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A335997
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Triangle read by rows: T(n,k) = Product_{i=n-k+1..n} i! for 0 <= k <= n.
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0
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1, 1, 1, 1, 2, 2, 1, 6, 12, 12, 1, 24, 144, 288, 288, 1, 120, 2880, 17280, 34560, 34560, 1, 720, 86400, 2073600, 12441600, 24883200, 24883200, 1, 5040, 3628800, 435456000, 10450944000, 62705664000, 125411328000, 125411328000
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OFFSET
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0,5
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COMMENTS
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Based on some integer sequence a(n), n>0, define triangular arrays A(a;n,k) by recurrence: A(a;0,0) = 1, and A(a;i,j) = 0 if j<0 or j>i, and A(a;n,k) = n! / (n-k)! * A(a;n-1,k) + a(n) * A(a;n-1,k-1) for 0<=k<=n. Then, Product_{i=1..n} (1 + (a(i) / i!) * x) = Sum_{k=0..n} A(a;n,k) / T(n,k) * x^k for n>=0 with empty product 1 (case n=0).
For the row reversed triangle R(n,k) = Product_{i=k+1..n} i! with empty product 1 (case k=n) the terms of the matrix inverse M are given by M(n,n) = 1 for n >= 0 and M(n,n-1) = -n! for n > 0 otherwise 0. - Werner Schulte, Oct 25 2022
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LINKS
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FORMULA
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T(n,k) = T(n,1) * T(n-1,k-1) for 0 < k <= n.
T(2*n,n) = A093002(n+1) for n >= 0.
T(n,k)/T(k,k) = A009963(n,k) for 0 <= k <= n.
(Sum_{k=0..n} T(n,k) * T(n,n-k))/T(n,n) = A193520(n) for n >= 0.
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EXAMPLE
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The triangle starts:
n\k : 0 1 2 3 4 5 6
============================================================
0 : 1
1 : 1 1
2 : 1 2 2
3 : 1 6 12 12
4 : 1 24 144 288 288
5 : 1 120 2880 17280 34560 34560
6 : 1 720 86400 2073600 12441600 24883200 24883200
etc.
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MATHEMATICA
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T[n_, k_] := Product[i!, {i, n - k + 1, n}]; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 08 2020 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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