A344381 revision #10
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A344381
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Triangle read by rows: T(n,k) = (n+2) * (Sum_{i=k..n} i!) / ((k+2) * k!) for 0 <= k <= n with T(i,j) = 0 if j < 0 or i < j.
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0
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1, 3, 1, 8, 4, 1, 25, 15, 5, 1, 102, 66, 24, 6, 1, 539, 357, 133, 35, 7, 1, 3496, 2328, 872, 232, 48, 8, 1, 26613, 17739, 6651, 1773, 369, 63, 9, 1, 231170, 154110, 57790, 15410, 3210, 550, 80, 10, 1, 2250127, 1500081, 562529, 150007, 31251, 5357, 781, 99, 11, 1
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OFFSET
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0,2
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COMMENTS
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There is a relationship to the left factorials A003422 (see formula).
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LINKS
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FORMULA
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T(n,k) = (n+2) * (A003422(n+1)-A003422(k)) / ((k+2) * k!) for 0 <= k <= n.
T(n,n) = 1 for n >= 0; T(n,n-1) = n+2 for n > 0.
T(n,k) = (n+2) * (T(n-1,k) - T(n-2,k)) for 0 < k+1 < n.
T(n,k) + T(n-2,k) - 2 * T(n-1,k) = (n+1)! / ((k+2) * k!) for 0 < k+1 < n.
T(n,k) * T(n-1,k-1) - T(n,k-1) * T(n-1,k) = (n+2)! / (k+2)! for 0 < k < n.
The row polynomials p(n; x) = Sum_{k=0..n} T(n,k) * x^k satisfy the recurrence equation p(n; x) = (n+2) * (p(n-1; x) - p(n-2; x)) + x^n for n > 1 with p(0; x) = 1 and p(1; x) = 3 + x.
Row sums are p(n; 1): 1, 4, 13, 46, 199, 1072, ....
Alternating row sums are p(n; -1): 1, 2, 5, 14, 55, 286, ....
With T as lower triangular matrix the matrix inverse M is defined by: M(n,n) = 1 for n >= 0, M(n,n-1) = -(n+2) for n > 0, M(n,n-2) = n+2 for n > 1, and M(i,j) = 0 otherwise.
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EXAMPLE
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The triangle T(n,k), 0 <= k <= n, begins:
n\k : 0 1 2 3 4 5 6 7 8 9
====================================================================
0 : 1
1 : 3 1
2 : 8 4 1
3 : 25 15 5 1
4 : 102 66 24 6 1
5 : 539 357 133 35 7 1
6 : 3496 2328 872 232 48 8 1
7 : 26613 17739 6651 1773 369 63 9 1
8 : 231170 154110 57790 15410 3210 550 80 10 1
9 : 2250127 1500081 562529 150007 31251 5357 781 99 11 1
etc.
The matrix inverse M(n,k), 0 <= k <= n, begins:
n\k : 0 1 2 3 4 5 6 7 8 9
===============================================
0 : 1
1 : -3 1
2 : 4 -4 1
3 : 0 5 -5 1
4 : 0 0 6 -6 1
5 : 0 0 0 7 -7 1
6 : 0 0 0 0 8 -8 1
7 : 0 0 0 0 0 9 -9 1
8 : 0 0 0 0 0 0 10 -10 1
9 : 0 0 0 0 0 0 0 11 -11 1
etc.
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MATHEMATICA
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T[n_, k_] := (n + 2) * Sum[i!, {i, k, n}]/((k + 2)*k!); Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 12 2021 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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editing
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