A154926 -id:A154926 - cp's OEIS frontend

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

Mats Granvik, Jan 17 2009

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

			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)

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

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.

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

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

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

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 *)

@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

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

New name by Peter Luschny, Feb 07 2015

A135494 Triangle read by rows: row n gives coefficients C(n,j) for a Sheffer sequence (binomial-type) with lowering operator (D-1)/2 + T{ (1/2) * exp[(D-1)/2] } where T(x) is Cayley's Tree function.

Original entry on oeis.org

1, -1, 1, -1, -3, 1, -1, -1, -6, 1, -1, 5, 5, -10, 1, -1, 19, 30, 25, -15, 1, -1, 49, 49, 70, 70, -21, 1, -1, 111, -70, -91, 70, 154, -28, 1, -1, 237, -883, -1218, -861, -126, 294, -36, 1, -1, 491, -4410, -4495, -3885, -2877, -840, 510, -45, 1

Offset: 1

Tom Copeland, Feb 08 2008

The lowering (or delta) operator for these polynomials is L = (D-1)/2 + T{ (1/2) * exp[(D-1)/2] } and the raising operator is R = 2t * { 1 - T[ (1/2) * exp[(D-1)/2] ] }, where T(x) is the tree function of A000169. In addition, L = E(D,1) = A(D) where E(x,t) is the e.g.f. of A134991 and A(x) is the e.g.f. of A000311, so L = sum(j=1,...) A000311(j) * D^j / j! also. The polynomials and operators can be generalized through A134991.
Also the Bell transform of A153881. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016
Exponential Riordan array [2 - exp(x), 1 + 2*x - exp(x)] belonging to the derivative subgroup of the exponential Riordan group. See the example section for a factorization of this array as an infinite product of arrays. - Peter Bala, Feb 13 2025

			The triangle begins:
  [1]  1;
  [2] -1,  1;
  [3] -1, -3,  1;
  [4] -1, -1, -6,   1;
  [5] -1,  5,  5, -10,   1;
  [6] -1, 19, 30,  25, -15,   1;
  [7] -1, 49, 49,  70,  70, -21, 1.
P(3,t) = [B(.,-t) + 2t]^3 = B(3,-t) + 3B(2,-t)2t + 3B(1,-t)(2t)^2 + (2t)^3 = (-t + 3t^2 - t^3) + 3(-t + t^2)(2t) + 3(-t)(2t)^2 + (2t)^3 = -t - 3t + t^3.
From _Peter Bala_, Feb 13 2025: (Start)
The array factorizes as an infinite product of lower triangular arrays:
  /  1               \    / 1             \ / 1             \ / 1             \
  | -1   1           |   | -1  1          | | 0 -1          | | 0  1          |
  | -1  -3   1       | = | -1 -2   1      | | 0 -1  1       | | 0  0  1       | ...
  | -1  -1  -6   1   |   | -1 -3  -3  1   | | 0 -1 -2  1    | | 0  0 -1  1    |
  | -1   5   5 -10  1|   | -1 -4  -6 -4  1| | 0 -1 -3 -3  1 | | 0  0 -1 -2  1 |
  |...               |   |...             | |...            | |...            |
where the first array in the product on the right-hand side is A154926. (End)

S. Roman, The Umbral Calculus, Academic Press, New York, 1984.
G. Rota, Finite Operator Calculus, Academic Press, New York, 1975.

Vincenzo Librandi, Rows n = 1..25
Peter Bala, Factorisations of some Riordan arrays as infinite products
J. Taylor, Formal group laws and hypergraph colorings, doctoral thesis, Univ. of Wash., 2016, p. 95.

Cf. A000169, A000311, A008277, A074051, A126617, A134991, A154926, A264428.

Cf. A298673 for the inverse matrix.

# The function BellMatrix is defined in A264428.
# Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> `if`(n=0,1,-1), 9); # Peter Luschny, Jan 27 2016

max = 8; s = Series[Exp[t*(-Exp[x]+2*x+1)], {x, 0, max}, {t, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {t, 0, k}]*n!; Table[t[n, k], {n, 0, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 23 2014 *)
BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
M = BellMatrix[If[# == 0, 1, -1] &, rows];
Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)

Row polynomials are P(n,t) = Sum_{j=1..n} C(n,j) * t^j = [ Bell(.,-t) + 2t ]^n, umbrally, where Bell(j,t) are the Touchard/Bell/exponential polynomials described in A008277, with P(0,t) = 1.
E.g.f.: exp{ t * [ -exp(x) + 2x + 1] } and [ P(.,t) + P(.,s) ]^n = P(n,s+t).
The lowering operator gives L[P(n,t)] = n * P(n-1,t) = (D-1)/2 * P(n,t) + Sum_{j>=1} j^(j-1) * 2^(-j) / j! * exp(-j/2) * P(n,t + j/2).
The raising operator gives R[P(n,t)] = P(n+1,t) = 2t * { P(n,t) - Sum_{j>=1} j^(j-1) * 2^(-j) / j! * exp(-j/2) * P(n,t + j/2) } .
Therefore P(n+1,t) = 2t * { [ (1+D)/2 * P(n,t) ] - n * P(n-1,t) }.
P(n,1) = (-1)^n * A074051(n) and P(n,-1) = A126617(n).
See Rota, Roman, Mathworld or Wikipedia on Sheffer sequences and umbral calculus for more formulas, including expansion theorems.
From Tom Copeland, Jan 20 2018: (Start)
Define Q(n,z;w) = [Bell(.,w)+z]^n. Then Q(n,z;w) are a sequence of Appell polynomials with e.g.f. exp[(exp(t)-1+z)*w], lowering operator D = d/dz, and raising operator R = z + w*exp(D), and exp[(exp(D)-1)w] z^n = exp[Bell(.,w)D] z^n = Q(n,z;w) = e^(-w) (w d/dw + z)^n e^w =  e^(-w) exp(a.w) = exp[(a. - 1)w] with (a.)^k = a_k = (k + z)^n and (a. - 1)^m = sum{k = 0,..,m} (-1)^k a^(m-k). Then P(n,t) = Q(n,2t;-t).
For example, exp[(a. - 1)w] = (a. - 1)^0 + (a. - 1)^1 w + (a. - 1)^2 w^2/2! + ... = a_0 + (a_1 - a_0) w + (a_2 - 2a_1 + a_0) w^2/2! + ... = z^n + [(1+z)^n - z^n] w + [(2+z)^n - 2(1+z)^n + z^n] w^2/2! + ... . (End)
T(n+1, k) = Sum_{i = 0..n} s(n,k)*binomial(n, i)*T(i, k-1), where s(n,i) = 1 if i = n else -1. - Peter Bala, Feb 13 2025

More terms from Vincenzo Librandi, Jan 21 2018

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

Extensions

A135494 Triangle read by rows: row n gives coefficients C(n,j) for a Sheffer sequence (binomial-type) with lowering operator (D-1)/2 + T{ (1/2) * exp[(D-1)/2] } where T(x) is Cayley's Tree function.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A373164 Triangle read by rows: the exponential almost-Riordan array ( 1 | 2 - exp(x), x ).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula