A228909 -id:A228909 - cp's OEIS frontend

A163626 Triangle read by rows: The n-th derivative of the logistic function written in terms of y, where y = 1/(1 + exp(-x)).

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

Offset: 0

Richard V. Scholtz, III, Aug 01 2009

Apart from signs and offset, same as A028246. - Joerg Arndt, Nov 06 2016
Triangle T(n,k), read by rows, given by (1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...) DELTA (-1,-1,-2,-2,-3,-3,-4,-4,-5,-5,-6,-6,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2011
The "Stirling-Bernoulli transform" maps a sequence b_0, b_1, b_2, ... to a sequence c_0, c_1, c_2, ..., where if B has o.g.f. B(x), c has e.g.f. exp(x)*B(1 - exp(x)). More explicity, c_n = Sum_{k = 0..n} A163626(n,k)*b_k. - Philippe Deléham, May 26 2015
Row sums of absolute values of terms give A000629. - Yahia DJEMMADA, Aug 16 2016
This is the triangle of connection constants for expressing the monomial polynomials (-x)^n as a linear combination of the basis polynomials {binomial(x+n,n)}n>=0, that is, (-x)^n = Sum_{k = 0..n} T(n,k)*binomial(x+k,k). Cf. A145901. - Peter Bala, Jun 06 2019
Row sums for n > 0 are zero. - Shel Kaphan, May 14 2024
The Akiyama-Tanigawa algorithm applied to a sequence yields the same result as the Stirling-Bernoulli Transform applied to the same sequence.  See Philippe Deléham's comment of May 26 2015. - Shel Kaphan, May 16 2024

			y = 1/(1+exp(-x))
y^(0) = y
y^(1) = y-y^2
y^(2) = y-3*y^2+2*y^3
y^(3) = y-7*y^2+12*y^3-6*y^4
Triangle begins :
n\k 0     1     2     3     4     5    6
----------------------------------------
0:  1
1:  1    -1
2:  1    -3     2
3:  1    -7    12    -6
4:  1   -15    50   -60    24
5:  1   -31   180  -390   360  -120
6:  1   -63   602 -2100  3360 -2520  720
7:  1  -127 ... - Reformatted by _Philippe Deléham_, May 26 2015
Change of basis constants: x^4 = 1 - 15*binomial(x+1,1) + 50*binomial(x+2,2) - 60*binomial(x+3,3) + 24*binomial(x+4,4). - _Peter Bala_, Jun 06 2019

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, Logistic function

Cf. A000629, A027642, A028246, A084938, A163626, A164555.
Columns k=0-10 give: A000012, A000225, A028243, A028244, A028245, A032180, A228909, A228910, A228911, A228912, A228913.
Cf. A278075.

A163626 := (n, k) -> add((-1)^j*binomial(k, j)*(j+1)^n, j = 0..k):
for n from 0 to 6 do seq(A163626(n, k), k = 0..n) od; # Peter Luschny, Sep 21 2017

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1-y[x]);
Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n-1][y][x], x];
row[n_] := CoefficientList[Derivative[n][y][x], y[x]] // Rest;
Table[row[n], {n, 0, 9}] // Flatten
(* or *) Table[(-1)^k*k!*StirlingS2[n+1, k+1], {n, 0, 9}, {k, 0, n}] // Flatten
(* Jean-François Alcover, Dec 16 2014 *)

from sympy.core.cache import cacheit
@cacheit
def T(n, k):return 1 if n==0 and k==0 else 0 if k>n or k<0 else (k + 1)*T(n - 1, k) - k*T(n - 1, k - 1)
for n in range(51): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Sep 11 2017

T(n, k) = (-1)^k*k!*Stirling2(n+1, k+1). - Jean-François Alcover, Dec 16 2014
T(n, k) = (k+1)*T(n-1,k) - k*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, May 29 2015
Worpitzky's representation of the Bernoulli numbers B(n, 1) = Sum_{k = 0..n} T(n,k)/(k+1) = A164555(n)/A027642(n) (Bernoulli numbers). - Philippe Deléham, May 29 2015
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(j+1)^n. - Peter Luschny, Sep 21 2017
Let W_n(x) be the row polynomials of this sequence and F_n(x) the row polynomials of A278075. Then W_n(1 - x) = F_n(x). Also Integral_{x=0..1} U_n(x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021

A228910 a(n) = 8^n - 77^n + 216^n - 355^n + 354^n - 213^n + 72^n - 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 5040, 181440, 3780000, 59875200, 801496080, 9574044480, 105398092800, 1091804313600, 10794490827120, 102896614941120, 952741767650400, 8617145057539200, 76461500619902160, 667855517349303360, 5757691363157764800, 49099453300298016000, 414884142077935345200

Offset: 0

Richard V. Scholtz, III, Sep 07 2013

Calculates the eighth column of coefficients with respect to the derivatives, d^n/dx^n(y), of the logistic equation when written as y=1/[1+exp(-x)].

Seiichi Manyama, Table of n, a(n) for n = 0..1107
Index entries for linear recurrences with constant coefficients, signature (36,-546,4536,-22449,67284,-118124,109584,-40320).

Cf. A008277, A049434.
The eighth column of results of A163626.
Cf. A000225, A028243, A028244, A028245, A032180, A228909.

[8^(n)-7*7^(n)+21*6^(n)-35*5^(n)+35*4^(n)-21*3^(n)+7*2^(n)-1: n in [0..30]]; // G. C. Greubel, Nov 19 2017

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1 - y[x]); Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n - 1][y][x], x]; row[n_] := CoefficientList[ Derivative[n][y][x], y[x]] // Rest; Join[{0, 0, 0, 0, 0, 0, 0}, Table[ -row[n], {n, 7, 23}] [[All, 8]]] (* Jean-François Alcover, Dec 16 2014 *)
Table[7!*StirlingS2[n + 1, 8], {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)
Table[8^n - 7*7^n + 21*6^n - 35*5^n + 35*4^n - 21*3^n + 7*2^n - 1, {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)
CoefficientList[Series[5040*x^7 / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 16 2014 after Colin Barker *)

a(n)=8^(n)-7*7^(n)+21*6^(n)-35*5^(n)+35*4^(n)-21*3^(n)+7*2^(n)-1.

for(n=0,30, print1(5040*stirling(n+1,8,2), ", ")) \\ G. C. Greubel, Nov 19 2017

a(n) = 5040 * S2(n+1,8), n>=0.
G.f.: 5040*x^7 / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)). - Colin Barker, Sep 20 2013
E.g.f.: Sum_{k=1..8} (-1)^(8-k)*binomial(8-1,k-1)*exp(k*x). - Wolfdieter Lang, May 03 2017

Offset corrected by Vaclav Kotesovec, Dec 16 2014

A249163 Triangle read by rows: the positive terms of A163626.

Original entry on oeis.org

1, 1, 1, 2, 1, 12, 1, 50, 24, 1, 180, 360, 1, 602, 3360, 720, 1, 1932, 25200, 20160, 1, 6050, 166824, 332640, 40320, 1, 18660, 1020600, 4233600, 1814400, 1, 57002, 5921520, 46070640, 46569600, 3628800, 1, 173052, 33105600, 451725120, 898128000, 239500800

Offset: 0

Paul Curtz, Dec 15 2014

We have two possibilities: with or without 0's.
Without 0's:
1,
1,
1,   2,
1,  12,
1,  50,  24,
1, 180, 360,
etc.
Sum of every row: A000670(n).
First two terms of successive columns: 1, 1, 2, 12, 24, 360, ... = A211374.
With 0's:
1,   0,    0,   0,
1,   0,    0,   0,
1,   2,    0,   0,
1,  12,    0,   0,
1,  50,   24,   0,
1, 180,  360,   0,
1, 602, 3360, 720,
etc.
The columns are essentially A000012, A028243, A028246, A228909, A228911, A228913, from Stirling numbers of the second kind S(n,3), S(n,5), S(n,7), S(n,9), S(n,11), ... .

Cf. A163626, A000670, A211374; also A000012, A000392, A000481, A000771, A049447, A028243, A028246, A091137, A228909, A163626, A228911, A228913 and Worpitzky numbers for the second Bernoulli numbers A164555(n)/A027642(n).

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1 - y[x]); Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n - 1][y][x], x]; row[n_] := CoefficientList[Derivative[n][y][x], y[x]] // Rest; Table[ Select[row[n], Positive] , {n, 0, 12}] // Flatten
(* or, simply: *) Table[(-1)^k*k!*StirlingS2[n+1, k+1], {n, 0, 12}, {k, 0, n}] // Flatten // Select[#, Positive]& (* Jean-François Alcover, Dec 16 2014 *)

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 2, 1, 0, 1, 10, 3, 7, 3, 0, 1, 15, 4, 25, 12, 2, 0, 1, 21, 5, 65, 30, 6, 1, 0, 1, 28, 6, 140, 60, 12, 15, 7, 0, 1, 36, 7, 266, 105, 20, 90, 50, 12, 0, 1, 45, 8, 462, 168, 30, 350, 195, 60, 6, 0, 1, 55, 9, 750, 252, 42, 1050, 560, 180, 24, 1, 0

Offset: 0

Peter Luschny, Oct 20 2017

			Array starts:
m\j| 0   1  2     3       4       5       6       7       8       9      10
---|-----------------------------------------------------------------------
m=0| 1,  0, 0,    0,      0,      0,      0,      0,      0,      0,      0
m=1| 1,  1, 1,    1,      3,      2,      1,      7,     12,      6,      1
m=2| 1,  3, 2,    7,     12,      6,     15,     50,     60,     24,     31
m=3| 1,  6, 3,   25,     30,     12,     90,    195,    180,     60,    301
m=4| 1, 10, 4,   65,     60,     20,    350,    560,    420,    120,   1701
m=5| 1, 15, 5,  140,    105,     30,   1050,   1330,    840,    210,   6951
m=6| 1, 21, 6,  266,    168,     42,   2646,   2772,   1512,    336,  22827
m=7| 1, 28, 7,  462,    252,     56,   5880,   5250,   2520,    504,  63987
m=8| 1, 36, 8,  750,    360,     72,  11880,   9240,   3960,    720, 159027
m=9| 1, 45, 9, 1155,    495,     90,  22275,  15345,   5940,    990, 359502
   A000217, A001296,A027480,A002378,A001297,A293475,A033486,A007531,A001298
.
m\j| ...      11      12      13      14
---|-----------------------------------------
m=0| ...,      0,      0,      0,      0, ... [A000007]
m=1| ...,     15,     50,     60,     24, ... [A028246]
m=2| ...,    180,    390,    360,    120, ... [A053440]
m=3| ...,   1050,   1680,   1260,    360, ... [A294032]
m=4| ...,   4200,   5320,   3360,    840, ...
m=5| ...,  13230,  13860,   7560,   1680, ...
m=6| ...,  35280,  31500,  15120,   3024, ...
m=7| ...,  83160,  64680,  27720,   5040, ...
m=8| ..., 178200, 122760,  47520,   7920, ...
m=9| ..., 353925, 218790,  77220,  11880, ...
         A293476,A293608,A293615,A052762, ...
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A053440, T(3, 2) is row 2 of A294032 (which is [25, 30, 12]) and T(3, 2, 1) = 30.
.
Remark: To adapt the sequences A028246 and A053440 to our enumeration use the exponential generating functions exp(x)/(1 - y*(exp(x) - 1)) and exp(x)*(2*exp(x) - y*exp(2*x) + 2*y*exp(x) - 1 - y)/(1 - y*(exp(x) - 1))^2 instead of those indicated in their respective entries.

Eric Weisstein's World of Mathematics, Nørlund Polynomial.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A027480(n) = T(n, 2, 1),
A002378(n) = T(n, 2, 2), A001297(n) = T(n, 3, 0), A293475(n) = T(n, 3, 1),
A033486(n) = T(n, 3, 2), A007531(n) = T(n, 3, 3), A001298(n) = T(n, 4, 0),
A293476(n) = T(n, 4, 1), A293608(n) = T(n, 4, 2), A293615(n) = T(n, 4, 3),
A052762(n) = T(n, 4, 4), A052787(n) = T(n, 5, 5), A000225(n) = T(1, n, 1),
A028243(n) = T(1, n, 2), A028244(n) = T(1, n, 3), A028245(n) = T(1, n, 4),
A032180(n) = T(1, n, 5), A228909(n) = T(1, n, 6), A228910(n) = T(1, n, 7),
A000225(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
A028246(n,k) = T(1, n, k), A053440(n,k) = T(2, n, k), A294032(n,k) = T(3, n, k),
A293926(n,k) = T(n, n, k), A124320(n,k) = T(n, k, k), A156991(n,k) = T(k, n, n).
Cf. A293616.

A293617 := proc(m, n, k) option remember:
if m = 0 then 0^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293617(m,n-1,k) + k*A293617(m,n-1,k-1) + A293617(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293617(m, n, k), k=0..n)) od od;
# Sample uses:
A027480 := n -> A293617(n, 2, 1): A293608 := n -> A293617(n, 4, 2):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293617(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

T[m_, n_, k_] := Pochhammer[m, k] StirlingS2[n + m, k + m];
For[m = 0, m < 7, m++, Print[Table[T[m, n, k], {n,0,6}, {k,0,n}]]]
A293617Row[m_, n_] := Table[T[m, n, k], {k,0,n}];
(* Sample use: *)
A293926Row[n_] := A293617Row[n, n];

T(m,n,k) = (k + m)*T(m, n-1, k) + k*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k<0 or k>n; and T(m, 0, k) = 0^k.

T(m,n,k) = Pochhammer(m, k)*binomial(n + m, k + m)*NorlundPolynomial(n - k, -k - m).

A298668 Number T(n,k) of set partitions of [n] into k blocks such that the absolute difference between least elements of consecutive blocks is always > 1; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 3, 0, 1, 7, 2, 0, 1, 15, 12, 0, 1, 31, 50, 6, 0, 1, 63, 180, 60, 0, 1, 127, 602, 390, 24, 0, 1, 255, 1932, 2100, 360, 0, 1, 511, 6050, 10206, 3360, 120, 0, 1, 1023, 18660, 46620, 25200, 2520, 0, 1, 2047, 57002, 204630, 166824, 31920, 720

Offset: 0

Alois P. Heinz, Jan 24 2018

			T(5,1) = 1: 12345.
T(5,2) = 7: 1234|5, 1235|4, 123|45, 1245|3, 124|35, 125|34, 12|345.
T(5,3) = 2: 124|3|5, 12|34|5.
T(7,4) = 6: 1246|3|5|7, 124|36|5|7, 124|3|56|7, 126|34|5|7, 12|346|5|7, 12|34|56|7.
T(9,5) = 24: 12468|3|5|7|9, 1246|38|5|7|9, 1246|3|58|7|9, 1246|3|5|78|9, 1248|36|5|7|9, 124|368|5|7|9, 124|36|58|7|9, 124|36|5|78|9, 1248|3|56|7|9, 124|38|56|7|9, 124|3|568|7|9, 124|3|56|78|9, 1268|34|5|7|9, 126|348|5|7|9, 126|34|58|7|9, 126|34|5|78|9, 128|346|5|7|9, 12|3468|5|7|9, 12|346|58|7|9, 12|346|5|78|9, 128|34|56|7|9, 12|348|56|7|9, 12|34|568|7|9, 12|34|56|78|9.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1;
  0, 1,    1;
  0, 1,    3;
  0, 1,    7,     2;
  0, 1,   15,    12;
  0, 1,   31,    50,     6;
  0, 1,   63,   180,    60;
  0, 1,  127,   602,   390,    24;
  0, 1,  255,  1932,  2100,   360;
  0, 1,  511,  6050, 10206,  3360,  120;
  0, 1, 1023, 18660, 46620, 25200, 2520;
  ...

Alois P. Heinz, Rows n = 0..200, flattened
Wenqin Cao, Emma Yu Jin, and Zhicong Lin, Enumeration of inversion sequences avoiding triples of relations, Discrete Applied Mathematics (2019); see also author's copy

Columns k=0-11 give (offsets may differ): A000007, A057427, A168604, A028243, A028244, A028245, A032180, A228909, A228910, A228911, A228912, A228913.
Row sums give A229046(n-1) for n>0.
T(2n+1,n+1) gives A000142.
T(2n,n) gives A001710(n+1).
Cf. A000629, A000670, A008277, A028246, A048993, A076726, A110654, A136011, A173018.

b:= proc(n, m, t) option remember; `if`(n=0, x^m, add(
      b(n-1, max(m, j), `if`(j>m, 1, 0)), j=1..m+1-t))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..14);
# second Maple program:
T:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0), (k-1)!*Stirling2(n-k+1, k)):
seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);
# third Maple program:
T:= proc(n, k) option remember; `if`(k<2, `if`(n=0 xor k=0, 0, 1),
      `if`(k>ceil(n/2), 0, add((k-j)*T(n-1-j, k-j), j=0..1)))
    end:
seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);

T[n_, k_] := T[n, k] = If[k < 2, If[Xor[n == 0, k == 0], 0, 1],
     If[k > Ceiling[n/2], 0, Sum[(k-j) T[n-1-j, k-j], {j, 0, 1}]]];
Table[Table[T[n, k], {k, 0, Ceiling[n/2]}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Mar 08 2021, after third Maple program *)

T(n,k) = (k-1)! * Stirling2(n-k+1,k) for k>0, T(n,0) = A000007(n).
T(n,k) = Sum_{j=0..k-1} (-1)^j*C(k-1,j)*(k-j)^(n-k) for k>0, T(n,0) = A000007(n).
T(n,k) = (k-1)! * A136011(n,k) for n, k >= 1.
Sum_{j>=0} T(n+j,j) = A076726(n) = 2*A000670(n) = A000629(n) + A000007(n).

A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)k! + S2(n, k-1)(k-1)! with the Stirling2 triangle S2 = A048993.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 7, 12, 0, 1, 15, 50, 60, 0, 1, 31, 180, 390, 360, 0, 1, 63, 602, 2100, 3360, 2520, 0, 1, 127, 1932, 10206, 25200, 31920, 20160, 0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 0, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 0, 1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 59875200, 46569600, 19958400

Offset: 0

Wolfdieter Lang, May 03 2017

This triangle T(n, k) appears in the e.g.f. of the sum of powers SP(n, m) = Sum_{j=0..m} j^n, n >= 0, m >= 0 with 0^0:=1 as ESP(n, t) = exp(t)*(Sum_{k=0..n} T(n, k)*t^k/k! + t^(n+1)/(n+1)), n >= 0.
The sub-triangle T(n, k) for 1 <= k <=n, see A028246(n+1,k) (diagonal not needed).
For S2(n, m)*m! see A131689.
The columns (starting sometimes with n=k) are A000007, A000012, A000225, A028243(n-1), A028244(n-1), A028245(n-1), A032180(n-1), A228909, A228910, A228911, A228912, A228913. See below for the e.g.f.s and o.g.f.s.
The row sums are 1 for n=1 and A000629(n) - n! for n >= 1, See A285868.

			The triangle T(n, k) begins:
n\k 0  1    2     3      4       5        6        7        8        9  ...
0:  1
1:  0  1
2:  0  1    3
3:  0  1    7    12
4:  0  1   15    50     60
5:  0  1   31   180    390     360
6:  0  1   63   602   2100    3360     2520
7:  0  1  127  1932  10206   25200    31920    20160
8:  0  1  255  6050  46620  166824   317520   332640   181440
9:  0  1  511 18660 204630 1020600  2739240  4233600  3780000  1814400
...

Cf. A000629, A028246, A048993, A131689, A285868.

Table[If[k == 0, Boole[n == 0], StirlingS2[n, k] k! + StirlingS2[n, k - 1] (k - 1)!], {n, 0, 10}, {k, 0, n}] (* Michael De Vlieger, May 08 2017 *)

T(n, k) =  A131689(n, k) + A131689(n, k-1), 0 <= k <= n, with A131689(n, -1) = 0.
T(0, 0) = 1 and T(n, k) = Stirling2(n+1, k)*(k-1)! for n >= k >= 1. For Stirling2 see A048993. Stirling2(n, k)*(k-1)! = A028246(n, k) for n >= k >= 1.
Recurrence: T(0, 0) = 1, T(n, n) = (n+1)!/2, T(n, -1) = 0, T(n, k) = 0 if n < k, and T(n, k) = (k-1)*T(n-1, k-1) + k*T(n-1, k), for n > k >= 0.
E.g.f. for column k=0 is 1, and for  k >= 1: Sum_{j=1..k}((-1)^(k-j) * binomial(k-1, j-1) * exp(j*x)) - x^(k-1).
O.g.f. for column k = 0 is 1, and for  k >= 1: ((k-1)!*x^(k-1) / Product_{j=1..k} (1-j*x)) - (k-1)!*x^(k-1).

cp's OEIS Frontend

A163626 Triangle read by rows: The n-th derivative of the logistic function written in terms of y, where y = 1/(1 + exp(-x)).

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A228910 a(n) = 8^n - 7*7^n + 21*6^n - 35*5^n + 35*4^n - 21*3^n + 7*2^n - 1.

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A249163 Triangle read by rows: the positive terms of A163626.

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A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

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A298668 Number T(n,k) of set partitions of [n] into k blocks such that the absolute difference between least elements of consecutive blocks is always > 1; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

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A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)*k! + S2(n, k-1)*(k-1)! with the Stirling2 triangle S2 = A048993.

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A228910 a(n) = 8^n - 77^n + 216^n - 355^n + 354^n - 213^n + 72^n - 1.

A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)k! + S2(n, k-1)(k-1)! with the Stirling2 triangle S2 = A048993.