A154921 Triangle read by rows, T(n, k) = binomial(n, k) * Sum_{j=0..n-k} E(n-k, j)*2^j, where E(n, k) are the Eulerian numbers A173018(n, k), for 0 <= k <= n.

1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 75, 52, 18, 4, 1, 541, 375, 130, 30, 5, 1, 4683, 3246, 1125, 260, 45, 6, 1, 47293, 32781, 11361, 2625, 455, 63, 7, 1, 545835, 378344, 131124, 30296, 5250, 728, 84, 8, 1, 7087261, 4912515, 1702548, 393372, 68166, 9450, 1092, 108, 9, 1

Offset: 0

Author

Mats Granvik, Jan 17 2009

Keywords

• nonn
• tabl

Comments

Previous name: Matrix inverse of A154926.

A000670 appears in the first column. A052882 appears in the second column. A000027 and A045943 appear as diagonals. An alternative to calculating the matrix inverse of A154926 is to move the term in the lower right corner to a position in the same column and calculate the determinant instead, which yields the same answer.

Matrix inverse of (2*I - P), where P is Pascal's triangle and I the identity matrix. See A162312 for the matrix inverse of (2*P - I) and some general remarks about arrays of the form M(a) := (I - a*P)^-1 and their connection with weighted sums of powers of integers. The present array equals (1/2)*M(1/2). - Peter Bala, Jul 01 2009

From Mats Granvik, Aug 11 2009: (Start)

The values in this triangle can be seen as permanents of the Pascal triangle analogous to the method in the Redheffer matrix. The elements satisfy (T(n,k)/T(n,k-1))*k = (T(n-1,k)/T(n,k))*n which converges to log(2) as n->oo and k->0. More generally to calculate log(x) multiply the negative values in A154926 by 1/(x-1) and calculate the matrix inverse. Then (T(n,k)/T(n,k-1))*k and (T(n-1,k)/T(n,k))*n in the resulting triangle converge to log(x).

This method for calculating log(x) converges faster than the Taylor series when x is greater than 5 or so. See chapter on Taylor series in Spiegel for comparison. (End)

Exponential Riordan array [1/(2-exp(x)),x]. - Paul Barry, Apr 06 2011

T(n,k) is the number of ordered set partitions of {1,2,...,n} such that the first block contains k elements. For k=0 the first block contains arbitrarily many elements. - Geoffrey Critzer, Jul 22 2013

A natural (signed) refinement of these polynomials is given by the Appell sequence e.g.f. e^(xt)/ f(t) = exp[tP.(x)] with the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... and with raising operator R = x - d[log(f(D)]/dD (cf. A263634). - Tom Copeland, Nov 06 2015

Examples

			From _Peter Bala_, Jul 01 2009: (Start)
Triangle T(n, k) begins:
n\k|     0     1     2     3     4     5     6
==============================================
0  |     1
1  |     1     1
2  |     3     2     1
3  |    13     9     3     1
4  |    75    52    18     4     1
5  |   541   375   130    30     5     1
6  |  4683  3246  1125   260    45     6     1
...
(End)
From _Mats Granvik_, Aug 11 2009: (Start)
Row 4 equals 75,52,18,4,1 because permanents of:
  1,0,0,0,1  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0
  1,1,0,0,0  1,1,0,0,1  1,1,0,0,0  1,1,0,0,0  1,1,0,0,0
  1,2,1,0,0  1,2,1,0,0  1,2,1,0,1  1,2,1,0,0  1,2,1,0,0
  1,3,3,1,0  1,3,3,1,0  1,3,3,1,0  1,3,3,1,1  1,3,3,1,0
  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,1
are:
     75         52         18          4          1
(End)
		

References

• Murray R. Spiegel, Mathematical handbook, Schaum's Outlines, p. 111.

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• R. B. Nelsen, Problem E3062, Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 1987), 376-377.
• R. B. Nelsen and H. Schmidt, Jr., Chains in Power Sets, Mathematics Magazine, Vol. 64, No. 1 (Feb., 1991), 23-31.

Crossrefs

Cf. A000629 (row sums), A000670, A007047, A052822 (column 1), A052841 (alt. row sums), A080253, A162312, A162313.

Cf. A263634, A099880 (T(2n,n)).

Programs

Maple

A154921_row := proc(n) local i,p; p := proc(n,x) option remember; local k;
if n = 0 then 1 else add(p(k,0)*binomial(n,k)*(1+x^(n-k)),k=0..n-1) fi end:
seq(coeff(p(n,x),x,i),i=0..n) end: for n from 0 to 5 do A154921_row(n) od;
# Peter Luschny, Jul 15 2012
T := (n,k) -> binomial(n,k)*add(combinat:-eulerian1(n-k,j)*2^j, j=0..n-k):
seq(print(seq(T(n,k), k=0..n)),n=0..6); # Peter Luschny, Feb 07 2015
# third Maple program:
b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=1..n)) end:
T:= (n, k)-> n!/k! *b(n-k):
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Feb 03 2019
# fourth Maple program:
p := proc(n, m) option remember; if n = 0 then 1 else
    (m + x)*p(n - 1, m) + (m + 1)*p(n - 1, m + 1) fi end:
row := n -> local k; seq(coeff(p(n, 0), x, k), k = 0..n):
for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 23 2023

Mathematica

nn = 8; a = Exp[x] - 1;
Map[Select[#, # > 0 &] &,
  Transpose[
   Table[Range[0, nn]! CoefficientList[
Series[x^n/n!/(1 - a), {x, 0, nn}], x], {n, 0, nn}]]] // Grid (* Geoffrey Critzer, Jul 22 2013 *)
E1[n_ /; n >= 0, 0] = 1; E1[n_, k_] /; k < 0 || k > n = 0; E1[n_, k_] := E1[n, k] = (n - k) E1[n - 1, k - 1] + (k + 1) E1[n - 1, k];
T[n_, k_] := Binomial[n, k] Sum[E1[n - k, j] 2^j, {j, 0, n - k}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 30 2018, after Peter Luschny *)

Sage

@CachedFunction
def Poly(n, x):
    return 1 if n == 0 else add(Poly(k,0)*binomial(n,k)*(x^(n-k)+1) for k in range(n))
R = PolynomialRing(ZZ, 'x')
for n in (0..6): print(R(Poly(n,x)).list()) # Peter Luschny, Jul 15 2012

Formula

From Peter Bala, Jul 01 2009: (Start)

TABLE ENTRIES

(1) T(n,k) = binomial(n,k)*A000670(n-k).

GENERATING FUNCTION

(2) exp(x*t)/(2-exp(t)) = 1 + (1+x)*t + (3+2*x+x^2)*t^2/2! + ....

PROPERTIES OF THE ROW POLYNOMIALS

The row generating polynomials R_n(x) form an Appell sequence. They appear in the study of the poset of power sets [Nelsen and Schmidt].

The first few values are R_0(x) = 1, R_1(x) = 1+x, R_2(x) = 3+2*x+x^2 and R_3(x) = 13+9*x+3*x^2+x^3.

The row polynomials may be recursively computed by means of

(3) R_n(x) = x^n + Sum_{k = 0..n-1} binomial(n,k)*R_k(x).

Explicit formulas include

(4) R_n(x) = (1/2)*Sum_{k >= 0} (1/2)^k*(x+k)^n,

(5) R_n(x) = Sum_{j = 0..n} Sum_{k = 0..j} (-1)^(j-k)*binomial(j,k) *(x+k)^n,

and

(6) R_n(x) = Sum_{j = 0..n} Sum_{k = j..n} k!*Stirling2(n,k) *binomial(x,k-j).

SUMS OF POWERS OF INTEGERS

The row polynomials satisfy the difference equation

(7) 2*R_m(x) - R_m(x+1) = x^m,

which easily leads to the evaluation of the weighted sums of powers of integers

(8) Sum_{k = 1..n-1} (1/2)^k*k^m = 2*R_m(0) - (1/2)^(n-1)*R_m(n).

For example, m = 2 gives

(9) Sum_{k = 1..n-1} (1/2)^k*k^2 = 6 - (1/2)^(n-1)*(n^2+2*n+3).

More generally we have

(10) Sum_{k=0..n-1} (1/2)^k*(x+k)^m = 2*R_m(x) - (1/2)^(n-1)*R_m(x+n).

RELATIONS WITH OTHER SEQUENCES

Sequences in the database given by particular values of the row polynomials are

(11) A000670(n) = R_n(0)

(12) A052841(n) = R_n(-1)

(13) A000629(n) = R_n(1)

(14) A007047(n) = R_n(2)

(15) A080253(n) = 2^n*R_n(1/2).

This last result is the particular case (x = 0) of the result that the polynomials 2^n*R_n(1/2+x/2) are the row generating polynomials for A162313.

The above formulas should be compared with those for A162312. (End)

From Peter Luschny, Jul 15 2012: (Start)

(16) A151919(n) = R_n(1/3)*3^n*(-1)^n

(17) A052882(n) = [x^1] R_n(x)

(18) A045943(n) = [x^(n-1)] R_n+1(x)

(19) A099880(n) = [x^n] R_2n(x). (End)

The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1} binomial(n,k)*p{k}(0)*(1+x^(n-k)). - Peter Luschny, Jul 15 2012

Extensions

New name by Peter Luschny, Feb 07 2015