OFFSET
0,4
COMMENTS
Special values: H(n,1)=0, H(2n,2)=0, H(n,-x) ~= ( x/log(1+x) )^(n+1), for x>0. H'(n,1) = -1/n!, where H'(n,x) = d/dx H(n,x).
The zeros of these polynomials are all positive reals >= 1. If we order the zeros of H(n,x), {r_k, k=0..(n-1)}, by magnitude so that r_0 = 1, r_k > r_(k-1), for 0 < k < n, then r_(n-k) = r_k/(r_k - 1) when 0 < k < n, n > 1, where r_(n/2) = 2 for even n.
Also Product_{k=0..(n-1)} r_k = n!, r_(n-1) ~ C 2^n.
I believe that these numbers are the coefficients of the Eulerian polynomials An(z) written in powers of z-1. That is, the sequence is: A0(1); A1(1), A1'(1); A2(1), A2'(1), A2''(1)/2!; A3(1), A3'(1), A3''(1)/2!, A3'''(1)/3!; A4(1), A4'(1), A4''(1)/2!, A4'''(1)/3!, A4''''(1)/4! etc. My convention: A0(z)=z, A1(z)=z, A2(z)=z+z^2, A3(z)=z+4z^2+z^3, A4(z)=z+11z^2+11z^3+z^4, etc. - Louis Zulli (zullil(AT)lafayette.edu), Jan 19 2005
H(n,2) gives 1,-1,0,2,0,-16,0,272,0,-7936,0,..., see A009006. - Philippe Deléham, Aug 20 2007
Row sums are zero except for first row. - Roger L. Bagula, Sep 11 2008
From Groux Roland, May 12 2011: (Start)
Let f(x) = (exp(x)+1)^(-1) then the n-th derivative of f equals Sum_{k=0..n} T(n,k)*(f(x))^(n+1-k).
T(n+1,0) = (n+1)*T(n,0); T(n+1,n+1) = -T(n,n) and for 0 < k < n T(n+1,k) = (n+1-k) * T(n,k) - (n-k+2)*T(n,k-1).
T(n,k) = Sum_{i=0..k} (-1)^(i+k)*(n-i)!*binomial(n-i,k-i)*S(n,n-i) where S(n,k) is a Stirling number of the second kind. (End)
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Nguyen-Huu-Bong, Some Combinatorial Properties of Summation Operators, J. Comb. Theory, Ser. A 11.3 (1971): 213-221. See Table on page 214.
FORMULA
Generated by [1, 1, 2, 2, 3, 3, ...] DELTA [ -1, 0, -2, 0, -3, 0, ...], where DELTA is the operator defined in A084938.
T(n, k) = Sum_{i=0..n-k} (-1)^(n-i)*C(n-k, i)*(i+1)^n; n >= 0, 0 <= k <= n. - Paul D. Hanna, Jul 21 2005
E.g.f.: A(x, y) = -log(1-(1-exp(-x*y))/y). - Paul D. Hanna, Jul 21 2005
p(x,n) = x^(n + 1)*Sum_{k>=0} k^n*(1 - x)^k; t(n,m) = Coefficients(p(x,n)). - Roger L. Bagula, Sep 11 2008
p(x,n) = x^(n + 1)*PolyLog(-n, 1 - x); t(n,m) = coefficients(p(x,n)) for n >= 1. - Roger L. Bagula and Gary W. Adamson, Sep 15 2008
EXAMPLE
H(0,x) = 1
H(1,x) = (1 - 1*x)/1!
H(2,x) = (2 - 3*x + 1*x^2)/2!
H(3,x) = (6 - 12*x + 7*x^2 - 1*x^3)/3!
H(4,x) = (24 - 60*x + 50*x^2 - 15*x^3 + 1*x^4)/4!
H(5,x) = (120 - 360*x + 390*x^2 - 180*x^3 + 31*x^4 - 1*x^5)/5!
H(6,x) = (720 - 2520*x + 3360*x^2 - 2100*x^3 + 602*x^4 - 63*x^5 + 1*x^5)/6!
Triangle begins:
1;
1, -1;
2, -3, 1;
6, -12, 7, -1;
24, -60, 50, -15, 1;
120, -360, 390, -180, 31, -1;
720, -2520, 3360, -2100, 602, -63, 1;
5040, -20160, 31920, -25200, 10206, -1932, 127, -1;
MAPLE
CL := f -> PolynomialTools:-CoefficientList(f, x):
T_row := n -> `if`(n=0, [1], CL(x^(n+1)*polylog(-n, 1-x))):
for n from 0 to 6 do T_row(n) od; # Peter Luschny, Sep 28 2017
MATHEMATICA
Table[CoefficientList[x^(n+1)*Sum[k^n*(1-x)^k, {k, 0, Infinity}], x], {n, 0, 10}]//Flatten (* Roger L. Bagula, Sep 11 2008 *)
p[x_, n_]:= x^(n+1)*PolyLog[-n, 1-x]; Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten (* Roger L. Bagula and Gary W. Adamson, Sep 15 2008 *)
PROG
(PARI) T(n, k)=if(k<0 || k>n, 0, n!*polcoeff((-x/log(1-x+x^2*O(x^n)))^(n+1), k))
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(PARI) T(n, k)=sum(i=0, n-k, (-1)^(n-i)*binomial(n-k, i)*(i+1)^n)
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(PARI) /* Using e.g.f. A(x, y): */
{T(n, k)=local(X=x+x*O(x^n), Y=y+y^2*O(y^(k))); n!*polcoeff(polcoeff(-log(1-(1-exp(-X*Y))/y), n, x), k, y)}
for(n=0, 10, for(k=0, n-1, print1(T(n, k), ", ")); print(""))
(PARI) /* Deléham's DELTA: T(n, k) = [x^(n-k)*y^k] P(n, 0) */
{P(n, k)=if(n<0||k<0, 0, if(n==0, 1, P(n, k-1)+(x*(k\2+1)+y*(-(k\2+1)*((k+1)%2)))*P(n-1, k+1)))}
{T(n, k)=polcoeff(polcoeff(P(n, 0), n-k, x), k, y)}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(Magma)
T:= func< n, k | &+[(-1)^(n-j)*Binomial(n-k, j)*(j+1)^n: j in [0..n-k]] >;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 27 2020
(Sage)
def T(n, k): return sum( (-1)^(n-j)*binomial(n-k, j)*(j+1)^n for j in (0..n-k))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 27 2020
(GAP) Flat(List([0..12], n-> List([0..n], k-> Sum([0..n-k], j->
(-1)^(n-j)*Binomial(n-k, j)*(j+1)^n )))); # G. C. Greubel, Jan 27 2020
CROSSREFS
KEYWORD
AUTHOR
Paul D. Hanna, Sep 13 2002
EXTENSIONS
Error in one term corrected by Benoit Cloitre, Aug 20 2007
STATUS
approved