A186020 -id:A186020 - cp's OEIS frontend

A186022 Diagonal sums of number triangle A186020.

1, 1, 3, 6, 19, 62, 239, 1013, 4723, 23870, 129779, 753784, 4651285, 30349577, 208575027, 1504582526, 11358446495, 89501837358, 734420232291, 6262593615501, 55392254339667, 507335283474158, 4804245661175599, 46971060027956608, 473534995965134505, 4916730522706507665

Offset: 0

Paul Barry, Feb 10 2011

nonn

A186021 a(n) = Bell(n)*(2 - 0^n).

Original entry on oeis.org

1, 2, 4, 10, 30, 104, 406, 1754, 8280, 42294, 231950, 1357140, 8427194, 55288874, 381798644, 2765917090, 20960284294, 165729739608, 1364153612318, 11665484410114, 103448316470744, 949739632313502, 9013431476894646, 88304011710168692, 891917738589610578, 9277180664459998706

Offset: 0

Paul Barry, Feb 10 2011

a(n) is the number of collections of subsets of {1,2,...,n-1} that are pairwise disjoint. a(n+1) = 2*Sum_{j=0..n} C(n,j)*Bell(j). For example a(3)=10 because we have: {}, {{}}, {{1}}, {{2}}, {{1,2}}, {{},{1}}, {{},{2}}, {{},{1,2}}, {{1},{2}}, {{},{1},{2}}. - Geoffrey Critzer, Aug 28 2014

a(n) is the number of collections of subsets of [n] that are pairwise disjoint and cover [n], with [0] = {}. For disjoint collections of nonempty subsets see A000110. For arbitrary collections of subsets see A000371. For arbitrary collections of nonempty subsets see A003465. - Manfred Boergens, May 02 2024 and Apr 09 2025

			a(4) = A060719(3) + 1 = 29 + 1 = 30.

G. C. Greubel, Table of n, a(n) for n = 0..575

Row sums of A186020 and A256894.
Main diagonal of A271466 (shifted) and A381682.
Cf. A000110, A000371, A003465.

[Bell(n)*(2-0^n): n in [0..50]]; // Vincenzo Librandi, Apr 06 2011

A186021List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([P[-1], op(P)]);
A := [op(A), P[-1]] od; A end: A186021List(26); # Peter Luschny, Mar 24 2022

Prepend[Table[2 Sum[Binomial[n, j] BellB[j], {j, 0, n}], {n, 0, 25}], 1] (* Geoffrey Critzer, Aug 28 2014 *)
With[{nmax = 50}, CoefficientList[Series[2*Exp[Exp[x] - 1] - 1, {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jul 24 2017 *)

x='x+O('x^50); Vec(serlaplace(2*exp(exp(x) - 1) -1)) \\ G. C. Greubel, Jul 24 2017

from itertools import accumulate
def A186021_list(size):
    if size < 1: return []
    L, accu = [1], [2]
    for _ in range(size-1):
        accu = list(accumulate([accu[-1]] + accu))
        L.append(accu[0])
    return L
print(A186021_list(26)) # Peter Luschny, Apr 25 2016

E.g.f.: 2*exp(exp(x)-1)-1. - Paul Barry, Apr 06 2011
a(n) = A000110(n)*A040000(n).
a(n+1) = 1 + Sum_{k=0..n} C(n,k)*a(k). - Franklin T. Adams-Watters, Oct 02 2011
From Sergei N. Gladkovskii, Nov 11 2012 to Mar 29 2013: (Start)
Continued fractions:
G.f.: A(x)= 1 + 2*x/(G(0)-x) where G(k)= 1 - x*(k+1)/(1 - x/G(k+1)).
G.f.: G(0)-1 where G(k) = 1-(x*k+1)/(x*k - 1 - x*(x*k - 1)/(x + (x*k + 1)/G(k+1))).
G.f.: (G(0)-2)/x - 1 where G(k) = 1 + 1/(1 - x/(x + (1 - x*k)/G(k+1))).
G.f.: (S-2)/x - 1 where S = 2*Sum_{k>=0} x^k/Product_{n=0..k-1}(1 - n*x).
G.f.: 1/(1-x) - x/(G(0)-x^2+x) where G(k) =x^2 + x - 1 + k*(2*x-x^2) - x^2*k^2 + x*(x*k - 1)*(x*k + 2*x - 1)^2/G(k+1).
E.g.f.: E(0) - 1 where E(k) = 1 + 1/(1 - 1/(1 + (k+1)/x*Bell(k)/Bell(k+1)/E(k+1))). (End)
a(n) = A060719(n-1) + 1, and the inverse binomial transform of A060719. - Gary W. Adamson, May 20 2013
G.f. A(x) satisfies: A(x) = 1 + (x/(1 - x)) * (1 + A(x/(1 - x))). - Ilya Gutkovskiy, Jun 30 2020

A121207 Triangle read by rows. The definition is by diagonals. The r-th diagonal from the right, for r >= 0, is given by b(0) = b(1) = 1; b(n+1) = Sum_{k=0..n} binomial(n+2,k+r)*a(k).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 9, 15, 1, 1, 5, 14, 31, 52, 1, 1, 6, 20, 54, 121, 203, 1, 1, 7, 27, 85, 233, 523, 877, 1, 1, 8, 35, 125, 400, 1101, 2469, 4140, 1, 1, 9, 44, 175, 635, 2046, 5625, 12611, 21147, 1, 1, 10, 54, 236, 952, 3488, 11226, 30846, 69161, 115975

Offset: 0

N. J. A. Sloane, based on email from R. J. Mathar, Dec 11 2006

From Paul D. Hanna, Dec 12 2006: (Start)
Consider the row reversal, which is A124496 with an additional left column (A000110 = Bell numbers). The matrix inverse of this triangle is very simple:
   1;
  -1,   1;
  -1,  -1,   1;
  -1,  -2,  -1,   1;
  -1,  -3,  -3,  -1,   1;
  -1,  -4,  -6,  -4,  -1,   1;
  -1,  -5, -10, -10,  -5,  -1,   1;
  -1,  -6, -15, -20, -15,  -6,  -1,   1;
  -1,  -7, -21, -35, -35, -21,  -7,  -1,   1;
  -1,  -8, -28, -56, -70, -56, -28,  -8,  -1,   1; ...
This gives the recurrence and explains why the Bell numbers appear. (End)
Triangle A160185 = reversal then deletes right border of 1's. - Gary W. Adamson, May 03 2009

			Triangle begins (compare also table 9.2 in the Gould-Quaintance reference):
  1;
  1, 1;
  1, 1,  2;
  1, 1,  3,  5;
  1, 1,  4,  9,  15;
  1, 1,  5, 14,  31, 52;
  1, 1,  6, 20,  54, 121, 203;
  1, 1,  7, 27,  85, 233, 523,  877;
  1, 1,  8, 35, 125, 400,1101, 2469,  4140;
  1, 1,  9, 44, 175, 635,2046, 5625, 12611, 21147;
  1, 1, 10, 54, 236, 952,3488,11226, 30846, 69161, 115975;
  1, 1, 11, 65, 309,1366,5579,20425, 65676,180474, 404663, 678570;
  1, 1, 12, 77, 395,1893,8494,34685,126817,407787,1120666,2512769,4213597;

Alois P. Heinz, Rows n = 0..140, flattened
Robert Dougherty-Bliss, Gosper's algorithm and Bell numbers, arXiv:2210.13520 [cs.SC], 2022.
Robert Dougherty-Bliss, Experimental Methods in Number Theory and Combinatorics, Ph. D. Dissertation, Rutgers Univ. (2024). See pp. 69-70.
H. W. Gould and Jocelyn Quaintance, A linear binomial recurrence and the Bell numbers and polynomials, Applicable Analysis and Discrete Mathematics, 1 (2007), 371-385.

Diagonals, reading from the right, are A000110, A040027, A045501, A045499, A045500.
A124496 is a very similar triangle, obtained by reversing the rows and appending a rightmost diagonal which is A000110, the Bell numbers. See also A046936, A298804, A186020, A160185.
T(2n,n) gives A297924.

function Gould_diag(diag, size)
    size < 1 && return []
    size == 1 && return [1]
    L = [1, 1]
    accu = ones(BigInt, diag)
    for _ in 1:size-2
        accu = cumsum(vcat(accu[end], accu))
        L = vcat(L, accu[end])
    end
L end # Peter Luschny, Mar 30 2022

# This is the Jovovic formula with general index 'd'
# where A040027, A045499, etc. use one explicit integer
# Index n+1 is shifted to n from the original formula.
Gould := proc(n, d) local k;
    if n <= 1 then return 1 else
    return add(binomial(n-1+d, k+d)*Gould(k, d), k=0..n-1);
    fi
end:
# row and col refer to the extrapolated super-table:
# working up to row, not row-1, shows also the Bell numbers
# at the end of each row.
for row from 0 to 13 do
    for col from 0 to row do
       # 'diag' is constant for one of A040027, A045499 etc.
       diag := row - col;
       printf("%4d, ", Gould(col, diag));
    od;
    print();
od; # R. J. Mathar
# second Maple program:
T:= proc(n, k) option remember; `if`(k=0, 1,
      add(T(n-j, k-j)*binomial(n-1, j-1), j=1..k))
    end:
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Jan 08 2018

g[n_ /; n <= 1, ] := 1; g[n, d_] := g[n, d] = Sum[ Binomial[n-1+d, k+d]*g[k, d], {k, 0, n-1}]; Flatten[ Table[ diag = row-col; g[col, diag], {row, 0, 13}, {col, 0, row}]] (* Jean-François Alcover, Nov 25 2011, after R. J. Mathar *)
T[n_, k_] := T[n, k] = If[k == 0, 1, Sum[T[n-j, k-j] Binomial[n-1, j-1], {j, 1, k}]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2018, after Alois P. Heinz *)

# Computes the n-th diagonal of the triangle reading from the right.
from itertools import accumulate
def Gould_diag(diag, size):
    if size < 1: return []
    if size == 1: return [1]
    L, accu = [1,1], [1]*diag
    for _ in range(size-2):
        accu = list(accumulate([accu[-1]] + accu))
        L.append(accu[-1])
    return L # Peter Luschny, Apr 24 2016

A124496 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} in which the size of the last block is k, 1<=k<=n; the blocks are ordered with increasing least elements.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 9, 4, 1, 1, 31, 14, 5, 1, 1, 121, 54, 20, 6, 1, 1, 523, 233, 85, 27, 7, 1, 1, 2469, 1101, 400, 125, 35, 8, 1, 1, 12611, 5625, 2046, 635, 175, 44, 9, 1, 1, 69161, 30846, 11226, 3488, 952, 236, 54, 10, 1, 1, 404663, 180474, 65676, 20425, 5579, 1366, 309, 65, 11, 1, 1

Offset: 1

Emeric Deutsch, Nov 14 2006

Number of restricted growth functions of length n with a multiplicity k of the maximum value. RGF's are here defined as f(1)=1, f(i) <= 1+max_{1<=jR. J. Mathar, Mar 18 2016

This is table 9.2 in the Gould-Quaintance reference. - Peter Luschny, Apr 25 2016

			T(4,2) = 4 because we have 13|24, 14|23, 12|34 and 1|2|34.
Triangle starts:
  1;
  1,1;
  3,1,1;
  9,4,1,1;
  31,14,5,1,1;
  121,54,20,6,1,1;
  523,233,85,27,7,1,1;
  2469,1101,400,125,35,8,1,1;
  12611,5625,2046,635,175,44,9,1,1;
  69161,30846,11226,3488,952,236,54,10,1,1;
  404663,180474,65676,20425,5579,1366,309,65,11,1,1;
  2512769,1120666,407787,126817,34685,8494,1893,395,77,12,1,1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
H. W. Gould and Jocelyn Quaintance, A linear binomial recurrence and the Bell numbers and polynomials. Applicable Analysis and Discrete Mathematics, 1 (2007), 371-385.

Row sums are the Bell numbers (A000110). It seems that T(n, 1), T(n, 2), T(n, 3) and T(n, 4) are given by A040027, A045501, A045499 and A045500, respectively. A121207 gives a very similar triangle.
T(2n,n) gives A297924.
Cf. A000110, A040027, A045501, A045499, A045500, A186020, A160185.

Q[1]:=t*s: for n from 2 to 12 do Q[n]:=expand(t*s*subs(t=1,Q[n-1])+s*diff(Q[n-1],s)+t*Q[n-1]-Q[n-1]) od:for n from 1 to 12 do P[n]:=sort(subs(s=1,Q[n])) od: for n from 1 to 12 do seq(coeff(P[n],t,j),j=1..n) od;
# second Maple program:
T:= proc(n, k) option remember; `if`(n=k, 1,
      add(T(n-j, k)*binomial(n-1, j-1), j=1..n-k))
    end:
seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Jul 05 2016

T[n_, k_] := T[n, k] = If[n == k, 1, Sum[T[n-j, k]*Binomial[n-1, j-1], {j, 1, n-k}]];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten; (* Jean-François Alcover, Jul 21 2016, after Alois P. Heinz *)

The row enumerating polynomial P[n](t)=Q[n](t,1), where Q[1](t,s)=ts and Q[n](t,s)=s*dQ[n-1](t,s)/ds +(t-1)Q[n-1](t,s)+tsQ[n-1](1,s) for n>=2.
A008275^-1*ONES*A008275 or A008277*ONES*A008277^-1 where ONES is a triangle with all entries = 1. [From Gerald McGarvey, Aug 20 2009]
Conjectures: T(n,n-3) = A000096(n). T(n,n-4)= A055831(n+1). - R. J. Mathar, Mar 13 2016

A160185 Triangle read by rows, (1 / ((-1)A129184 A007318 + I)) - I, I = Identity matrix.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 15, 9, 4, 1, 52, 31, 14, 5, 1, 203, 121, 54, 20, 6, 1, 877, 523, 233, 85, 27, 7, 1, 4140, 2469, 1101, 400, 125, 35, 8, 1, 21147, 12611, 5625, 2046, 635, 175, 44, 9, 1, 115975, 69161, 30846, 11226, 3488, 952, 236, 54, 10, 1

Offset: 0

Gary W. Adamson, May 03 2009

Inverse binomial transform of the triangle shifts to left (= adding I as right border, I = Identity matrix); resulting in reversed rows of A121207.
Left border = Bell numbers, A000110 = eigensequence of Pascal's triangle.
Successive columns from left to right = eigensequences of Pascal's triangle deleting columns one at a time.
Row sums of the triangle = A060719: (1, 3, 9, 29, 103, ...). - Gary W. Adamson, May 20 2013
From Gary W. Adamson, Jul 18 2019: (Start)
Rows are eigensequences of triangles exemplified by the following arrangement of binomial sequences. Example: row 5 is (1, 5, 14, 31, 52, 0, 0, 0, ...), the eigensequence of:
  1;
  4, 1;
  6, 3, 1;
  4, 3, 2, 1;
  1, 1, 1, 1, 1;
  ... and the rest zeros.
Similarly, the production matrix for (1, 6, 20, 54, 121, 203, 0, 0, 0, ...) is:
   1;
   5, 1;
  10, 4, 1;
  10, 6, 3, 1;
   5, 4, 3, 2, 1;
   1, 1, 1, 1, 1, 1;
  ... and the rest zeros. (End)

			First few rows of the triangle:
       1;
       2,     1;
       5,     3,     1;
      15,     9,     4,     1;
      52,    31,    14,     5,    1;
     203,   121,    54,    20,    6,   1;
     877,   523,   233,    85,   27,   7,   1;
    4140,  2469,  1101,   400,  125,  35,   8,  1;
   21147, 12611,  5625,  2046,  635, 175,  44,  9,  1;
  115975, 69161, 30846, 11226, 3488, 952, 236, 54, 10, 1;
  ...

Cf. A121207, A124496, A186020.

Cf. A060719.

Triangle read by rows, 1 / ((-1)*A129184 * A051731 + I), I = Identity matrix.

Equals reversal by rows of triangle A121207, then delete right border. A121207 begins: 1; 1, 1; 1, 1, 2 1, 1, 3, 5; ...

Corrected by Alois P. Heinz, Apr 18 2013

A212431 Triangle read by rows: row sums, right and left borders are the Bell sequence, or a shifted variant. See Comments for precise definition.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 5, 3, 2, 5, 15, 9, 8, 5, 15, 52, 31, 28, 25, 15, 52, 203, 121, 108, 100, 90, 52, 203, 877, 523, 466, 425, 405, 364, 203, 877, 4140, 2469, 2202, 2000, 1875, 1820, 1624, 877, 4140, 21147, 12611, 11250, 10230, 9525, 9100, 8932, 7893, 4140, 21147

Offset: 0

Gary W. Adamson, Jun 21 2012

Consider A186020 as an infinite lower triangular matrix, and multiply the columns successively by the Bell numbers A000110, (1, 1, 2, 5, 15, 52,...).
Right and left borders = the Bell numbers, A000110: (1, 1, 2, 5, 15,...). Row sums = the shifted Bell numbers, (1, 2, 5, 15, 52,...).
Sum of n-th row terms = rightmost and leftmost terms of next row.

			First few rows of the triangle are:
1;
1, 1
2, 1, 2;
5, 3, 2, 5;
15, 9, 8, 5, 15;
52, 31, 28, 25, 15, 52;
203, 121, 108, 100, 90, 52, 203;
877, 523, 466, 425, 405, 364, 203, 877;
4140, 2469, 2202, 2000, 1875, 1820, 1624, 877, 4140;
21147, 12611, 11250, 10230, 9525, 9100, 8932, 7893, 4140, 21147;
...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A000110, A186020.

b:= proc(n) option remember; `if`(n=0, [1, 0],
      add((p-> p+[0, p[1]*x^(n-j)])(b(n-j)*
      binomial(n-1, j-1)), j=1..n))
    end:
T:= n-> (p-> seq(`if`(i=n, p[1], coeff(
         p[2], x, i)), i=0..n))(b(n)):
seq(T(n), n=0..12);  # Alois P. Heinz, May 16 2017

b[n_] := b[n] = If[n == 0, {1, 0}, Sum[Function[p, p + {0, p[[1]]*x^(n - j)}][b[n - j]*Binomial[n - 1, j - 1]], {j, 1, n}]];
T[n_] := Function[p, Table[If[i == n, p[[1]], Coefficient[p[[2]], x, i]], {i, 0, n}]][b[n]];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 12 2018, after Alois P. Heinz *)

Edited by N. J. A. Sloane, Jun 22 2012

A343234 Triangle T read by rows: lower triangular Riordan matrix of the Toeplitz type with first column A067687.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 5, 2, 1, 1, 12, 5, 2, 1, 1, 29, 12, 5, 2, 1, 1, 69, 29, 12, 5, 2, 1, 1, 165, 69, 29, 12, 5, 2, 1, 1, 393, 165, 69, 29, 12, 5, 2, 1, 1, 937, 393, 165, 69, 29, 12, 5, 2, 1, 1, 2233, 937, 393, 165, 69, 29, 12, 5, 2, 1, 1

Offset: 0

Wolfdieter Lang, Apr 16 2021

This infinite lower triangular Riordan matrix T is the so-called L-eigen-matrix of the infinite lower triangular Riordan matrix A027293 (but with offset 0 for rows and columns). Such eigentriangles have been considered by Paul Barry in the paper given as a link in A186020.
This means that E is the L-eigen-matrix of an infinite lower triangular matrix M if M*E = L*(E - I), with the unit matrix I and the matrix L with elements L(i, j) = delta_{i, j-1} (Kronecker's delta-symbol; first upper diagonal with 1's).
Therefore, this notion is analogous to calling sequence S an L-eigen-sequence of matrix M if M*vec(S) = L.vec(S) (or vec(S) is an eigensequence of M - L with eigenvalue 0), used by Bernstein and Sloane, see the links in A155002.
L*(E - I) is the E matrix after elimination of the main diagonal and then the first row, and starting with offset 0. Because for infinite lower triangular matrices L^{tr}.L = I (tr stands for transposed), this leads to M = L*(I - E^{-1}) or E = (I - L^{tr}*M)^{-1}.
Note that Gary W. Adamson uses a different notion: E is the eigentriangle of a triangle T if the columns of E are the columns j of T multiplied by the sequence elements B_j of B with o.g.f. x/(1 - x*G(x)), with the o.g.f. G(x) of column no. 1 of T.  Or E(i, j) = T(i, j)*B(j). In short: sequence B is the L-eigen-sequence of the infinite lower triangular matrix T (but with offset 1): T*vec(B) = L.vec(B). See, e.g., A143866.
Thanks to Gary W. Adamson for motivating my occupation with such eigentriangles and eigensequences.
The first column of the present triangle T is A067687, which is then shifted downwards (Riordan of Toeplitz type).

			The triangle T begins:
n \ m   0   1   2   3  4  5  6  7  8  9 ...
-----------------------------------------
0:      1
1:      1   1
2:      2   1   1
3:      5   2   1   1
4:     12   5   2   1  1
5:     29  12   5   2  1  1
6:     69  29  12   5  2  1  1
7:    165  69  29  12  5  2  1  1
8:    393 165  69  29 12  5  2  1  1
9:    937 393 165  69 29 12  5  2  1  1
...

Cf. A000041, A027293, A067687, A143866, A155002, A186020.

Matrix elements: T(n, m) = A067687(n-m), for n >= m >= 0, and 0 otherwise.
O.g.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n, m)*x^m is
G(z, x) = 1/((1 - z*P(z))*(1 - x*z)), with the o.g.f. P of A000041 (number of partitions).
O.g.f. column m: G_m(x) = x^m/(1 - x*P(x)), for m >= 0.

A176663 T(n, k) = [x^k] Sum_{j=0..n} j!*binomial(x, j), for 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 2, -2, 1, 1, -4, 9, -5, 1, 1, 20, -41, 30, -9, 1, 1, -100, 233, -195, 76, -14, 1, 1, 620, -1531, 1429, -659, 161, -20, 1, 1, -4420, 11537, -11703, 6110, -1799, 302, -27, 1, 1, 35900, -98047, 106421, -61174, 20650, -4234, 519, -35, 1

Offset: 0

Roger L. Bagula, Apr 23 2010

			Triangle starts:
  {1},
  {1,       1},
  {1,       0,      1},
  {1,       2,     -2,        1},
  {1,      -4,      9,       -5,      1},
  {1,      20,    -41,       30,     -9,       1},
  {1,    -100,    233,     -195,     76,     -14,     1},
  {1,     620,  -1531,     1429,   -659,     161,   -20,     1},
  {1,   -4420,  11537,   -11703,   6110,   -1799,   302,   -27,   1},
  {1,   35900, -98047,   106421, -61174,   20650, -4234,   519, -35,   1},
  {1, -326980, 928529, -1066279, 662506, -248675, 59039, -8931, 835, -44, 1}

Row sums are A040000. Alternating row sums are A058006, which are also T(n,1).

Cf. A186020.

with(PolynomialTools):
T_row := n -> CoefficientList(expand(add(k!*binomial(x, k), k=0..n)), x):
ListTools:-Flatten([seq(T_row(n), n=0..9)]); # Peter Luschny, Jul 02 2019

p[x_, n_] := Sum[k! Binomial[x, k], {k, 0, n}];
Table[CoefficientList[FunctionExpand[p[x, n]], x], {n, 0, 10}] // Flatten
(* Alternative: *)
Table[CoefficientList[FunctionExpand[Sum[FactorialPower[x, k], {k, 0, n}]], x], {n, 0, 10}] // Flatten (* Peter Luschny, Jul 02 2019 *)

From Peter Luschny, Jul 02 2019: (Start)
Sum_{k=0..n} T(n, k)*x^k = Sum_{k=0..n} (x)_k, where (x)_k denotes the falling factorial.
Let T be the lower triangular matrix associated to the T(n, k) and S the lower triangular matrix associated to the Stirling set numbers S2(n, k). Then S*T = A186020 (seen as a matrix) and T*S = A000012 (seen as a matrix). (End)
T(n, k) = Sum_{i=0..n-k} Stirling1(i+k, k). - Igor Victorovich Statsenko, May 25 2024

Edited by Peter Luschny, Jul 02 2019

A186023 Eigentriangle of triangle A085478.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 6, 4, 1, 1, 23, 15, 6, 1, 1, 106, 68, 28, 8, 1, 1, 567, 365, 145, 45, 10, 1, 1, 3434, 2215, 877, 262, 66, 12, 1, 1, 23137, 14917, 5936, 1750, 427, 91, 14, 1, 1, 171174, 110324, 43936, 13020, 3108, 648, 120, 16, 1, 1, 1376525, 887232, 353067, 104904, 25125, 5091, 933, 153, 18, 1, 1

Offset: 0

Paul Barry, Feb 10 2011

First column is A125273. Inverse of A186024. Rows sums are A125273(n)*(2-0^n).

			Triangle begins
       1;
       1,      1;
       2,      1,     1;
       6,      4,     1,     1;
      23,     15,     6,     1,    1;
     106,     68,    28,     8,    1,   1;
     567,    365,   145,    45,   10,   1,   1;
    3434,   2215,   877,   262,   66,  12,   1,  1;
   23137,  14917,  5936,  1750,  427,  91,  14,  1, 1;
  171174, 110324, 43936, 13020, 3108, 648, 120, 16, 1, 1;

Paul Barry, Invariant number triangles, eigentriangles and Somos-4 sequences, arXiv preprint arXiv:1107.5490 [math.CO], 2011.

Cf. A125273, A186020, A186024,

cp's OEIS Frontend

A186022 Diagonal sums of number triangle A186020.

Views

Author

Keywords

A186021 a(n) = Bell(n)*(2 - 0^n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

Formula

A121207 Triangle read by rows. The definition is by diagonals. The r-th diagonal from the right, for r >= 0, is given by b(0) = b(1) = 1; b(n+1) = Sum_{k=0..n} binomial(n+2,k+r)*a(k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Maple

Mathematica

Python

A124496 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} in which the size of the last block is k, 1<=k<=n; the blocks are ordered with increasing least elements.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A160185 Triangle read by rows, (1 / ((-1)*A129184 * A007318 + I)) - I, I = Identity matrix.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A212431 Triangle read by rows: row sums, right and left borders are the Bell sequence, or a shifted variant. See Comments for precise definition.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A343234 Triangle T read by rows: lower triangular Riordan matrix of the Toeplitz type with first column A067687.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A176663 T(n, k) = [x^k] Sum_{j=0..n} j!*binomial(x, j), for 0 <= k <= n, triangle read by rows.

Views

Author

Keywords

Examples

Crossrefs

A160185 Triangle read by rows, (1 / ((-1)A129184 A007318 + I)) - I, I = Identity matrix.