0,5

This expansion is based on the partial fraction identity: 1/product(x+j,j=1..m)= (1 + sum(((-1)^j)*binomial(m,j)*x/(x+j),j=1..m))/m!, e.g., p. 37 of the Ch. Jordan reference.

Another version of this triangle (without a column of 1's) is A008969.

The column sequences are, for m=1..10: A000012 (powers of 1), A000225, A001240, A001241, A001242, A111886-A111888.

From Gottfried Helms, Dec 11 2001. (Start)

The triangle occurs as U-factor in the LDU-decomposition of the matrix M defined by m(r,c)=1/(1+r)^c (r,c,beginning at 0).

Then

  a(r,c)= m(r,c) * (1+r)!^(c-r)

An explicite expansion based on this can be made by defining a "recursice harmonic number" (rhn). (This representation is just a heuristic pattern-interpretation, no analytic proof yet available).

Consider

  h(k,0)=1      for k>0      as rhn of order zero(0).

Then consider

  h(1,1)=1*h(1,0)

  h(2,1)=1*h(1,0) + 1/2*h(2,0)

  h(3,1)=1*h(1,0) + 1/2*h(2,0) + 1/3*h(3,0) = h(2,1)+1/3*h(3,0)

  ...

and recursively

  h(1,r)=1*h(1,r-1)

  h(2,r)=1*h(1,r-1) + 1/2*h(2,r-1)

  h(3,r)=1*h(1,r-1) + 1/2*h(2,r-1) + 1/3*h(3,r-1) = h(2,r)+1/3*h(3,r-1)

  ...

  h(k,r)=h(k-1,r)+1/k*h(k,r-1)

then the upper triangular triangle A:=a(r,c) for c-r>0

a(r,c) = h(r,c-r) *(1+r)!^(c-r)

(End)

Charles Jordan, Calculus of Finite Differences, Chelsea, 1965.

Table of n, a(n) for n=0..47.

W. Lang, First 10 rows.

L. M. Smiley, Completion of a Rational Function Sequence of Carlitz, arXiv:0006106 [math.CO]

G.f. for column m>=1: (x^m)/product(1-m!*x/j, j=1..m).

a(n, m)= -(m!^(n-m+1))*sum(((-1)^j)*binomial(m, j)/j^(n-m+1), j=1..m), m>=1. a(n, m)=0 if n+1<m.

G.f. of column k: x^k/Product_{j=0..k} (j+1 - x) = Sum_{n>=k} T(n,k)*x^k/(k+1)!^(n-k+1). - Paul D. Hanna, Oct 20 2012

T(n,k) = (k+1)!^(n-k+1) * [x^n] x^k / Product_{j=0..k} (j+1 - x). - Paul D. Hanna, Oct 20 2012

G.f. of row n: Sum_{j>=0} (j+1)^(j-n-1) * exp((j+1)*x) * (-x)^j/j! = Sum_{k>=0} T(n,k)*x^k/(k+1)!^(n-k+1). - Paul D. Hanna, Oct 20 2012

T(n,k) = (k+1)!^(n-k+1) * [x^k] Sum_{j>=0} (j+1)^(j-n-1) * exp((j+1)*x) * (-x)^j/j!. - Paul D. Hanna, Oct 20 2012

Triangle begins:

1;

1, 1;

1, 3, 1;

1, 7, 11, 1;

1, 15, 85, 50, 1;

1, 31, 575, 1660, 274, 1;

1, 63, 3661, 46760, 48076, 1764, 1;

1, 127, 22631, 1217776, 6998824, 1942416, 13068, 1;

1, 255, 137845, 30480800, 929081776, 1744835904, 104587344, 109584, 1; ...

The g.f.s for the rows are illustrated by:

Sum_{n>=0} (n+1)^(n-1)*exp((n+1)*x)*(-x)^n/n! = 1;

Sum_{n>=0} (n+1)^(n-2)*exp((n+1)*x)*(-x)^n/n! = 1 + 1*x/2!;

Sum_{n>=0} (n+1)^(n-3)*exp((n+1)*x)*(-x)^n/n! = 1 + 3*x/2!^2 + 1*x^2/3!;

Sum_{n>=0} (n+1)^(n-4)*exp((n+1)*x)*(-x)^n/n! = 1 + 7*x/2!^3 + 11*x^2/3!^2 + 1*x^3/4!;

Sum_{n>=0} (n+1)^(n-5)*exp((n+1)*x)*(-x)^n/n! = 1 + 15*x/2!^4 + 85*x^2/3!^3 + 50*x^3/4!^2 + 1*x^4/5!; ...

which are derived from a LambertW() identity. - Paul D. Hanna, Oct 20 2012

a[_, 0] = 1; a[n_, m_] := -m!^(n - m + 1)*Sum[(-1)^j*Binomial[m, j]/j^(n - m + 1), {j, 1, m}]; Table[a[n, m], {n, 1, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2013, from 2nd formula *)

(PARI): {h(n, recurse=1) = if(recurse == 0, return(1)); ;

return( sum(k=0, n, h(k, recurse-1) / (1+k) )); }

a(r, c) = h(r-1, c-r) * r!^(c-r) - Gottfried Helms, Dec 11 2001

(PARI) /* From g.f. for column k: */

{T(n, k) = (k+1)!^(n-k+1)*polcoeff(prod(j=0, k, 1/(j+1-x +x*O(x^(n-k)))), n-k)} \\ Paul D. Hanna, Oct 20 2012

for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))

(PARI) /* From g.f. for row n: */

{T(n, k) = (k+1)!^(n-k+1)*polcoeff(sum(j=0, k, (j+1)^(j-n-1)*exp((j+1)*x +x*O(x^k))*(-x)^j/j!), k)} \\ Paul D. Hanna, Oct 20 2012

for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))

Row sums give A111885.

Sequence in context: A220555 A075440 A137470 * A210574 A049290 A147990

Adjacent sequences:  A112489 A112490 A112491 * A112493 A112494 A112495

nonn,easy,tabl

Wolfdieter Lang, Sep 12 2005

approved