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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, 747497920, 117550462624, 1413470290176, 673781602752, 7245893376, 1026576, 1
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
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<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; ...
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
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 *)
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
(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
(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
CROSSREFS
Row sums give A111885.
Sequence in context: A373398 A075440 A137470 * A210574 A353532 A049290
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Sep 12 2005
EXTENSIONS
Terms a(48) onward added by G. C. Greubel, Nov 12 2017
STATUS
approved