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%I #12 Jun 24 2021 02:16:56
%S 1,3,1,8,4,1,25,15,5,1,102,66,24,6,1,539,357,133,35,7,1,3496,2328,872,
%T 232,48,8,1,26613,17739,6651,1773,369,63,9,1,231170,154110,57790,
%U 15410,3210,550,80,10,1,2250127,1500081,562529,150007,31251,5357,781,99,11,1
%N 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.
%C There is a relationship to the left factorials A003422 (see formula).
%F T(n,k) = (n+2) * (A003422(n+1)-A003422(k)) / ((k+2) * k!) for 0 <= k <= n.
%F T(n,n) = 1 for n >= 0; T(n,n-1) = n+2 for n > 0.
%F T(n,k) = (n+2) * (T(n-1,k) - T(n-2,k)) for 0 < k+1 < n.
%F T(n,k) + T(n-2,k) - 2 * T(n-1,k) = (n+1)! / ((k+2) * k!) for 0 < k+1 < n.
%F T(n,k) * T(n-1,k-1) - T(n,k-1) * T(n-1,k) = (n+2)! / (k+2)! for 0 < k < n.
%F 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.
%F Row sums are p(n; 1): 1, 4, 13, 46, 199, 1072, ....
%F Alternating row sums are p(n; -1): 1, 2, 5, 14, 55, 286, ....
%F 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.
%e The triangle T(n,k), 0 <= k <= n, begins:
%e n\k : 0 1 2 3 4 5 6 7 8 9
%e ====================================================================
%e 0 : 1
%e 1 : 3 1
%e 2 : 8 4 1
%e 3 : 25 15 5 1
%e 4 : 102 66 24 6 1
%e 5 : 539 357 133 35 7 1
%e 6 : 3496 2328 872 232 48 8 1
%e 7 : 26613 17739 6651 1773 369 63 9 1
%e 8 : 231170 154110 57790 15410 3210 550 80 10 1
%e 9 : 2250127 1500081 562529 150007 31251 5357 781 99 11 1
%e etc.
%e The matrix inverse M(n,k), 0 <= k <= n, begins:
%e n\k : 0 1 2 3 4 5 6 7 8 9
%e ===============================================
%e 0 : 1
%e 1 : -3 1
%e 2 : 4 -4 1
%e 3 : 0 5 -5 1
%e 4 : 0 0 6 -6 1
%e 5 : 0 0 0 7 -7 1
%e 6 : 0 0 0 0 8 -8 1
%e 7 : 0 0 0 0 0 9 -9 1
%e 8 : 0 0 0 0 0 0 10 -10 1
%e 9 : 0 0 0 0 0 0 0 11 -11 1
%e etc.
%t 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 *)
%Y Cf. A003422, A051403 (column 0).
%K nonn,easy,tabl
%O 0,2
%A _Werner Schulte_, May 16 2021
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