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A112492 Triangle from inverse scaled Pochhammer symbols. 11
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, 1, 511, 833375 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

This expansion is based on the partial fraction identity: 1/Product_{j=1..m}(x+j) = (1 + Sum_{j=1..m} (-1)^j*binomial(m,j)*x/(x+j))/m!, e.g., p. 37 of the 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 explicit expansion based on this can be made by defining a "recursive 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)

REFERENCES

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

LINKS

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], 2000.

FORMULA

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

T(n,0) = T(n,n) = 1 and T(n,k) = (k+1)^(n-k)*T(n-1,k-1)+(k!)*T(n-1,k) for 0<k<n. - Werner Schulte, Dec 14 2016

EXAMPLE

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

MATHEMATICA

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-Fran├žois Alcover, Jul 09 2013, from 2nd formula *)

PROG

(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)

for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Oct 20 2012

(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)

for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Oct 20 2012

CROSSREFS

Row sums give A111885.

Sequence in context: A220555 A075440 A137470 * A210574 A049290 A147990

Adjacent sequences:  A112489 A112490 A112491 * A112493 A112494 A112495

KEYWORD

nonn,easy,tabl

AUTHOR

Wolfdieter Lang, Sep 12 2005

STATUS

approved

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Last modified June 26 06:03 EDT 2017. Contains 288754 sequences.