cp's OEIS Frontend

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.

E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.

S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.

Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.

S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.

A019538(n, k) = k! * S2(n, k).

A028248(n, k) = (k-1)! * S2(n, k).

For asymptotics see Hsu (1948), among other sources.

Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).

Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005

Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006

Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007

Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007

From Tom Copeland, Oct 10 2007: (Start)

Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.

For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).

Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).

(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)

n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007

The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011

Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012

S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012

O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012

n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012

G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013

From Tom Copeland, Apr 17 2014: (Start)

Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).

With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].

In particular, for f(y) = (1+y),

A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],

B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and

C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].

D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).

E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.

(End)

As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014

O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014

Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015

From Daniel Forgues, Jan 16 2016: (Start)

Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.

For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)

Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022

O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022

S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022

S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024

From Mikhail Kurkov, Mar 05 2025: (Start)

For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.

Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).

Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

G.f.: x/((1-2*x)*(1-3*x)).

a(n) = 5*a(n-1) - 6*a(n-2).

a(n) = 3*a(n-1) + 2^(n-1). - Jon Perry, Aug 23 2002

Starting 0, 0, 1, 5, 19, ... this is 3^n/3 - 2^n/2 + 0^n/6, the binomial transform of A086218. - Paul Barry, Aug 18 2003

a(n) = A083323(n)-1 = A056182(n)/2 = (A002783(n)-1)/2 = (A003063(n+2)-A003063(n+1))/2. - Ralf Stephan, Jan 12 2004

Binomial transform of A000225. - Ross La Haye, Feb 07 2005

a(n) = Sum_{k=0..n-1} binomial(n, k)*2^k. - Ross La Haye, Aug 20 2005

a(n) = 2^(2n) - A083324(n). - Ross La Haye, Sep 10 2005

a(n) = A112626(n, 1). - Ross La Haye, Jan 11 2006

E.g.f.: exp(3*x) - exp(2*x). - Mohammad K. Azarian, Jan 14 2009

a(n) = A217764(n,1). - Ross La Haye, Mar 27 2013

a(n) = 2*a(n-1) + 3^(n-1). - Toby Gottfried, Mar 28 2013

a(n) = A000244(n) - A000079(n). - Omar E. Pol, Mar 28 2013

a(n) = Sum_{k=0..2} Stirling1(2,k)*(k+1)^n = c_2^{(-n)}, poly-Cauchy numbers. - Takao Komatsu, Mar 28 2013

a(n) = A227048(n,A098294(n)). - Reinhard Zumkeller, Jun 30 2013

a(n+1) = Sum_{k=0..n} 2^k*3^(n-k). - J. M. Bergot, Mar 27 2018

Sum_{n>=1} 1/a(n) = A329064. - Amiram Eldar, Nov 20 2020

a(n) = (1/2)*Sum_{k=0..n} binomial(n, k)*(2^(n-k) + 2^k - 2).

a(n) = A001117(n) + 2*A000918(n) + 1. - Ambrosio Valencia-Romero, Mar 08 2022

a(n) = A000225(n) + A028243(n+1). - Ambrosio Valencia-Romero, Mar 09 2022

From Peter Bala, Jun 27 2025: (Start)

exp(Sum_{n >=1} a(2*n)/a(n)*x^n/n) = Sum_{n >= 0} a(n+1)*x^n.

exp(Sum_{n >=1} a(3*n)/a(n)*x^n/n) = 1 + 19*x + 247*x^2 + ... is the g.f. of A019443.

exp(Sum_{n >=1} a(4*n)/a(n)*x^n/n) = 1 + 65*x + 2743*x^2 + ... is the g.f. of A383754.

The following are all examples of telescoping series:

Sum_{n >= 1} 6^n/(a(n)*a(n+1)) = 2, since 6^n/(a(n)*a(n+1)) = b(n) - b(n+1), where b(n) = 2^n/a(n);

Sum_{n >= 1} 18^n/(a(n)*a(n+1)*a(n+2)) = 22/75, since 18^n/(a(n)*a(n+1)*a(n+2)) = c(n) - c(n+1), where c(n) = (5*6^n - 2*4^n)/(15*a(n)*a(n+1));

Sum_{n >= 1} 54^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = 634/48735 since 54^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = d(n) - d(n+1), where d(n) = (57*18^n - 38*12^n + 8*8^n)/(513*a(n)*a(n+1)*a(n+2)).

Sum_{n >= 1} 6^n/(a(n)*a(n+2)) = 14/25; Sum_{n >= 1} (-6)^n/(a(n)*a(n+2)) = -6/25.

Sum_{n >= 1} 6^n/(a(n)*a(n+3)) = 306/1805.

Sum_{n >= 1} 6^n/(a(n)*a(n+4)) = 4282/80275; Sum_{n >= 1} (-6)^n/(a(n)*a(n+4)) = -1698/80275. (End)

a(n-1) = Sum_{k=1..n} k*Stirling2(n, k) for n>=1.

E.g.f.: exp(exp(x) + 2*x - 1). First differences of Bell numbers (if offset 1). - Michael Somos, Oct 09 2002

G.f.: Sum_{k>=0} (x^k/Product_{l=1..k} (1-(l+1)x)). - Ralf Stephan, Apr 18 2004

a(n) = Sum_{i=0..n} 2^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n). - Fred Lunnon, Aug 04 2007 [Written umbrally, a(n) = (B+2)^n. - N. J. A. Sloane, Feb 07 2009]

Representation as an infinite series: a(n-1) = Sum_{k>=2} (k^n*(k-1)/k!)/exp(1), n=1, 2, ... This is a Dobinski-type summation formula. - Karol A. Penson, Mar 14 2002

Row sums of A011971 (Aitken's array, also called Bell triangle). - Philippe Deléham, Nov 15 2003

a(n) = exp(-1)*Sum_{k>=0} ((k+2)^n)/k!. - Gerald McGarvey, Jun 03 2004

Recurrence: a(n+1) = 1 + Sum_{j=1..n} (1+binomial(n, j))*a(j). - Jon Perry, Apr 25 2005

a(n) = A000296(n+3) - A000296(n+1). - Philippe Deléham, Jul 31 2005

a(n) = B(n+2) - B(n+1), where B(n) are Bell numbers (A000110). - Franklin T. Adams-Watters, Jul 13 2006

a(n) = A123158(n,2). - Philippe Deléham, Oct 06 2006

Binomial transform of Bell numbers 1, 2, 5, 15, 52, 203, 877, 4140, ... (see A000110).

Define f_1(x), f_2(x), ... such that f_1(x)=x*e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n-1) = e^(-1)*f_n(1). - Milan Janjic, May 30 2008

Representation of numbers a(n), n=0,1..., as special values of hypergeometric function of type (n)F(n), in Maple notation: a(n)=exp(-1)*2^n*hypergeom([3,3...3],[2.2...2],1), n=0,1..., i.e., having n parameters all equal to 3 in the numerator, having n parameters all equal to 2 in the denominator and the value of the argument equal to 1. Examples: a(0)= 2^0*evalf(hypergeom([],[],1)/exp(1))=1 a(1)= 2^1*evalf(hypergeom([3],[2],1)/exp(1))=3 a(2)= 2^2*evalf(hypergeom([3,3],[2,2],1)/exp(1))=10 a(3)= 2^3*evalf(hypergeom([3,3,3],[2,2,2],1)/exp(1))=37 a(4)= 2^4*evalf(hypergeom([3,3,3,3],[2,2,2,2],1)/exp(1))=151 a(5)= 2^5*evalf(hypergeom([3,3,3,3,3],[2,2,2,2,2],1)/exp(1)) = 674. - Karol A. Penson, Sep 28 2007

Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i <= j), and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = (-1)^(n)charpoly(A,-2). - Milan Janjic, Jul 08 2010

a(n) = D^(n+1)(x*exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A003128, A052852 and A009737. - Peter Bala, Nov 25 2011

From Sergei N. Gladkovskii, Oct 11 2012 to Jan 26 2014: (Start)

Continued fractions:

G.f.: 1/U(0) where U(k) = 1 - x*(k+3) - x^2*(k+1)/U(k+1).

G.f.: 1/(U(0)-x) where U(k) = 1 - x - x*(k+1)/(1 - x/U(k+1)).

G.f.: G(0)/(1-x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )).

G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*x-k*x)/(1-x/(x-1/G(k+1) )).

G.f.: -G(0)/x where G(k) = 1 - 1/(1-k*x-x)/(1-x/(x-1/G(k+1) )).

G.f.: 1 - 2/x + (1/x-1)*S where S = sum(k>=0, ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k / prod(i=0..k-1, (1-x-x*i)/(1-x) ) ).

G.f.: (1-x)/x/(G(0)-x) - 1/x where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ).

G.f.: (1/G(0) - 1)/x^3 where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )).

G.f.: 1/Q(0), where Q(k)= 1 - 2*x - x/(1 - x*(k+1)/Q(k+1)).

G.f.: G(0)/(1-3*x), where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1 - x*(k+3))*(1 - x*(k+4))/G(k+1) ). (End)

a(n) ~ exp(n/LambertW(n) + 3*LambertW(n)/2 - n - 1) * n^(n + 1/2) / LambertW(n)^(n+1). - Vaclav Kotesovec, Jun 09 2020

a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Ilya Gutkovskiy, Jul 02 2020

a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021

a(n) = Sum_{k=0..n} 3^k*A124323(n, k). - Mélika Tebni, Jun 02 2022

G.f.: 1/((1-2*x)*(1-3*x)*(1-4*x)).

a(n-2) = (2^n)*(2^n - 1)/2 - 3^n + 2^n.

From Hieronymus Fischer, Jun 25 2007: (Start)

a(n) = Sum_{0<=i,j,k,<=n, i+j+k=n} 2^i*3^j*4^k.

a(n) = 2^(n+1)*(1+2^(n+2))-3^(n+2). (End)

a(n) = 3*StirlingS2(n+3,4) + StirlingS2(n+3,3). - Ross La Haye, Jan 10 2008

If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-2) = f(n,2,2), (n >= 2). - Milan Janjic, Apr 26 2009

E.g.f.: (d^2/dx^2) (exp(2*x)*((exp(x)-1)^2)/2!). See the Sheffer comment given above. - Wolfdieter Lang, Oct 08 2011

a(n) = A006516(n+2) - A001047(n+2). - Ross La Haye, Jan 26 2016

a(n) = A006516(n+1) + 3*a(n-1), n>=1, a(0)=1. - Carlos A. Rico A., Jun 22 2019

T(n,k) = Sum_{m=n..k} |S1(n,m)|*S2(m,k), k>=n>=1, with Stirling triangles S2(n,m):=A048993 and S1(n,m):=A048994.

T(n,k) = C(n,k)*(n-1)!/(k-1)!.

Sum_{k=1..n} T(n,k) = A000262(n).

n*Sum_{k=1..n} T(n,k) = A103194(n) = Sum_{k=1..n} T(n,k)*k^2.

E.g.f. column k: (x^(k-1)/(1-x)^(k+1))/(k-1)!, k>=1.

Recurrence from Sheffer (here Jabotinsky) a-sequence [1,1,0,...] (see the W. Lang link under A006232): T(n,k)=(n/k)*T(n-1,m-1) + n*T(n-1,m). - Wolfdieter Lang, Jun 29 2007

The e.g.f. is, umbrally, exp[(.)!* L(.,-t,1)*x] = exp[t*x/(1-x)]/(1-x)^2 where L(n,t,1) = Sum_{k=0..n} T(n+1,k+1)*(-t)^k = Sum_{k=0..n} binomial(n+1,k+1)* (-t)^k / k! is the associated Laguerre polynomial of order 1. - Tom Copeland, Nov 17 2007

For this Lah triangle, the n-th row polynomial is given umbrally by

n! C(B.(x)+1+n,n) = (-1)^n C(-B.(x)-2,n), where C(x,n)=x!/(n!(x-n)!),

the binomial coefficient, and B_n(x)= exp(-x)(xd/dx)^n exp(x), the n-th Bell / Touchard / exponential polynomial (cf. A008277). E.g.,

2! C(-B.(-x)-2,2) = (-B.(x)-2)(-B.(x)-3) = B_2(x) + 5*B_1(x) + 6 = 6 + 6x + x^2.

n! C(B.(x)+1+n,n) = n! e^(-x) Sum_{j>=0} C(j+1+n,n)x^j/j! is a corresponding Dobinski relation. See the Copeland link for the relation to inverse Mellin transform. - Tom Copeland, Nov 21 2011

The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A008277 (D = (1+x)*d/dx), A035342 (D = (1+x)^3*d/dx), A035469 (D = (1+x)^4*d/dx) and A049029 (D = (1+x)^5*d/dx). - Peter Bala, Nov 25 2011

T(n,k) = Sum_{i=k..n} A130534(n-1,i-1)*A008277(i,k). - Reinhard Zumkeller, Mar 18 2013

Let E(x) = Sum_{n >= 0} x^n/(n!*(n+1)!). Then a generating function is exp(t)*E(x*t) = 1 + (2 + x)*t + (6 + 6*x + x^2)*t^2/(2!*3!) + (24 + 36*x + 12*x^2 + x^3)*t^3/(3!*4!) + ... . - Peter Bala, Aug 15 2013

P_n(x) = L_n(1+x) = n!*Lag_n(-(1+x);1), where P_n(x) are the row polynomials of A059110; L_n(x), the Lah polynomials of A105278; and Lag_n(x;1), the Laguerre polynomials of order 1. These relations follow from the relation between the iterated operator (x^2 D)^n and ((1+x)^2 D)^n with D = d/dx. - Tom Copeland, Jul 23 2018

Dividing each n-th diagonal by n!, where the main diagonal is n=1, generates the Narayana matrix A001263. - Tom Copeland, Sep 23 2020

T(n,k) = A089231(n,n-k). - Ron L.J. van den Burg, Dec 12 2021

T(n,k) = T(n-1,k-1) + (n+k-1)*T(n-1,k). - Bérénice Delcroix-Oger, Jun 25 2025

E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).

T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019

T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.

T(n,k) = (1/(2^k*k!)) * A145901(n,k). - Peter Bala

The row polynomials R(n,x) satisfy the Dobinski-type identity:

R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011

Let f(x) = exp((1/2)*exp(2*x)+x). Then the row polynomials R(n,x) are given by R(n,exp(2*x)) = (1/f(x))*(d/dx)^n(f(x)). Similar formulas hold for A008277, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012

From Peter Bala, Jul 20 2012: (Start)

The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details.

An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).

(End)

From Tom Copeland, Dec 31 2015: (Start)

T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts.

The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2).

The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].

(End)

Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016

From Wolfdieter Lang, May 26 2017: (Start)

G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.

T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)

With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds:

(1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k);

(2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024

a(n, m) = A008949(n, n-m), if n > m >= 0.

a(n, m) = Sum_{k=m..n} A007318(n, k) (partial row sums in columns m).

Column m recursion: a(n, m) = Sum_{j=m..n-1} a(j, m) + A007318(n, m) if n >= m >= 0, a(n, m) := 0 if n

G.f. for column m: (1/(1-2*x))*(x/(1-x))^m, m >= 0.

a(n, m) = Sum_{j=0..n} binomial(n, m+j). - Paul Barry, Feb 03 2005

Inverse binomial transform (by columns) of A112626. - Ross La Haye, Dec 31 2006

T(2n,n) = A032443(n). - Philippe Deléham, Sep 16 2009

From Peter Bala, Dec 23 2014: (Start)

Exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(8 + 7*x + 4*x^2/2! + x^3/3!) = 8 + 15*x + 26*x^2/2! + 42*x^3/3! + 64*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).

Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143494 (but with a different offset). See the Example section. Cf. A106516. (End)

a(n,m) = Sum_{p=m..n} 2^(n-p)*binomial(p-1,m-1), n >= m >= 0, else 0. - Wolfdieter Lang, Jan 09 2015

T(n, k) = 2^n - (1/2)*binomial(n, k-1)*hypergeom([1, n+1], [n-k+2], 1/2). - Peter Luschny, Oct 10 2019

T(n, k) = binomial(n, k)*hypergeom([1, k - n], [k + 1], -1). - Peter Luschny, Oct 06 2023

n-th row polynomial R(n, x) = (2^n - x*(1 + x)^n)/(1 - x). These polynomials can be used to find series acceleration formulas for the constants log(2) and Pi. - Peter Bala, Mar 03 2025