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A112492
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Triangle from inverse scaled Pochhammer symbols.
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11
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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
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OFFSET
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0,5
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COMMENTS
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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 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)
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REFERENCES
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Charles Jordan, Calculus of Finite Differences, Chelsea, 1965.
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LINKS
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FORMULA
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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<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
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EXAMPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(PARI): {h(n, recurse=1) = if(recurse == 0, return(1)); ;
return( sum(k=0, n, h(k, recurse-1) / (1+k) )); }
(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
(Magma)
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
(SageMath)
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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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