A112492 Triangle from inverse scaled Pochhammer symbols.

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, 747497920, 117550462624, 1413470290176, 673781602752, 7245893376, 1026576, 1

Offset: 0

Author

Wolfdieter Lang, Sep 12 2005

Keywords

• nonn
• easy
• tabl

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)

Examples

			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; ...
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
		

References

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

Links

• G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
• W. Lang, First 10 rows.
• L. M. Smiley, Completion of a Rational Function Sequence of Carlitz, arXiv:0006106 [math.CO], 2000.

Crossrefs

Row sums give A111885.

Programs

Magma

function T(n,k) // T = A112492
  if k eq 0 or k eq n then return 1;
  else return (k+1)^(n-k)*T(n-1,k-1) + Factorial(k)*T(n-1,k);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 24 2023

Mathematica

T[, 0]=1; T[n, m_]:= -m!^(n-m+1)*Sum[(-1)^j*Binomial[m, j]/j^(n-m+ 1), {j,m}]; Table[T[n, m], {n,10}, {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)
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

SageMath

def T(n,k): # T = A112492
    if (k==0 or k==n): return 1
    else: return (k+1)^(n-k)*T(n-1,k-1) + factorial(k)*T(n-1,k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 24 2023

Formula

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

T(n, m) = -(m!^(n-m+1))*Sum_{j=1..m} (-1)^j*binomial(m, j)/j^(n-m+1), m>=1. T(n, m)=0 if n+1

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 0Werner Schulte, Dec 14 2016

Extensions

Terms a(48) onward added by G. C. Greubel, Nov 12 2017

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