cp's OEIS Frontend

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.

The unsigned numbers (read from right to left) also give the number of permutations of 1..n with complexity k, where the complexity of a permutation is defined to be the sum of the lengths of the cycles minus the number of cycles. In other words, the complexity equals the sum of (length of cycle)-1 over all cycles. For n=5, the numbers of permutations with complexity 0,1,2,3,4 are 1, 10, 35, 50, 24. - N. J. A. Sloane, Feb 08 2019

The unsigned numbers are also the number of permutations of 1..n with k left to right maxima (see Khovanova and Lewis, Smith).

With P(n) = the number of integer partitions of n, T(i,n) = the number of parts of the i-th partition of n, D(i,n) = the number of different parts of the i-th partition of n, p(j,i,n) = the j-th part of the i-th partition of n, m(j,i,n) = multiplicity of the j-th part of the i-th partition of n, Sum_[T(i,n)=k]{i=1}^{P(n)} = sum running from i=1 to i=p(n) but taking only partitions with T(i,n)=k parts into account, Product{j=1..T(i,n)} = product running from j=1 to j=T(i,n), Product_{j=1..D(i,n)} = product running from j=1 to j=D(i,n) one has S1(n,k) = Sum_[T(i,n)=k]{i=1}^{P(n)} (n!/Product{j=1..T(i,n)} p(j,i,n))* (1/Product_{j=1..D(i,n)} m(j,i,n)!). For example, S1(6,3) = 225 because n=6 has the following partitions with k=3 parts: (114), (123), (222). Their complexions are: (114): (6!/1*1*4)*(1/2!*1!) = 90, (123): (6!/1*2*3)*(1/1!*1!*1!) = 120, (222): (6!/2*2*2)*(1/3!) = 15. The sum of the complexions is 90+120+15 = 225 = S1(6,3). - Thomas Wieder, Aug 04 2005

Row sums equal 0. - Jon Perry, Nov 14 2005

|s(n,k)| enumerates unordered n-vertex forests composed of k increasing non-plane (unordered) trees. Proof from the e.g.f. of the first column and the F. Bergeron et al. reference, especially Table 1, last row (non-plane "recursive"), given in A049029. - Wolfdieter Lang, Oct 12 2007

|s(n,k)| enumerates unordered increasing n-vertex k-forests composed of k unary trees (out-degree r from {0,1}) whose vertices of depth (distance from the root) j >= 0 come in j+1 colors (j=0 for the k roots). - Wolfdieter Lang, Oct 12 2007, Feb 22 2008

A refinement of the unsigned array is A036039. For an association to forests of "naturally grown" rooted non-planar trees, dispositions of flags on flagpoles, and colorings of the vertices of the complete graphs K_n, see A130534. - Tom Copeland, Mar 30 and Apr 05 2014

The Stirling numbers of the first kind were related to the falling factorial and the convolved, or generalized, Bernoulli numbers B_n by Norlund in 1924 through Sum_{k=1..n+1} T(n+1,k) * x^(k-1) = (x-1)!/(x-1-n)! = (x + B.(0))^n = B_n(x), umbrally evaluated with (B.(0))^k = B_k(0) and the associated Appell polynomial B_n(x) defined by the e.g.f. (t/(exp(t) - 1))^(n+1) * exp(x*t) = exp(B.(x)t). - Tom Copeland, Sep 29 2015

With x = e^z, D_x = d/dx, D_z = d/dz, and p_n(x) the row polynomials of this entry, x^n (D_x)^n = p_n(D_z) = (D_z)! / (D_z - n)! = (xD_x)! / (xD_x - n)!. - Tom Copeland, Nov 27 2015

From the operator relation z + Psi(1) + sum_{n > 0} (-1)^n (-1/n) binomial(D,n) = z + Psi(1+D) with D = d/dz and Psi the digamma function, Zeta(n+1) = Sum_{k > n-1} (1/k) |S(k,n)| / k! for n > 0 and Zeta the Riemann zeta function. - Tom Copeland, Aug 12 2016

Let X_1,...,X_n be i.i.d. random variables with exponential distribution having mean = 1. Let Y = max{X_1,...,X_n}. Then (-1)^n*n!/( Sum_{k=1..n+1} a(n+1,k) t^(k-1) ) is the moment generating function of Y. The expectation of Y is the n-th harmonic number. - Geoffrey Critzer, Dec 25 2018

In the Ewens sampling theory describing the multivariate probability distribution of the sizes of the allelic classes in a sample of size n under the Infinite Alleles Model, |s(n,k)| gives the coefficient in the formula for the probability that a sample of n alleles has exactly k distinct types. - Noah A Rosenberg, Feb 10 2019

Named by Nielson (1906) after the Scottish mathematician James Stirling (1692-1770). - Amiram Eldar, Jun 11 2021 and Oct 02 2023

The first few row polynomials along with a recursion formula are found in a manuscript by Newton written in 1664 or 1665 (p. 169 of Turnbull) giving a geometric presentation of the binomial theorem for rational powers. - Tom Copeland, Dec 10 2022

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.

Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006

Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012

From Wolfdieter Lang, Nov 14 2018: (Start)

This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.

This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.

The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)

is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.

The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)

Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

a(n) is the number of ways to seat n people at an unspecified number of circular tables and then linearly order the nonempty tables. - Geoffrey Critzer, Mar 18 2009

The terms of this sequence for n >= 1 are the row sums of A008275^2, the unsigned version of A039814. - Peter Bala, Jul 22 2014

Signed sequence is the base for an Appell sequence of polynomials with the e.g.f. e^(x*t)/[log(1+t) + 1] = exp(P(.,x),t) that is the umbral compositional inverse for A238385, reverse of A111492, i.e., umbrally evaluated UP(n,P(.,t))= x^n = P(n,UP(.,t)) where UP(n,t) are the polynomials of A238385. Umbrally evaluated means letting (A(.,t))^n = A(n,t) after substituting A for the independent variable of the polynomial. - Tom Copeland, Nov 15 2014

a(n) is the number of unimodal rooted forests on n labeled nodes (i.e., those forests that avoid the patterns 213 and 312). - Kassie Archer, Aug 30 2018

Number of permutations of [n] where fixed points at index j are j-colored and all other points are unicolored. - Alois P. Heinz, Apr 24 2020

a(n+1) is the permanent of the n X n matrix M with M(i,i) = i+1, other entries 1. - Philippe Deléham, Nov 03 2005

Supernecklaces of type III (cycles of cycles). - Ricardo Bittencourt, May 05 2013

Unsigned coefficients for the raising / creation operator R for the Appell sequence of polynomials A238385: R = x + 1 - 2 D + 7 D^2/2! - 35 D^3/3! + ... . - Tom Copeland, May 09 2016

This triangle groups certain generalized Stirling numbers of the second kind A000558, A000559, ... They can also be interpreted in terms of trees of height 3 with n leaves and constraints on the order of the root.

From Peter Bala, Jul 19 2014: (Start)

The (n,k)-th entry in this table gives the number of double partitions of the set [n] = {1,2,...,n} into k blocks. To form a double partition of [n] we first write [n] as a disjoint union X_1 U...U X_k of k nonempty subsets (blocks) X_i of [n]. Then each block X_i is further partitioned into sub-blocks to give a double partition. For instance, {1,2,4} U {3,5} is a partition of [5] into 2 blocks and {{1,4},{2}} U {{3},{5}} is a refinement of this partition to a double partition of [5] into 2 blocks (and 4 sub-blocks).

Compare the above interpretation for the (n,k)-th entry of this table with the interpretation of the (n,k)-th entry of A013609 (the square of Pascal's triangle but with the rows read in reverse order) as counting the pairs (X,Y) of subsets of [n] such that |Y| = k and X is contained in Y. (End)

Also the Bell transform of the shifted Bell numbers B(n+1) without column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 28 2016

T(n,k) is the number of partitions of an n-set into colored blocks, such that exactly k colors are used and the colors are introduced in increasing order. T(3,2) = 6: 1a|23b, 13a|2b, 12a|3b, 1a|2a|3b, 1a|2b|3a, 1a|2b|3b. - Alois P. Heinz, Aug 27 2019

Triangle is the matrix product of the binomial coefficients with the Stirling numbers of the first kind.

Triangle T(n,k) giving coefficients in expansion of n!*C(x,n) in powers of x. E.g., 3!*C(x,3) = x^3-3*x^2+2*x.

The matrix product of binomial coefficients with the Stirling numbers of the first kind results in the Stirling numbers of the first kind again, but the triangle is shifted by (1,1).

Essentially [1,0,1,0,1,0,1,0,...] DELTA [0,-1,-1,-2,-2,-3,-3,-4,-4,...] where DELTA is the operator defined in A084938; mirror image of the Stirling-1 triangle A048994. - Philippe Deléham, Dec 30 2006

From Doudou Kisabaka, Dec 18 2009: (Start)

The sum of the entries on each row of the triangle, starting on the 3rd row, equals 0. E.g., 1+(-3)+2+0 = 0.

The entries on the triangle can be computed as follows. T(n,r) = T(n-1,r) - (n-1)*T(n-1,r-1). T(n,r) = 0 when r equals 0 or r > n. T(n,r) = 1 if n==1. (End)