A129186 -id:A129186 - cp's OEIS frontend

A153869 Triangle read by rows, A129186 * A128064(unsigned).

1, 1, 0, 1, 2, 0, 0, 2, 3, 0, 0, 0, 3, 4, 0, 0, 0, 0, 4, 5, 0, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 0, 6, 7, 0, 0, 0, 0, 0, 0, 0, 7, 8, 0, 0, 0, 0, 0, 0, 0, 0, 8, 9, 0

Offset: 1

Gary W. Adamson, Jan 03 2009

Lim_{k->inf} A153869^n = A000255: (1, 1, 3, 11, 53, 309, 2119,...).
Row sums = (1, 1, 3, 5, 7, 9,...).
A153869 * (1, 2, 3,...) = A001844 prefaced with a 1: (1, 1, 5, 13, 25, 41,...).

			First few rows of the triangle =
1;
1, 0;
1, 2, 0;
0, 2, 3, 0;
0, 0, 3, 4, 0;
0, 0, 0, 4, 5, 0;
0, 0, 0, 0, 5, 6, 0;
0, 0, 0, 0, 0, 6, 7, 0;
0, 0, 0, 0, 0, 0, 7, 8, 0;
...

Cf. A128064, A129186, A000255, A001844

Triangle read by rows, A129186 * A128064; where A129186 = a shift operator, shifting down triangle A128064(unsigned) one row and inserting a "1" at (1,1).

A154559 Triangle read by rows, A007318 * (A129186 * (A001006 * 0^(n-k))).

Original entry on oeis.org

1, 2, 0, 3, 2, 0, 4, 6, 4, 0, 5, 12, 16, 9, 0, 6, 20, 40, 45, 21, 0, 7, 30, 80, 135, 126, 51, 0, 8, 42, 140, 315, 441, 357, 127, 0, 9, 56, 224, 630, 1176, 1428, 1016, 323, 0, 10, 72, 336, 1134, 2646, 4284, 4572, 2907, 835, 0

Offset: 1

Gary W. Adamson, Jan 11 2009

Row sums = the Catalan numbers, A000108, starting with offset 1:

(1, 2, 5, 14, 42,...).

			First few rows of the triangle =
1;
2, 0;
3, 2, 0;
4, 6, 4, 0;
5, 12, 16, 9, 0;
6, 20, 40, 45, 21, 0;
7, 30, 80, 135, 126, 51, 0;
8, 42, 140, 315, 441, 357, 127, 0;
9, 56, 224, 630, 1176, 1428, 1016, 323, 0;
10, 72, 336, 1134, 2646, 4284, 4572, 2907, 835, 0;
11, 90, 480, 1890, 5292, 10710, 15240, 14535, 8350, 2188;
12, 110, 660, 2970, 9702, 23562, 41910, 53295, 45925, 24068, 5798;
...

Cf. A001006, A000108

Triangle read by rows, A007318 * (A129186 * (A001006 * 0^(n-k)))
Binomial transform of a bidiagonal matrix with (1,0,0,0,..,.) as the main
diagonal and A001006 as the subddiagonal starting (1, 2, 4, 9, 21, 51,...).

A182703 Triangle read by rows: T(n,k) = number of occurrences of k in the last section of the set of partitions of n.

Original entry on oeis.org

1, 1, 1, 2, 0, 1, 3, 2, 0, 1, 5, 1, 1, 0, 1, 7, 4, 2, 1, 0, 1, 11, 3, 2, 1, 1, 0, 1, 15, 8, 3, 3, 1, 1, 0, 1, 22, 7, 6, 2, 2, 1, 1, 0, 1, 30, 15, 6, 5, 3, 2, 1, 1, 0, 1, 42, 15, 10, 5, 4, 2, 2, 1, 1, 0, 1, 56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1

Offset: 1

Omar E. Pol, Nov 28 2010

For the definition of "section" of the set of partitions of n see A135010.
Also, column 1 gives the number of partitions of n-1. For k >= 2, row n lists the number of k's in all partitions of n that do not contain 1 as a part.
From Omar E. Pol, Feb 12 2012: (Start)
It appears that reversed rows converge to A002865.
It appears that row n is also the base of an isosceles triangle in which the column sums give the partition numbers A000041 in descending order starting with p(n-1) = A000041(n-1). Example for n = 7:
.
.         1,
.      1, 0, 1,
.   4, 2, 1, 0, 1,
11, 3, 2, 1, 1, 0, 1,
---------------------
11, 7, 5, 3, 2, 1, 1,
.
It appears that in row n starts an infinite trapezoid in which column sums always give the number of partitions of n-1. Example for n = 7:
.
11, 3, 2, 1, 1, 0, 1,
.   8, 3, 3, 1, 1, 0, 1,
.      6, 2, 2, 1, 1, 0, 1,
.         5, 3, 2, 1, 1, 0, 1,
.            4, 2, 2, 1, 1, 0, 1,
.               5, 2, 2, 1, 1, 0,...
.                  4, 2, 2, 1, 1,...
.                     4, 2, 2, 1,...
.                        4, 2, 2,...
.                           4, 2,...
.                              4,...
.
The sum of any column is always p(7-1) = p(6) = A000041(6) = 11.
It appears that the first term of row n is one of the vertices of an infinite isosceles triangle in which column sums give the partition numbers A000041 in ascending order starting with p(n-1) = A000041(n-1). Example for n = 7:
11,
.    8,
.    7,  6,
.        6,  5,
.       10,  5, ...
.           10, ...
.           10, ...
-------------------
11, 15, 22, 30, ...
(End)
It appears that row n lists the first differences of the row n of triangle A207031 together with 1 (as the final term of row n). - Omar E. Pol, Feb 26 2012
More generally T(n,k) is the number of occurrences of k in the n-th section of the set of partitions of any integer >= n. - Omar E. Pol, Oct 21 2013

			Illustration of three arrangements of the last section of the set of partitions of 7, or more generally the 7th section of the set of partitions of any integer >= 7:
.                                        _ _ _ _ _ _ _
.     (7)                    (7)        |_ _ _ _      |
.     (4+3)                (4+3)        |_ _ _ _|_    |
.     (5+2)                (5+2)        |_ _ _    |   |
.     (3+2+2)            (3+2+2)        |_ _ _|_ _|_  |
.       (1)                  (1)                    | |
.         (1)                (1)                    | |
.         (1)                (1)                    | |
.           (1)              (1)                    | |
.         (1)                (1)                    | |
.           (1)              (1)                    | |
.           (1)              (1)                    | |
.             (1)            (1)                    | |
.             (1)            (1)                    | |
.               (1)          (1)                    | |
.                 (1)        (1)                    |_|
.    ----------------
.     19,8,5,3,2,1,1 --> Row 7 of triangle A207031.
.      |/|/|/|/|/|/|
.     11,3,2,1,1,0,1 --> Row 7 of this triangle.
.
Note that the "head" of the last section is formed by the partitions of 7 that do not contain 1 as a part. The "tail" is formed by A000041(7-1) parts of size 1. The number of rows (or zones) is A000041(7) = 15. The last section of the set of partitions of 7 contains eleven 1's, three 2's, two 3's, one 4, one 5, there are no 6's and it contains one 7. So, for k = 1..7, row 7 gives: 11, 3, 2, 1, 1, 0, 1.
Triangle begins:
   1;
   1,  1;
   2,  0,  1;
   3,  2,  0,  1;
   5,  1,  1,  0, 1;
   7,  4,  2,  1, 0, 1;
  11,  3,  2,  1, 1, 0, 1;
  15,  8,  3,  3, 1, 1, 0, 1;
  22,  7,  6,  2, 2, 1, 1, 0, 1;
  30, 15,  6,  5, 3, 2, 1, 1, 0, 1;
  42, 15, 10,  5, 4, 2, 2, 1, 1, 0, 1;
  56, 27, 14, 10, 5, 5, 2, 2, 1, 1, 0, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Omar E. Pol, Illustration of the section model of partitions, Figure 1 (n = 1..6), Figure 2 (2D view, n = 1..10), Figure 3 (3D view, n = 1..9)

Row sums give A138137. Where records occur is A134869.
Columns 1-10: A000041, A182712-A182714, A206555-A206560.
Sub-triangles (1-11): A023531, A129186, A194702-A194710
Cf. A066633, A135010, A182742, A182743, A194812, A206563, A207031, A207032, A206437, A211025.

p:= (f, g)-> zip((x, y)-> x+y, f, g, 0):
b:= proc(n,i) option remember; local g;
      if n=0        then [1]
    elif n<2 or i<2 then [0]
    else g:=   `if`(i>n, [0],  b(n-i, i));
         p(p([0$j=2..i, g[1]], b(n, i-1)), g)
      fi
    end:
h:= proc(n) option remember;
      `if`(n=0, 1, b(n, n)[1]+h(n-1))
    end:
T:= proc(n) h(n-1), b(n, n)[2..n][] end:
seq(T(n), n=1..20);  # Alois P. Heinz, Feb 19 2012

p[f_, g_] := Plus @@ PadRight[{f, g}]; b[n_, i_] := b[n, i] = Module[{g}, Which[n == 0, {1}, n<2 || i<2, {0}, True, g = If [i>n, {0}, b[n-i, i]]; p[p[Append[Array[0&, i-1], g[[1]]], b[n, i-1]], g]]]; h[n_] := h[n] = If[n == 0, 1, b[n, n][[1]] + h[n-1]]; t[n_] := {h[n-1], Sequence @@ b[n, n][[2 ;; n]]}; Table[t[n], {n, 1, 20}] // Flatten (* Jean-François Alcover, Jan 16 2014, after Alois P. Heinz's Maple code *)
Table[{PartitionsP[n-1]}~Join~Table[Count[Flatten@Cases[IntegerPartitions[n], x_ /; Last[x] != 1], k], {k,2,n}], {n,1,12}]  // Flatten (* Robert Price, May 15 2020 *)

It appears that T(n,k) = A207032(n,k) - A207032(n,k+2). - Omar E. Pol, Feb 26 2012

A071919 Number of monotone nondecreasing functions [n]->[m] for n >= 0, m >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 0

Offset: 0

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 14 2002

Sometimes called a Riordan array.
Number of different partial sums of 1 + [2,3] + [3,4] + [4,5] + ... - Jon Perry, Jan 01 2004
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 05 2005
T(n,k)=abs(A110555(n,k)), A110555(n,k)=T(n,k)*(-1)^k. - Reinhard Zumkeller, Jul 27 2005
(1,0)-Pascal triangle. - Philippe Deléham, Nov 21 2006
A129186*A007318 as infinite lower triangular matrices. - Philippe Deléham, Mar 07 2009
Let n>=0 index the rows and m>=0 index the columns of this rectangular array. R(n,m) is "m multichoose n", the number of multisets of length n on m symbols. R(n,m) = Sum_{i=0..n} R(i,m-1). The summation conditions on the number of members in a size n multiset that are not the element m (an arbitrary element in the set of m symbols). R(n,m) = Sum_{i=1..m} R(n-1,i). The summation conditions on the largest element in a size n multiset on {1,2,...,m}. - Geoffrey Critzer, Jun 03 2009
Sum_{k=0..n} T(n,k)*B(k) = B(n), n>=0, with the Bell numbers B(n):=A000110(n) (eigensequence). See, e.g., the W. Lang link, Corollary 4. - Wolfdieter Lang, Jun 23 2010
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			   1,    1,    1,    1,    1,    1,    1,    1,    1, ...
   0,    1,    2,    3,    4,    5,    6,    7,    8, ...
   0,    1,    3,    6,   10,   15,   21,   28,   36, ...
   0,    1,    4,   10,   20,   35,   56,   84,  120, ...
   0,    1,    5,   15,   35,   70,  126,  210,  330, ...
   0,    1,    6,   21,   56,  126,  252,  462,  792, ...
   0,    1,    7,   28,   84,  210,  462,  924, 1716, ...
   0,    1,    8,   36,  120,  330,  792, 1716, 3432, ...
   0,    1,    9,   45,  165,  495, 1287, 3003, 6435, ...

G. C. Greubel, Table of n, a(n) for the first 101 antidiagonals, flattened
D. Merlini, F. Uncini and M. C. Verri, A unified approach to the study of general and palindromic compositions, Integers 4 (2004), A23, 26 pp.
Wolfdieter Lang, Simple proofs of some facts related to the Bell sequence and triangles A007318 (Pascal) and A071919 (enlarged Pascal). [From _Wolfdieter Lang_, Jun 23 2010]

Cf. A000110, A007318, A228196, A228576.
Columns are A000007, A000012, A000027, A000217, A000292, A000332, A000389, ...
Main diagonal gives A088218.

A:= (n, m)-> binomial(n+m-1, n):
seq(seq(A(n, d-n), n=0..d), d=0..14);  # Alois P. Heinz, Jan 13 2017

Table[Table[Binomial[m - 1 + n, n], {m, 0, 10}], {n, 0, 10}] // Grid (* Geoffrey Critzer, Jun 03 2009 *)
a[n_, m_] := Binomial[m - 1 + n, n]; Table[Table[a[n, m - n], {n, 0, m}], {m, 0, 10}] // Flatten (* G. C. Greubel, Nov 22 2017 *)

{ n=20; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+i; v[i][j+k]=v[i-1][j]+i+1)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } \\ Jon Perry

{a(n) = my(m); if( n<1, n==0, m = (sqrtint(8*n+1) - 1)\2; binomial(m-1, n - m*(m+1)/2))}; /* Michael Somos, Aug 20 2006 */

Limit_{k->infinity} A071919^k = (A000110,0,0,0,0,...) with the Bell numbers in the first column. For a proof see, e.g., the W. Lang link, proposition 12.
A(n,k) = binomial(n+k-1,n). - Reinhard Zumkeller, Jul 27 2005
G.f.: 1 + x + x^3(1+x) + x^6(1+x)^2 + x^10(1+x)^3 + ... . - Michael Somos, Aug 20 2006
G.f. of the triangular interpretation: (-1+x*y)/(-1+x*y+x). - R. J. Mathar, Aug 11 2015

A129185 Shift operator, left.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Gary W. Adamson, Apr 01 2007

Let A129185 = M, then M*V (V a vector), shifts V to the left. Example M*V, V = [1, 2, 3, ...] = [2, 3, 4, ...]. A129184 = right shift operator.

			First few rows of the matrix:
  0, 1;
  0, 0, 1;
  0, 0, 0, 1;
  0, 0, 0, 0, 1;
  ...

Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Cf. A129184, A129186.

Table[PadLeft[{1},n,0],{n,20}]//Flatten//Rest (* Harvey P. Dale, Jul 11 2020 *)

As an infinite matrix, all 1's in the superdiagonal and the rest zeros.

A194702 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (2 + m).

Original entry on oeis.org

2, 0, 2, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 2. For further information see A182703 and A135010.

			Triangle begins:
2,
0, 2,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 0, 1, 0, 1,
...
For k = 1 and  m = 1; T(1,1) = 2 because there are two parts of size 1 in the last section of the set of partitions of 3, since 2 + m = 3, so a(1) = 2. For k = 2 and m = 1; T(2,1) = 0 because there are no parts of size 2 in the last section of the set of partitions of 3, since 2 + m = 3, so a(2) = 0.

Always the sum of row k = p(2) = A000041(n) = 2.
The first (0-10) members of this family of triangles are A023531, A129186, this sequence, A194703-A194710.
Cf. A135010, A138121, A182712-A182714, A194812.

T(k,m) = A182703(2+m,k), with T(k,m) = 0 if k > 2+m.

T(k,m) = A194812(2+m,k).

A194710 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (10 + m).

Original entry on oeis.org

42, 15, 27, 10, 14, 18, 5, 10, 10, 17, 4, 5, 8, 10, 15, 2, 5, 4, 8, 9, 14, 2, 2, 4, 5, 7, 9, 13, 1, 2, 2, 4, 4, 8, 8, 13, 1, 1, 2, 2, 4, 4, 7, 9, 12, 0, 1, 1, 2, 2, 4, 4, 7, 8, 13, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of row k is also the number of partitions of 10. For further information see A182703 and A135010.

			Triangle begins:
  42;
  15, 27;
  10, 14, 18;
   5, 10, 10, 17;
   4,  5,  8, 10, 15;
   2,  5,  4,  8,  9, 14;
   2,  2,  4,  5,  7,  9, 13;
   1,  2,  2,  4,  4,  8,  8, 13;
   1,  1,  2,  2,  4,  4,  7,  9, 12;
   0,  1,  1,  2,  2,  4,  4,  7,  8, 13;
   1,  0,  1,  1,  2,  2,  4,  4,  7,  8, 12;
   0,  1,  0,  1,  1,  2,  2,  4,  4,  7,  8, 12;
   0,  0,  1,  0,  1,  1,  2,  2,  4,  4,  7,  8, 12;
   0,  0,  0,  1,  0,  1,  1,  2,  2,  4,  4,  7,  8, 12;
  ...
For k = 1 and m = 1; T(1,1) = 42 because there are 42 parts of size 1 in the last section of the set of partitions of 11, since 10 + m = 11, so a(1) = 42. For k = 2 and m = 1; T(2,1) = 15 because there are 15 parts of size 2 in the last section of the set of partitions of 11, since 10 + m = 11, so a(2) = 15.

Always the sum of row k = p(10) = A000041(10) = 42.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194709, this sequence.
Cf. A002865, A135010, A138121, A194812.

T(k,m) = A182703(10+m,k), with T(k,m) = 0 if k > 10+m.
T(k,m) = A194812(10+m,k).
Beginning with row k=11 each row starts with (k-11) 0's and ends with the subsequence 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12, the initial terms of A002865. - Alois P. Heinz, Feb 15 2012

A129184 Shift operator, right.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Gary W. Adamson, Apr 01 2007

Let A129184 = matrix M, then M*V, (V a vector); shifts V to the right, preceded by zeros. Example: M*V, V = [1, 2, 3, ...] = [0, 1, 2, 3, ...]. A129185 = left shift operator.

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x)= n * P_(n-1)(x) and R P_n(x)= P_(n+1)(x), the matrix T represents the action of R in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x)= x^n/n!, L= DxD and R=D^(-1). - Tom Copeland, Nov 10 2012

			First few rows of the triangle:
  0;
  1, 0;
  0, 1, 0;
  0, 0, 1, 0;
  0, 0, 0, 1, 0;
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A129185, A129186.

Cf. A010054.

Infinite lower triangular matrix with all 1's in the subdiagonal and the rest zeros.
From Tom Copeland, Nov 10 2012: (Start)
Let M(t) = I/(I-t*T) = I + t*T + (t*T)^2 + ... where T is the shift operator matrix and I the Identity matrix. Then the inverse matrix is MI(t)=(I-tT) and M(t) is A000012 with each n-th diagonal multiplied by t^n. M(1)=A000012 with inverse MI(1)=A167374. Row sums of M(2), M(3), and M(4) are A000225, A003462, and A002450.
Let E(t)=exp(t*T) with inverse E(-t). Then E(t) is A000012 with each n-th diagonal multiplied by t^n/n! and each row represents e^t truncated at the n+1 term.
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x):
  1) b(0) = 0, b(n) = a(n-1),
  2) B(x) = x A(x), or
  3) EB(x) = D^(-1)  EA(x), where D^(-1)x^j/j! = x^(j+1)/(j+1)!.
The operator M(t) can be characterized as
  4)M(t)EA(x)= sum(n>=0)a(n)[e^(x*t)-[1+x*t+...+ (x*t)^(n-1)/(n-1)!]]/t^n
    = exp(a*D_y)[t*e^(x*t)-y*e(x*y)]/(t-y) 
    = [t*e^(x*t)-a*e(x*a)]/(t-a), umbrally where (a)^k=a_k,
  5)[M(t) * a]_n = a(0)t^n +a(1)t^(n-1)+a(2)t^(n-2)+...+a(n).
The exponentiated operator can be characterized as
  6) E(t) A(x) = exp(t*x) A(x),
  7) E(t) EA(x) = exp(t*D^(-1)) EA(x)
  8) [E(t) * a]_n = a(0)t^n/n! + a(1)t^(n-1)/(n-1)! + ... + a(n).
  (End)
a(n) = A010054(n+1). - Andrew Howroyd, Feb 02 2020

Terms a(46) and beyond from Andrew Howroyd, Feb 02 2020

A292604 Triangle read by rows, coefficients of generalized Eulerian polynomials F_{2}(x).

Original entry on oeis.org

1, 1, 0, 5, 1, 0, 61, 28, 1, 0, 1385, 1011, 123, 1, 0, 50521, 50666, 11706, 506, 1, 0, 2702765, 3448901, 1212146, 118546, 2041, 1, 0, 199360981, 308869464, 147485535, 24226000, 1130235, 8184, 1, 0

Offset: 0

Peter Luschny, Sep 20 2017

The generalized Eulerian polynomials F_{m}(x) are defined F_{m; 0}(x) = 1 for all m >= 0 and for n > 0:
F_{0; n}(x) = Sum_{k=0..n} A097805(n, k)*(x-1)^(n-k) with coeffs. in A129186.
F_{1; n}(x) = Sum_{k=0..n} A131689(n, k)*(x-1)^(n-k) with coeffs. in A173018.
F_{2; n}(x) = Sum_{k=0..n} A241171(n, k)*(x-1)^(n-k) with coeffs. in A292604.
F_{3; n}(x) = Sum_{k=0..n} A278073(n, k)*(x-1)^(n-k) with coeffs. in A292605.
F_{4; n}(x) = Sum_{k=0..n} A278074(n, k)*(x-1)^(n-k) with coeffs. in A292606.
The case m = 1 are the Eulerian polynomials whose coefficients are the Eulerian numbers which are displayed in Euler's triangle A173018.
Evaluated at x in {-1, 1, 0} these families of polynomials give for the first few m:
F_{m} :   F_{0}    F_{1}    F_{2}    F_{3}    F_{4}
x = -1:  A165326  A155585  A002105  A292609  A292607
x =  1:  A000012  A000142  A000680  A014606  A014608  ... (m*n)!/m!^n
x =  0:    --     A000012  A000364  A002115  A211212  ... m-alternating permutations of length m*n.
Note that the constant terms of the polynomials are the generalized Euler numbers as defined in A181985. In this sense generalized Euler numbers are also generalized Eulerian numbers.

			Triangle starts:
[n\k][    0        1        2       3     4  5  6]
--------------------------------------------------
[0][      1]
[1][      1,       0]
[2][      5,       1,       0]
[3][     61,      28,       1,      0]
[4][   1385,    1011,     123,      1,    0]
[5][  50521,   50666,   11706,    506,    1, 0]
[6][2702765, 3448901, 1212146, 118546, 2041, 1, 0]

G. Frobenius. Über die Bernoullischen Zahlen und die Eulerschen Polynome. Sitzungsber. Preuss. Akad. Wiss. Berlin, pages 200-208, 1910.

F_{0} = A129186, F_{1} = A173018, F_{2} is this triangle, F_{3} = A292605, F_{4} = A292606.
First column: A000364. Row sums: A000680. Alternating row sums: A002105.
Cf. A181985, A241171.

Coeffs := f -> PolynomialTools:-CoefficientList(expand(f), x):
A292604_row := proc(n) if n = 0 then return [1] fi;
add(A241171(n, k)*(x-1)^(n-k), k=0..n); [op(Coeffs(%)), 0] end:
for n from 0 to 6 do A292604_row(n) od;

T[n_, k_] /; 1 <= k <= n := T[n, k] = k (2 k - 1) T[n - 1, k - 1] + k^2 T[n - 1, k]; T[, 1] = 1; T[, _] = 0;
F[2, 0][] = 1; F[2, n][x_] := Sum[T[n, k] (x - 1)^(n - k), {k, 0, n}];
row[n_] := If[n == 0, {1}, Append[CoefficientList[ F[2, n][x], x], 0]];
Table[row[n], {n, 0, 7}] (* Jean-François Alcover, Jul 06 2019 *)

def A292604_row(n):
    if n == 0: return [1]
    S = sum(A241171(n, k)*(x-1)^(n-k) for k in (0..n))
    return expand(S).list() + [0]
for n in (0..6): print(A292604_row(n))

F_{2; n}(x) = Sum_{k=0..n} A241171(n, k)*(x-1)^(n-k) for n>0 and F_{2; 0}(x) = 1.

A194704 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).

Original entry on oeis.org

5, 1, 4, 1, 2, 2, 0, 1, 1, 3, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 4. For further information see A182703 and A135010.

			Triangle begins:
  5,
  1, 4,
  1, 2, 2,
  0, 1, 1, 3,
  1, 0, 1, 1, 2,
  ...
For k = 1 and m = 1: T(1,1) = 5 because there are five parts of size 1 in the last section of the set of partitions of 5, since 4 + m = 5, so a(1) = 5.
For k = 2 and m = 1: T(2,1) = 1 because there is only one part of size 2 in the last section of the set of partitions of 5, since 4 + m = 5, so a(2) = 1.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(4) = A000041(4) = 5.
The first (0-10) members of this family of triangles are A023531, A129186, A194702, A194703, this sequence, A194705-A194710.
Cf. A135010, A138121, A182712-A182714, A194812.

P(n)={my(M=matrix(n,n), d=4); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(4+m,k), with T(k,m) = 0 if k > 4+m.

T(k,m) = A194812(4+m,k).

Terms a(16) and beyond from Andrew Howroyd, Feb 19 2020

cp's OEIS Frontend

A153869 Triangle read by rows, A129186 * A128064(unsigned).

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A154559 Triangle read by rows, A007318 * (A129186 * (A001006 * 0^(n-k))).

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A182703 Triangle read by rows: T(n,k) = number of occurrences of k in the last section of the set of partitions of n.

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A071919 Number of monotone nondecreasing functions [n]->[m] for n >= 0, m >= 0, read by antidiagonals.

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PARI

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A129185 Shift operator, left.

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A194702 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (2 + m).

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A194710 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (10 + m).

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A129184 Shift operator, right.

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