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

Author

Richard V. Scholtz, III, Aug 01 2009

Keywords

• easy
• sign
• tabl

Comments

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

Examples

			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
		

Links

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

Crossrefs

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.

Programs

Maple

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

Mathematica

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 *)

Python

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

Formula

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