

A028246


Triangular array a(n,k) = (1/k)*Sum_{i=0..k} (1)^(ki)*binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.


60



1, 1, 1, 1, 3, 2, 1, 7, 12, 6, 1, 15, 50, 60, 24, 1, 31, 180, 390, 360, 120, 1, 63, 602, 2100, 3360, 2520, 720, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880
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OFFSET

1,5


COMMENTS

Let M = n X n matrix with (i,j)th entry a(n+1j, n+1i), e.g., if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n1)], let b = [b(0)..b(n1)] be its inverse binomial transform and let c = [c(0)..c(n1)] = M^(1)*transpose(b). Then s(k) = Sum_{i=0..n1} b(i)*binomial(k,i) = Sum_{i=0..n1} c(i)*k^i, k=0..n1.  Gary W. Adamson, Nov 11 2001
Julius Worpitzky's 1883 algorithm generates Bernoulli numbers.
By way of example [Wikipedia]:
B0 = 1;
B1 = 1/1  1/2;
B2 = 1/1  3/2 + 2/3;
B3 = 1/1  7/2 + 12/3  6/4;
B4 = 1/1  15/2 + 50/3  60/4 + 24/5;
B5 = 1/1  31/2 + 180/3  390/4 + 360/5  120/6;
B6 = 1/1  63/2 + 602/3  2100/4 + 3360/5  2520/6 + 720/7;
...
Note that in this algorithm, odd n's for the Bernoulli numbers sum to 0, not 1, and the sum for B1 = 1/2 = (1/1  1/2). B3 = 0 = (1  7/2 + 13/3  6/4) = 0. The summation for B4 = 1/30. (End)
Pursuant to Worpitzky's algorithm and given M = A028246 as an infinite lower triangular matrix, M * [1/1, 1/2, 1/3, ...] (i.e., the Harmonic series with alternate signs) = the Bernoulli numbers starting [1/1, 1/2, 1/6, ...].  Gary W. Adamson, Mar 22 2012
G(x,t) = 1/(1 + (1exp(x*t))/t) = 1 + 1 x + (2 + t)*x^2/2! + (6 + 6t + t^2)*x^3/3! + ... gives row polynomials for A090582, the fpolynomials for the permutohedra (see A019538).
G(x,t1) = 1 + 1*x + (1 + t)*x^2 / 2! + (1 + 4t + t^2)*x^3 / 3! + ... gives row polynomials for A008292, the hpolynomials for permutohedra.
G[(t+1)x,1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ... gives row polynomials for the present triangle. (End)
The Worpitzky triangle seems to be an apt name for this triangle.  Johannes W. Meijer, Jun 18 2009
If Pascal's triangle is written as a lower triangular matrix and multiplied by A028246 written as an upper triangular matrix, the product is a matrix where the (i,j)th term is (i+1)^j. For example,
1,0,0,0 1,1,1, 1 1,1, 1, 1
1,1,0,0 * 0,1,3, 7 = 1,2, 4, 8
1,2,1,0 0,0,2,12 1,3, 9,27
1,3,3,1 0,0,0, 6 1,4,16,64
So, numbering all three matrices' rows and columns starting at 0, the (i,j) term of the product is (i+1)^j.  Jack A. Cohen (ProfCohen(AT)comcast.net), Aug 03 2010
The Fi1 and Fi2 triangle sums are both given by sequence A000670. For the definition of these triangle sums see A180662. The mirror image of the Worpitzky triangle is A130850.  Johannes W. Meijer, Apr 20 2011
Let S_n(m) = 1^m + 2^m + ... + n^m. Then, for n >= 0, we have the following representation of S_n(m) as a linear combination of the binomial coefficients:
S_n(m) = Sum_{i=1..n+1} a(i+n*(n+1)/2)*C(m,i). E.g., S_2(m) = a(4)*C(m,1) + a(5)*C(m,2) + a(6)*C(m,3) = C(m,1) + 3*C(m,2) + 2*C(m,3).  Vladimir Shevelev, Dec 21 2011
Given the set X = [1..n] and 1 <= k <= n, then a(n,k) is the number of sets T of size k of subset S of X such that S is either empty or else contains 1 and another element of X and such that any two elemements of T are either comparable or disjoint.  Michael Somos, Apr 20 2013
Working with the row and column indexing starting at 1, a(n,k) gives the number of kdimensional faces in the first barycentric subdivision of the standard ndimensional simplex (apply Brenti and Welker, Lemma 2.1). For example, the barycentric subdivision of the 2simplex (a triangle) has 1 empty face, 7 vertices, 12 edges and 6 triangular faces giving row 4 of this triangle as (1,7,12,6). Cf. A053440.  Peter Bala, Jul 14 2014
See A074909 and above g.f.s for associations among this array and the Bernoulli polynomials and their umbral compositional inverses.  Tom Copeland, Nov 14 2014
An e.g.f. G(x,t) = exp[P(.,t)x] = 1/t  1/[t+(1t)(1e^(xt^2))] = (1t) * x + (2t + 3t^2  t^3) * x^2/2! + (6t^2  12t^3 + 7t^4  t^5) * x^3/3! + ... for the shifted, reverse, signed polynomials with the first element nulled, is generated by the infinitesimal generator g(u,t)d/du = [(1u*t)(1(1+u)t)]d/du, i.e., exp[x * g(u,t)d/du] u eval. at u=0 generates the polynomials. See A019538 and the G. Rzadkowski link below for connections to the Bernoulli and Eulerian numbers, a Ricatti differential equation, and a soliton solution to the KdV equation. The inverse in x of this e.g.f. is Ginv(x,t) = (1/t^2)*log{[1t(1+x)]/[(1t)(1tx)]} = [1/(1t)]x + [(2tt^2)/(1t)^2]x^2/2 + [(3t^23t^3+t^4)/(1t)^3]x^3/3 + [(4t^36t^4+4t^5t^6)/(1t)^4]x^4/4 + ... . The numerators are signed, shifted A135278 (reversed A074909), and the rational functions are the columns of A074909. Also, dG(x,t)/dx = g(G(x,t),t) (cf. A145271). (Analytic G(x,t) added, and Ginv corrected and expanded on Dec 28 2015.)  Tom Copeland, Nov 21 2014
The operator R = x + (1 + t) + t e^{D} / [1 + t(1e^(D))] = x + (1+t) + t  (t+t^2) D + (t+3t^2+2t^3) D^2/2!  ... contains an e.g.f. of the reverse row polynomials of the present triangle, i.e., A123125 * A007318 (with row and column offset 1 and 1). Umbrally, R^n 1 = q_n(x;t) = (q.(0;t)+x)^n, with q_m(0;t) = (t+1)^(m+1)  t^(m+1), the row polynomials of A074909, and D = d/dx. In other words, R generates the Appell polynomials associated with the base sequence A074909. For example, R 1 = q_1(x;t) = (q.(0;t)+x) = q_1(0;t) + q__0(0;t)x = (1+2t) + x, and R^2 1 = q_2(x;t) = (q.(0;t)+x)^2 = q_2(0:t) + 2q_1(0;t)x + q_0(0;t)x^2 = 1+3t+3t^2 + 2(1+2t)x + x^2. Evaluating the polynomials at x=0 regenerates the base sequence. With a simple sign change in R, R generates the Appell polynomials associated with A248727.  Tom Copeland, Jan 23 2015
The e.g.f. E(n, x) for {S(n, m)}_{m>=0} with S(n, m) = Sum_{k=1..m} k^n, n >= 0, (with undefined sum put to 0) is exp(x)*R(n+1, x) with the exponential row polynomials R(n, x) = Sum_{k=1..n} a(n, k)*x^k/k!. E.g., e.g.f. for n = 2, A000330: exp(x)*(1*x/1!+3*x^2/2!+2*x^3/3!).
The o.g.f. G(n, x) for {S(n, m)}_{m >=0} is then found by Laplace transform to be G(n, 1/p) = p*Sum_{k=1..n} a(n+1, k)/(p1)^(2+k).
Hence G(n, x) = x/(1  x)^(n+2)*Sum_{k=1..n} A008292(n,k)*x^(k1).
E.g., n=2: G(2, 1/p) = p*(1/(p1)^2 + 3/(p1)^3 + 2/(p1)^4) = p^2*(1+p)/(p1)^4; hence G(2, x) = x*(1+x)/(1x)^4.
This works also backwards: from the o.g.f. to the e.g.f. of {S(n, m)}_{m>=0}. (End)
a(n,k) is the number of ktuples of pairwise disjoint and nonempty subsets of a set of size n.  Dorian Guyot, May 21 2019
a(n1,k) is the number of chains of length k in a partially ordered set formed from subsets of an nelement set ordered by inclusion such that the first term of the chains is either the empty set or an nelement set.
Also, a(n1,k) is the number of distinct klevel rooted fuzzy subsets of an nset ordered by set inclusion. (End)
The relations on p. 34 of Hasan (also p. 17 of Franco and Hasan) agree with the relation between A019538 and this entry given in the formula section.  Tom Copeland, May 14 2020
T(n,k) is the size of the Green's Lclasses in the Dclasses of rank (k1) in the semigroup of partial transformations on an (n1)set.  Geoffrey Critzer, Jan 09 2023


LINKS

A. Riskin and D. Beckwith, Problem 10231, Amer. Math. Monthly, 102 (1995), 175176.


FORMULA

a(n, k) = S2(n, k)*(k1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.
Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938, but the notation is different.
The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t) = t(t+1)(d/dt)P(n, t) for n >= 1 (see the Riskin and Beckwith reference).  Emeric Deutsch, Aug 09 2005
Deltamatrix as can be read from H. Hasse's proof of a connection between the zetafunction and Bernoulli numbers (see link below).
Let P = lower triangular matrix with entries P[row,col] = binomial(row,col).
Let J = unit matrix with alternating signs J[r,r]=(1)^r.
Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)> natural numbers.
Let V = Vandermonde matrix with V[r,c] = (r+1)^c.
V is then also N(0)N(1)N(2)N(3)... (indices r,c always beginning at 0).
Then Delta = P*J * V and B' = N(1)' * Delta, where B is the column matrix of Bernoulli numbers and ' means transpose, or for the single kth Bernoulli number B_k with the appropriate column of Delta,
B_k = N(1)' * Delta[ *,k ] = N(1)' * P*J * N(k).
Using a single column instead of V and assuming infinite dimension, H. Hasse showed that in x = N(1) * P*J * N(s), where s can be any complex number and s*zeta(1s) = x.
His theorem reads: s*zeta(1s) = Sum_{n>=0..inf} (n+1)^1*delta(n,s), where delta(n,s) = Sum_{j=0..n} (1)^j * binomial(n,j) * (j+1)^s.
(End)
a(n,k) = k*a(n1,k) + (k1)*a(n1,k1) with a(n,1) = 1 and a(n,n) = (n1)!.  Johannes W. Meijer, Jun 18 2009
Rephrasing the Meijer recurrence above: Let M be the (n+1)X(n+1) bidiagonal matrix with M(r,r) = M(r,r+1) = r, r >= 1, in the two diagonals and the rest zeros. The row a(n+1,.) of the triangle is row 1 of M^n.  Gary W. Adamson, Jun 24 2011
With e.g.f.. A(x,t) = G[(t+1)x,1/(t+1)]1 (from 2008 comment) = 1 + 1/[1(1+t)(1e^(x))] = (1+t)x + (1+3t+2t^2)x^2/2! + ..., the comp. inverse in x is
B(x,t)= log(t/(1+t)+1/((1+t)(1+x))) = (1/(1+t))x  ((1+2t)/(1+t)^2)x^2/2 + ((1+3t+3t^2)/(1+t)^3)x^3/3 + .... The numerators are the row polynomials of A074909, and the rational functions are (omitting the initial constants) signed columns of the reindexed Pascal triangle A007318.
Let h(x,t)= 1/(dB/dx) = (1+x)(1+t(1+x)), then the row polynomial P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated 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(1,t)=1+t. (Series added Dec 29 2015.)(End)
Let <n,k> denote the Eulerian numbers A173018(n,k), then T(n,k) = Sum_{j=0..n} <n,j>*binomial(nj,nk).  Peter Luschny, Jul 12 2013
Matrix product A007318 * A131689. The nth row polynomial R(n,x) = Sum_{k >= 1} k^(n1)*(x/(1 + x))^k, valid for x in the open interval (1/2, inf). Cf A038719. R(n,1/2) = (1)^(n1)*(2^n  1)*Bernoulli(n)/n.  Peter Bala, Jul 14 2014
nth row polynomial R(n,x) = (1+x) o (1+x) o ... o (1+x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E11 in the Bala link.  Peter Bala, Jan 12 2018
Sum_{i=0..k} binomial(k,i) * a(n,i) = (k+1)^n.
Sum_{k=0..n} a(n,k) = 2*A000670(n).
(End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry, A028246, are given by x^n * A_n(1 + 1/x;0). Other specializations of A_n(x;y) give A046802, A090582, A119879, A130850, and A248727.  Tom Copeland, Jan 24 2020
The row generating polynomials R(n,x) = Sum_{i=1..n} a(n,i) * x^i satisfy the recurrence equation R(n+1,x) = R(n,x) + Sum_{k=0..n1} binomial(n1,k) * R(k+1,x) * R(nk,x) for n >= 1 with initial value R(1,x) = x.  Werner Schulte, Jun 17 2021


EXAMPLE

The triangle a(n, k) starts:
n\k 1 2 3 4 5 6 7 8 9
1: 1
2: 1 1
3: 1 3 2
4: 1 7 12 6
5: 1 15 50 60 24
6: 1 31 180 390 360 120
7: 1 63 602 2100 3360 2520 720
8: 1 127 1932 10206 25200 31920 20160 5040
9: 1 255 6050 46620 166824 317520 332640 181440 40320

Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.
Also, for power sums, we have
S_0(n) = C(n,1);
S_1(n) = C(n,1) + C(n,2);
S_2(n) = C(n,1) + 3*C(n,2) + 2*C(n,3);
S_3(n) = C(n,1) + 7*C(n,2) + 12*C(n,3) + 6*C(n,4);
S_4(n) = C(n,1) + 15*C(n,2) + 50*C(n,3) + 60*C(n,4) + 24*C(n,5); etc.
(End)
For X = [1,2,3], the sets T are {{}}, {{},{1,2}}, {{},{1,3}}, {{},{1,2,3}}, {{},{1,2},{1,2,3}}, {{},{1,3},{1,2,3}} and so a(3,1)=1, a(3,2)=3, a(3,3)=2.  Michael Somos, Apr 20 2013


MAPLE

a := (n, k) > add((1)^(ki)*binomial(k, i)*i^n, i=0..k)/k;
seq(print(seq(a(n, k), k=1..n)), n=1..10);
T := (n, k) > add(eulerian1(n, j)*binomial(nj, nk), j=0..n):
seq(print(seq(T(n, k), k=0..n)), n=0..9); # Peter Luschny, Jul 12 2013


MATHEMATICA

a[n_, k_] = Sum[(1)^(ki) Binomial[k, i]*i^n, {i, 0, k}]/k; Flatten[Table[a[n, k], {n, 10}, {k, n}]] (* JeanFrançois Alcover, May 02 2011 *)


PROG

(PARI) {T(n, k) = if( k<0  k>n, 0, n! * polcoeff( (x / log(1 + x + x^2 * O(x^n) ))^(n+1), nk))}; /* Michael Somos, Oct 02 2002 */
(PARI) {T(n, k) = stirling(n, k, 2)*(k1)!}; \\ G. C. Greubel, May 31 2019
(Sage)
x = polygen(ZZ, 'x')
A = []
for m in range(0, n, 1) :
A.append((x)^m)
for j in range(m, 0, 1):
A[j  1] = j * (A[j  1]  A[j])
return list(A[0])
(Sage) [[stirling_number2(n, k)*factorial(k1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, May 30 2019
(Magma) [[StirlingSecond(n, k)*Factorial(k1): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019
(GAP) Flat(List([1..10], n> List([1..n], k> Stirling2(n, k)* Factorial(k1) ))) # G. C. Greubel, May 30 2019
(Python) # Assuming offset (n, k) = (0, 0).
def T(n, k):
if k > n: return 0
if k == 0: return 1
return k*T(n  1, k  1) + (k + 1)*T(n  1, k)
for n in range(9):
print([T(n, k) for k in range(n + 1)]) # Peter Luschny, Apr 26 2022


CROSSREFS

Dropping the column of 1's gives A053440.
Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.
Row sums give A000629(n1) for n >= 1.
Cf. A007318, A008292, A046802, A074909, A090582, A123125, A130850, A135278, A141618, A145271, A163626, A248727, A263634.
Diagonal gives A000142(n1), for n >=1.
Nexttolast diagonal gives A001710,


KEYWORD



AUTHOR



EXTENSIONS

Definition corrected by Li Guo, Dec 16 2006


STATUS

approved



