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%I #54 Jul 25 2023 09:15:21
%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="/A112492/a112492.txt">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 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<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 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 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 *)
%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
%o (Magma)
%o function T(n,k) // T = A112492
%o if k eq 0 or k eq n then return 1;
%o else return (k+1)^(n-k)*T(n-1,k-1) + Factorial(k)*T(n-1,k);
%o end if;
%o end function;
%o [T(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jul 24 2023
%o (SageMath)
%o def T(n,k): # T = A112492
%o if (k==0 or k==n): return 1
%o else: return (k+1)^(n-k)*T(n-1,k-1) + factorial(k)*T(n-1,k)
%o flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # _G. C. Greubel_, Jul 24 2023
%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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