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A112492 Triangle from inverse scaled Pochhammer symbols. 11

%I

%S 1,1,1,1,3,1,1,7,11,1,1,15,85,50,1,1,31,575,1660,274,1,1,63,3661,

%T 46760,48076,1764,1,1,127,22631,1217776,6998824,1942416,13068,1,1,255,

%U 137845,30480800,929081776,1744835904,104587344,109584,1,1,511,833375,747497920,117550462624,1413470290176,673781602752,7245893376,1026576,1

%N Triangle from inverse scaled Pochhammer symbols.

%C 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.

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

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

%C From _Gottfried Helms_, Dec 11 2001: (Start)

%C 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).

%C Then

%C a(r,c)= m(r,c) * (1+r)!^(c-r).

%C 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).

%C Consider

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

%C Then consider

%C h(1,1)=1*h(1,0)

%C h(2,1)=1*h(1,0) + 1/2*h(2,0)

%C 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)

%C ...

%C and recursively

%C h(1,r)=1*h(1,r-1)

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

%C 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)

%C ...

%C h(k,r)=h(k-1,r)+1/k*h(k,r-1)

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

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

%C (End)

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

%H G. C. Greubel, <a href="/A112492/b112492.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H W. Lang, <a href="http://www.itp.kit.edu/~wl/EISpub/A112492.text">First 10 rows.</a>

%H L. M. Smiley, <a href="http://arxiv.org/abs/math.CO/0006106">Completion of a Rational Function Sequence of Carlitz</a>, arXiv:0006106 [math.CO], 2000.

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

%F 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.

%F 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

%F 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

%F 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

%F 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

%F 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

%e Triangle begins:

%e 1;

%e 1, 1;

%e 1, 3, 1;

%e 1, 7, 11, 1;

%e 1, 15, 85, 50, 1;

%e 1, 31, 575, 1660, 274, 1;

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

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

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

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

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

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

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

%e 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!;

%e 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!; ...

%e which are derived from a LambertW() identity. - _Paul D. Hanna_, Oct 20 2012

%t 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 *)

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

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

%o a(r,c) = h(r-1,c-r) * r!^(c-r) \\ _Gottfried Helms_, Dec 11 2001

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

%o T(n,k) = (k+1)!^(n-k+1)*polcoeff(prod(j=0,k,1/(j+1-x +x*O(x^(n-k)))),n-k)

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ _Paul D. Hanna_, Oct 20 2012

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

%o 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)

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ _Paul D. Hanna_, Oct 20 2012

%Y Row sums give A111885.

%K nonn,easy,tabl

%O 0,5

%A _Wolfdieter Lang_, Sep 12 2005

%E Terms a(48) onward added by _G. C. Greubel_, Nov 12 2017

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Last modified June 25 12:01 EDT 2019. Contains 324352 sequences. (Running on oeis4.)