverflow, Formula for n-th iteration of dx/dt=B(x), a question on MathOverflow posed by the user resolvent and answered by Tom Copeland, 2021.
H. Scherk, De evolvenda functione (yd.yd.yd...ydX/dxn) disquisitiones nonnullae analyticae, PhD thesis, Berlin, 1823. Scanned copy at Göttinger Digitalisierungszentrum (GDZ).
E. Schröder, Ueber unendlich viele Algorithmen zur Auflösung der Gleichungen, Mathematische Annalen vol. 2, 317-365, 1870.
G. Stewart, On infinitely many algorithms for solving equations, 1993, (translation into English of Schröder's paper above).

Cf. A000070 (number of distinct terms for each order).
Cf. A130534 (sum of numerical coefficients of the derivatives).
Cf. A008275, A048994, A080635, A124796, A129185, A139002, A145271.

row[n_] := With[{pn = CoefficientRules[Nest[g[x] D[#, x] &, f[x], n], Derivative[#][f][x] & /@ Range[n]][[;; , 2]] /. {Derivative[k_][g][x] :> h[k], g[x] -> 1}}, Table[Coefficient[pn[[k]], Product[h[x], {x, p}]], {k, n - 1}, {p, Sort[Sort /@ IntegerPartitions[n - k]]}]~Join~{{1}}];
Table[row[n], {n, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by f_n = (d/dx)^n f(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) f_(n-k). Then R^n = (1, 0, 0, 0,..) [VP * S]^n (1, D, D^2, ..)^T, where S is the shift matrix A129185, representing differentiation in the basis x^n/n!. Cf. A145271. - Tom Copeland, Jul 17 2016
A formula for the coefficients of this matrix is presented in Ihara, obtained from Comtet. - Tom Copeland, Mar 25 2020
Elaborating on my 2011 comments: Let exp[x F(t)] = exp[t p.(x)] be the e.g.f. for the binomial Sheffer sequence of polynomials (p.(x))^n = p_n(x). Then, evaluated at x = t = 0, the coefficient p_(n,k) = (D_x^k/k!) p_n(x) = D_t^n [F(t)]^k/k! = (f(x)D_x)^n x^k/k! = R^n x^k/k!, and so p_(n,k) is the coefficient of D^k of the operator R^n below evaluated at x=0. - Tom Copeland, May 14 2020
Per earlier comments, the action of the differentials of this entry on an exponential is exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)}(x) the compositional inverse of H(x). With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind (see A008275, A048994, and A130534). Consequently, summing the coefficients of this entry over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+4+1) D + (7+4) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. The Connes-Moscovici weights A139002 are a refinement of this entry. - Tom Copeland, Jul 14 2021

Title expanded by Tom Copeland, Mar 17 2014
Sequence terms rearranged in Abramowitz and Stegun order by Michael Somos, Mar 23 2014
Title expanded by Tom Copeland, Jul 14 2021

A163934 Triangle related to the asymptotic expansion of E(x,m=4,n).

Original entry on oeis.org

1, 6, 4, 35, 40, 10, 225, 340, 150, 20, 1624, 2940, 1750, 420, 35, 13132, 27076, 19600, 6440, 980, 56, 118124, 269136, 224490, 90720, 19110, 2016, 84, 1172700, 2894720, 2693250, 1265460, 330750, 48720, 3780, 120

Offset: 1

Johannes W. Meijer, Aug 13 2009

The higher order exponential integrals E(x,m,n) are defined in A163931 while the general formula for their asymptotic expansion can be found in A163932.
We used the latter formula and the asymptotic expansion of E(x,m=3,n), see A163932, to determine that E(x,m=4,n) ~ (exp(-x)/x^4)*(1 - (6+4*n)/x + (35+40*n+ 10*n^2)/x^2 - (225+340*n+ 150*n^2+20*n^3)/x^3 + ... ). This formula leads to the triangle coefficients given above.
The asymptotic expansion leads for the values of n from one to five to known sequences, see the cross-references.
The numerators of the o.g.f.s. of the right hand columns of this triangle lead for z=1 to A000457, see A163939 for more information.
The first Maple program generates the sequence given above and the second program generates the asymptotic expansion of E(x,m=4,n).

			The first few rows of the triangle are:
1;
6, 4;
35, 40, 10;
225, 340, 150, 20;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A163931 (E(x,m,n)), A163932 and A163939.
Cf. A048994 (Stirling1), A000454 (row sums).
A000399, 4*A000454, 10*A000482, 20*A001233, 35*A001234 equal the first five left hand columns.
A000292, A027777 and A163935 equal the first three right hand columns.
The asymptotic expansion leads to A000454 (n=1), A001707 (n=2), A001713 (n=3), A001718 (n=4) and A001723 (n=5).
Cf. A130534 (m=1), A028421 (m=2), A163932 (m=3).

with(combinat): A163934 := proc(n,m): (-1)^(n+m)* binomial(m+2, 3) *stirling1(n+2, m+2) end: seq(seq(A163934(n,m), m=1..n), n=1..8);
with(combinat): imax:=6; EA:=proc(x,m,n) local E, i; E:=0: for i from m-1 to imax+2 do E:=E + sum((-1)^(m+k+1)*binomial(k,m-1)*n^(k-m+1)* stirling1(i, k), k=m-1..i)/x^(i-m+1) od: E:= exp(-x)/x^(m)*E: return(E); end: EA(x,4,n);
# Maple programs revised by Johannes W. Meijer, Sep 11 2012

a[n_, m_] /; n >= 1 && 1 <= m <= n = (-1)^(n+m)*Binomial[m+2, 3] * StirlingS1[n+2, m+2]; Flatten[Table[a[n, m], {n, 1, 8}, {m, 1, n}]][[1 ;; 36]] (* Jean-François Alcover, Jun 01 2011, after formula *)

a(n,m) = (-1)^(n+m)*C(m+2,3)*stirling1(n+2,m+2) for n >= 1 and 1<= m <= n.

A094645 Triangle of generalized Stirling numbers of the first kind.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 0, -1, 0, 1, 0, -2, -1, 2, 1, 0, -6, -5, 5, 5, 1, 0, -24, -26, 15, 25, 9, 1, 0, -120, -154, 49, 140, 70, 14, 1, 0, -720, -1044, 140, 889, 560, 154, 20, 1, 0, -5040, -8028, -64, 6363, 4809, 1638, 294, 27, 1, 0, -40320, -69264, -8540, 50840, 44835, 17913, 3990, 510, 35, 1

Offset: 0

Vladeta Jovovic, May 17 2004

From Wolfdieter Lang, Jun 20 2011: (Start)
The row polynomials s(n,x) := Sum_{j=0..n} T(n,k)*x^k satisfy risefac(x-1,n) = s(n,x), with the rising factorials risefac(x-1,n) := Product_{j=0..n-1} (x-1+j), n >= 1, risefac(x-1,0) = 1. Compare with the formula risefac(x,n) = s1(n,x), with the row polynomials s1(n,x) of A132393 (unsigned Stirling1).
This is the lower triangular Sheffer array with e.g.f.
  T(x,z) = (1-z)*exp(-x*log(1-z)) (the rewritten e.g.f. from the formula section). See the W. Lang link under A006232 for Sheffer matrices and the Roman reference. In the notation which indicates the column e.g.f.s this is Sheffer (1-z,-log(1-z)). In the umbral notation (cf. Roman) this is called Sheffer for (exp(t),1-exp(-t)).
The row polynomials satisfy s(n,x) = (x+n-1)*s(n-1,x), s(0,x)=1, and s(n,x) = (x-1)*s1(n-1,x), n >= 1, s1(0,x) = 1, with the unsigned Stirling1 row polynomials s1(n,x).
The row polynomials also satisfy
  s(n,x) - s(n,x-1) = n*s(n-1,x), n > 1, s(0,x) = 1
  (from the Meixner identity, see the Meixner reference given at A060338).
The row polynomials satisfy as well (from corollary 3.7.2. p. 50 of the Roman reference)
  s(n,x) = (x-1)*s(n-1,x+1), n >= 1, s(0,n) = 1.
The exponential convolution identity is
  s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,y)*s1(n-k,x),
  n >= 0, with symmetry x <-> y.
The row sums are 1 for n=0 and 0 otherwise, and the alternating row sums are 1,-2,2, followed by zeros, with e.g.f. (1-x)^2.
The Sheffer a-sequence Sha(n) = A164555(n)/A027642(n) with e.g.f. x/(1-exp(-x)), and the z-sequence is Shz(n) = -1 with e.g.f. -exp(x).
The inverse Sheffer matrix is ((-1)^(n-k))*A105794(n,k) with e.g.f. exp(z)*exp(x*(1-exp(-z))). (End)
Triangle T(n,k), read by rows, given by (-1, 1, 0, 2, 1, 3, 2, 4, 3, 5, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 16 2012
Also coefficients of t in t*(t-1)*Sum[(-1)^(n+m) t^(m-1) StirlingS1[n,m], {m,n}] in which setting t^k equal to k gives n!, from this follows that the dot product of row n with [0,...,n] equals (n-1)!. - Wouter Meeussen, May 15 2012

			Triangle begins
   1;
  -1,   1;
   0,  -1,   1;
   0,  -1,   0,   1;
   0,  -2,  -1,   2,   1;
   0,  -6,  -5,   5,   5,   1;
   0, -24, -26,  15,  25,   9,   1;
   ...
Recurrence:
  -2 = T(4,1) = T(3,0) + (4-2)*T(3,1) = 0 + 2*(-1).
Row polynomials:
  s(3,x) = -x+x^3 = (x-1)*s1(2,x) = (x-1)*(x+x^2).
  s(3,x) = (x-1)*s(2,x+1) = (x-1)*(-(x+1)+(x+1)^2).
  s(3,x) - s(3,x-1) = -x+x^3 -(-(x-1)+(x-1)^3) = 3*(-x+x^2) = 3*s(2,x).

S. Roman, The Umbral Calculus, Academic Press, New York, 1984.

M. W. Coffey and M. C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.
Igor Victorovich Statsenko, On the ordinal numbers of triangles of generalized special numbers, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.

Cf. A049444, A049458, A094646, A132393, A105794.

A094645_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x-n+2, n)), x, k), k=0..n): seq(print(A094645_row(n)), n=0..6); # Peter Luschny, May 16 2013

t[n_, k_] /; n >= k >= 0 := t[n, k] = t[n-1, k-1] + (n-2)*t[n-1, k]; t[n_, k_] /; n < k = 0; t[, -1] = 0; t[0, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, 10}, {k, 0, n}] ] (* _Jean-François Alcover, Sep 29 2011, after recurrence *);
Table[CoefficientList[t*(t-1)*Sum[(-1)^(n+m)*t^(m-1)*StirlingS1[n,m],{m,n}],t],{n,1,7}] (* Wouter Meeussen, May 15 2012 *)

E.g.f.: (1-y)^(1-x).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively. - Philippe Deléham, Nov 13 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then |T(n,i)| = |f(n,i,-1)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Wolfdieter Lang, Jun 20 2011: (Start)
T(n,k) = |S1(n-1,k-1)| - |S1(n-1,k)|, n >= 1, k >= 1, with |S1(n,k)| = A132393(n,k) (unsigned Stirling1).
Recurrence: T(n,k) = T(n-1,k-1) + (n-2)*T(n-1,k) if n >= k >= 0; T(n,k) = 0 if n < k; T(n,-1) = 0; T(0,0) = 1.
E.g.f. column k: (1-x)*((-log(1-x))^k)/k!. (End)
T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(m,n,i) where TC(m,n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^i, m = 1 for n >= 0. See A130534, A370518 for m=0 and m=2. - Igor Victorovich Statsenko, Feb 27 2024

A053567 Stirling numbers of first kind, s(n+5, n).

Original entry on oeis.org

-120, 1764, -13132, 67284, -269325, 902055, -2637558, 6926634, -16669653, 37312275, -78558480, 156952432, -299650806, 549789282, -973941900, 1672280820, -2792167686, 4546047198, -7234669596, 11276842500, -17247104875, 25922927745, -38343278610, 55880640270

Offset: 1

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 17 2000

a(n) is equal to (-1)^n times the sum of the products of each distinct grouping of 5 members of the set {1, 2, 3, ..., n + 4}. So, a(1) = (-1)*1*2*3*4*5 = -120, and a(2) = 1*2*3*4*5 + 1*2*3*4*6 + 1*2*3*5*6 + 1*2*4*5*6 + 1*3*4*5*6 + 2*3*4*5*6 = 1764. See comment at A001303. - Greg Dresden, Aug 26 2019

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
G. C. Greubel, A Note on Jain basis functions, arXiv:1612.09385 [math.CA], 2016.
Index entries for linear recurrences with constant coefficients, signature (-11,-55,-165,-330,-462,-462,-330,-165,-55,-11,-1).

Next |Stirling1| diagonal A112002, 5th diagonal of A130534.

[(-1)^n*Binomial(n+5, 6)*Binomial(n+5, 2)*(3*n^2+23*n+38)/8: n in [1..30]]; // Vincenzo Librandi, Jun 09 2011

A053567 := proc(n) (-1)^(n+1)*combinat[stirling1](n+5,n) ; end proc: # R. J. Mathar, Jun 08 2011

Table[StirlingS1[n+5,n](-1)^(n-1),{n,30}] (* Harvey P. Dale, Sep 21 2011 *)
(* or *)
CoefficientList[Series[-x*(120 - 444*x + 328*x^2 - 52*x^3 + x^4)/(1+x)^11, {x, 0, 27}], x] (* Georg Fischer, May 19 2019 *)

a(n) = (-1)^(n-1)*stirling(n+5, n, 1); \\ Michel Marcus, Aug 29 2017

[stirling_number1(n,n-5)*(-1)^(n+1) for n in range(6, 26)] # Zerinvary Lajos, May 16 2009

a(n) = (-1)^n*binomial(n+5, 6)*binomial(n+5, 2)*(3*n^2 + 23*n + 38)/8.
G.f.: -x*(120 - 444*x + 328*x^2 - 52*x^3 + x^4)/(1+x)^11. See row k=4 of triangle A112007 for the coefficients. [G.f. corrected by Georg Fischer, May 19 2019]
E.g.f. with offset 5: exp(x)*(Sum_{m=0..5} A112486(5, m)*(x^(5+m)/(5+m)!).
a(n) = (f(n+4, 5)/10!)*Sum_{m=0..min(5, n-1)} A112486(5, m)*f(10, 5-m)*f(n-1, m)), with the falling factorials f(n, m):=n*(n-1)*, ..., *(n-(m-1)). From the e.g.f.

Definition edited by Eric M. Schmidt, Aug 29 2017

Incorrect formula removed by Greg Dresden, Aug 26 2019

A055505 Numerators in expansion of (1-x)^(-1/x)/e.

Original entry on oeis.org

1, 1, 11, 7, 2447, 959, 238043, 67223, 559440199, 123377159, 29128857391, 5267725147, 9447595434410813, 1447646915836493, 225037938358318573, 29911565062525361, 3651003047854884043877, 38950782815463986767

Offset: 0

N. J. A. Sloane, Jul 11 2000

From Miklos Kristof, Nov 04 2007: (Start) This is also the sequence of numerators associated with expansion of (1+x)^(1/x).
(1 + x)^(1/x) = exp(1)*(1 - 1/2*x + 11/24*x^2 - 7/16*x^3 + 2447/5760*x^4 - 959/2304*x^5 + 238043/580608*x^6 - ...).
(1+x)^(1/x) = exp(log(1+x)/x) = exp(1)*exp(-x/2)*exp(x^2/3)*exp(x^3/4)*...
Let a(n) be this sequence, let b(n) be A055535. Then (1+x)^(1/x)=exp(1)*a(n)/b(n) x^n.
a(n)/b(n) = Sum_{i>=n} s(i,i-n)/i! where s(n,m) is a Stirling number of the first kind.
exp(1) = 1 + Sum_{i>=1} s(i,i)/i!, for the n=1 case.
a(1)/b(1) = 1/1 because 1+1/1!+1/2!+1/3!+1/4!+... = exp(1)
a(2)/b(2) = 1/2 because 1/2!+3/3!+6/4!+10/5!+... = 1/2*exp(1)
a(3)/b(3) = 11/24 because 2/3!+11/4!+35/5!+85/6!+... = 11/24*exp(1)
a(4)/b(4) = 7/16 because 6/4!+50/5!+225/6!+735/7!+... = 7/16*exp(1) (End)

			1+1/2*x+11/24*x^2+7/16*x^3+2447/5760*x^4+...
1, -1/2, 11/24, -7/16, 2447/5760, -959/2304, 238043/580608, -67223/165888, ...

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.1.

G. C. Greubel, Table of n, a(n) for n = 0..250
Markus Brede, On the convergence of the sequence defining Euler's number, Math. Intelligencer, 27, no. 3 (2005), 6-7.
Chao-Ping Chen and Junesang Choi, An Asymptotic Formula for (1+1/x)^x Based on the Partition Function, Amer. Math. Monthly 121 (2014), no. 4, 338--343. MR3183017.
Branko Malesevic, Yue Hu, and Cristinel Mortici, Accurate Estimates of (1+x)^{1/x} Involved in Carleman Inequality and Keller Limit, arXiv:1801.04963 [math.CA], 2018.

Cf. A094638, A130534, A055535 (denominators).
See also A239897/A239898.
Cf. A276977.

T:= proc(u) local k, l; add( Stirling1(u+k,k)*((u+k)!)^(-1)* add( (-1)^l/l!, l=0..u-k), k=0..u); end;

a[n_] := Sum[StirlingS1[n+k, k]/(n+k)!*Sum[(-1)^j/j!, {j, 0, n-k}], {k, 0, n}]; Table[a[n] // Numerator // Abs, {n, 0, 17}] (* Jean-François Alcover, Mar 04 2014, after Maple *)
Numerator[((1-x)^(-1/x)/E + O[x]^20)[[3]]] (* or *)
Numerator[Table[Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}] (-1)^n/(2n)!, {n, 0, 10}]] (* Vladimir Reshetnikov, Sep 23 2016 *)

See Maple line for formula.

Edited by N. J. A. Sloane, Jul 01 2008 at the suggestion of R. J. Mathar

A055535 Denominators in expansion of (1-x)^(-1/x)/e.

Original entry on oeis.org

1, 2, 24, 16, 5760, 2304, 580608, 165888, 1393459200, 309657600, 73574645760, 13377208320, 24103053950976000, 3708162146304000, 578473294823424000, 77129772643123200, 9440684171518279680000, 100969884187361280000

Offset: 0

N. J. A. Sloane, Jul 11 2000

Or, equally, denominators in expansion of (1+x)^(1/x)/e.

			(1-x)^(-1/x) = exp(1)*(1 + 1/2*x + 11/24*x^2 + 7/16*x^3 + 2447/5760*x^4 + 959/2304*x^5 + 238043/580608*x^6 + ...).

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 293, Problem 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.1.

Robert Israel, Table of n, a(n) for n = 0..377
Chao-Ping Chen and Junesang Choi, An Asymptotic Formula for (1+1/x)^x Based on the Partition Function, Amer. Math. Monthly 121 (2014), no. 4, 338--343. MR3183017.

Cf. A094638, A130534, A055505 (numerators), A276977.

G:= (1-x)^(-1/x)/exp(1):
S:= series(G,x,32):
seq(denom(coeff(S,x,j)),j=0..30); # Robert Israel, Sep 23 2016

a[n_] := Sum[StirlingS1[n+k, k]/(n+k)!*Sum[(-1)^j/j!, {j, 0, n-k}], {k, 0, n}]; Table[a[n] // Denominator, {n, 0, 17}] (* Jean-François Alcover, Mar 04 2014 *)
Denominator[((1+x)^(1/x)/E + O[x]^20)[[3]]] (* or *)
Denominator[Table[Sum[StirlingS1[n+k, k] Subfactorial[n-k] Binomial[2n, n+k], {k, 0, n}]/(2n)!, {n, 0, 10}]] (* Vladimir Reshetnikov, Sep 23 2016 *)

From Miklos Kristof, Nov 04 2007 (Start):
(1+x)^(1/x) = exp(log(1+x)/x) = exp(1)*exp(-x/2)*exp(x^2/3)*exp(x^3/4)*...
Let a(n) be A055505, let b(n) be this sequence. Then (1+x)^(1/x) = exp(1)*a(n)/b(n) x^n.
a(n)/b(n) = Sum_{i>=n} s(i,i-n)/i! where s(n,m) is a Stirling number of the first kind.
exp(1) = 1 + Sum_{i>=1} s(i,i)/i! for the n = 1 case.
a(1)/b(1) = 1/1 because 1+1/1!+1/2!+1/3!+1/4!+... = exp(1)
a(2)/b(2) = 1/2 because 1/2!+3/3!+6/4!+10/5!+... = 1/2*exp(1)
a(3)/b(3) = 11/24 because 2/3!+11/4!+35/5!+85/6!+... = 11/24*exp(1)
a(4)/b(4) = 7/16 because 6/4!+50/5!+225/6!+735/7!+... = 7/16*exp(1) (End)

Edited by N. J. A. Sloane, Jul 25 2008 at the suggestion of R. J. Mathar and Eric Rowland

A286416 Number T(n,k) of entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 3, 1, 8, 6, 1, 24, 25, 10, 1, 83, 98, 63, 15, 1, 324, 399, 338, 135, 21, 1, 1400, 1746, 1727, 980, 257, 28, 1, 6609, 8271, 8874, 6426, 2455, 448, 36, 1, 33758, 42284, 47191, 40334, 20506, 5474, 730, 45, 1, 185136, 231939, 263458, 250839, 158827, 57239, 11128, 1128, 55, 1

Offset: 1

Alois P. Heinz, May 08 2017

			T(3,2) = 6 because the number of entries in the second last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+2+2+1+1 = 6.
Triangle T(n,k) begins:
     1;
     3,    1;
     8,    6,    1;
    24,   25,   10,    1;
    83,   98,   63,   15,    1;
   324,  399,  338,  135,   21,   1;
  1400, 1746, 1727,  980,  257,  28,  1;
  6609, 8271, 8874, 6426, 2455, 448, 36, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Wikipedia, Partition of a set

Columns k=1-2 give: A038561 (for n>1), A286433.
Main diagonal and first lower diagonal give: A000012, A000217.
Row sums give A070071.
Cf. A000110, A130534, A270236, A286232.

A056856 Triangle of numbers related to rooted trees and unrooted planar trees.

Original entry on oeis.org

1, 1, 2, 2, 9, 9, 6, 44, 96, 64, 24, 250, 875, 1250, 625, 120, 1644, 8100, 18360, 19440, 7776, 720, 12348, 79576, 252105, 420175, 352947, 117649, 5040, 104544, 840448, 3465728, 8028160, 10551296, 7340032, 2097152

Offset: 1

F. Chapoton, Aug 31 2000

The rows sum to A006963: (2*n - 1)!/n!.
The main diagonal is A000169: n^(n-1).
The left column is A000142: (n - 1)!.
The alternating sum in row n is (-1)^(n-1)*(n - 1)!
If Y := X * (1 - X)^(z-1), then (1 - z*X)^(-1) = 1 + Sum_{n>=1} Y^n/(n-1)! * (Sum_{k=1..n} (-1)^(n-k) * z^k * T(n, k)). Note that if Y = y^(z-1) and X = x^(z-1) then y = x - x^z, dy/dx = 1 - z*x^(z-1) = 1 - z*X, and dx/dy = (1 - z*X)^(-1). Also x = y + x^z = y + y^z + z*y^(2*z-1) + ... = y * (1 + Sum_{n>=1} Y^n/(n-1)! * (1+(z-1)*n)^(-1) * (Sum_{k=1..n} (-1)^(n-k) * z^k * T(n, k))). - Michael Somos, Aug 01 2019

			Triangle begins:
{1},
{1, 2},
{2, 9, 9},
{6, 44, 96, 64},
{24, 250, 875, 1250, 625},
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998.

Cf. A006963, A000142, A000169, A203904.

seq(seq(coeff(product(n*x + k, k = 1..n-1), x, i), i = 0..n-1), n = 1..8); # Peter Bala, Nov 08 2015

T[n_, m_] := (n^(m-1)*Binomial[n-1, m-1]*Sum[((-1)^(n-m-k)*Binomial[n+k-1, k]*StirlingS2[n-m+k, k]*Binomial[2*n-m, n-m-k])/Binomial[n-m+k, k], {k, 0, n-m}]); Table[T[n, m], {n, 1, 8}, {m, 1, n}] // Flatten (* Jean-François Alcover, Feb 23 2017, after Vladimir Kruchinin *)
T[ n_, k_] := If[ n < 1 || k < 1, 0, Coefficient[ (-1)^(n - k) Binomial[n z, n] (n - 1)!, z, k]]; (* Michael Somos, Aug 01 2019 *)

T(n,m):=(n^(m-1)*binomial(n-1,m-1)*sum(((-1)^(n-m-k)*binomial(n+k-1,k)*stirling2(n-m+k,k)*binomial(2*n-m,n-m-k))/binomial(n-m+k,k),k,0,n-m)); /* Vladimir Kruchinin, Apr 05 2016 */

{T(n, k) = if( n < 1 || k < 1, 0, polcoeff( (-1)^(n-k) * binomial(n*x, n)*(n-1)!, k))}; /* Michael Somos, Aug 01 2019 */

Formula for row n: Sum_{k = 0..n-1} T(n,k)*y^k = Product_{k = 1..n-1} (k + n*y)
E.g.f.: A(x,t) = Sum_{n >= 1} 1/(n*t)*binomial(n*t + n - 1, n)*x^n = log(B_(t+1)(x)), where B_t(x) = Sum_{n >= 0} 1/(n*t + 1)*binomial(n*t + 1, n)*x^n is Lambert's generalized binomial series - see Graham et al., Section 5.4. - Peter Bala, Nov 08 2015
T(n,m) = n^(m-1)*binomial(n-1,m-1)*Sum_{k=0..n-m} ((-1)^(n-m-k)*binomial(n+k-1,k)*stirling2(n-m+k,k)*binomial(2*n-m,n-m-k))/binomial(n-m+k,k). - Vladimir Kruchinin, Apr 05 2016
Conjecture: T(n,k) = A130534(n,k)* n^(k-1). - R. J. Mathar, Mar 31 2023

a(29)-a(36) from Peter Bala, Nov 08 2015

cp's OEIS Frontend

A164652 Triangle read by rows: Hultman numbers: a(n,k) is the number of permutations of n elements whose cycle graph (as defined by Bafna and Pevzner) contains k cycles for n >= 0 and 1 <= k <= n+1.

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A051431 a(n) = (n+10)!/10!.

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A139605 Weights for expansion of iterated derivatives, powers of a Lie derivative, or infinitesimal generator in vector form, (f(x)D_x)^n; coefficients of A-polynomials of Comtet; Scherk partition polynomials.

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A163934 Triangle related to the asymptotic expansion of E(x,m=4,n).

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A094645 Triangle of generalized Stirling numbers of the first kind.

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A053567 Stirling numbers of first kind, s(n+5, n).

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