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A348482 Triangle read by rows: T(n,k) = (Sum_{i=k..n} i!)/(k!) for 0 <= k <= n. 1

%I #16 Jul 11 2024 17:12:42

%S 1,2,1,4,3,1,10,9,4,1,34,33,16,5,1,154,153,76,25,6,1,874,873,436,145,

%T 36,7,1,5914,5913,2956,985,246,49,8,1,46234,46233,23116,7705,1926,385,

%U 64,9,1,409114,409113,204556,68185,17046,3409,568,81,10,1

%N Triangle read by rows: T(n,k) = (Sum_{i=k..n} i!)/(k!) for 0 <= k <= n.

%C The matrix inverse M = T^(-1) has terms M(n,n) = 1 for n >= 0, M(n,n-1) = -(n+1) for n > 0, and M(n,n-2) = n for n > 1, otherwise 0.

%H Sela Fried, <a href="/A348482/a348482.pdf">On a sum involving factorials</a>, 2024.

%F T(n,n) = 1 and T(2*n,n) = A109398(n) for n >= 0; T(n,n-1) = n+1 for n > 0; T(n,n-2) = n^2 for n > 1.

%F T(n,k) - T(n-1,k) = (n!) / (k!) = A094587(n,k) for 0 <= k < n.

%F T(n,k) = (k+2) * (T(n,k+1) - T(n,k+2)) for 0 <= k < n-1.

%F T(n,k) = (T(n,k-1) - 1) / k for 0 < k <= n.

%F T(n,k) * T(n-1,k-1) - T(n-1,k) * T(n,k-1) = (n!) / (k!) for 0 < k < n.

%F T(n,1) = T(n,0)-1 = Sum_{k=0..n-1} T(n,k)/(k+2) for n > 0 (conjectured).

%F Sum_{k=0..n} binomial(k+r,k) * (1-k) * T(n+r,k+r) = binomial(n+r+1,n) for n >= 0 and r >= 0.

%F Sum_{k=0..n} (-1)^k * (k+1) * T(n,k) = (1 + (-1)^n) / 2 for n >= 0.

%F Sum_{k=0..n} (-1)^k * (k!) * T(n,k) = Sum_{k=0..n} (k!) * (1+(-1)^k) / 2 for n >= 0.

%F The row polynomials p(n,x) = Sum_{k=0..n} T(n,k) * x^k for n >= 0 satisfy the following equations:

%F (a) p(n,x) - p'(n,x) = (x^(n+1)-1) / (x-1) for n >= 0, where p' is the first derivative of p;

%F (b) p(n,x) - (n+1) * p(n-1,x) + n * p(n-2,x) = x^n for n > 1.

%F (c) p(n,x) = (x+1) * p(n-1,x) + 1 + Sum_{i=1..n-1} (d/dx)^i p(n-1,x) for n > 0 (conjectured).

%F Row sums p(n,1) equal A002104(n+1) for n >= 0.

%F Alternating row sums p(n,-1) equal A173184(n) for n >= 0 (conjectured).

%F The three conjectures stated above are true. See links. - _Sela Fried_, Jul 11 2024.

%F From _Peter Luschny_, Jul 11 2024: (Start)

%F T(n, k) = (t(k) - t(n + 1)) / k!, where t(n) = (-1)^(n + 1) * Gamma(n + 1) * Subfactorial(-(n + 1)).

%F T(n, k) = A143122(n, k) / k!. (End)

%e The triangle T(n,k) for 0 <= k <= n starts:

%e n\k : 0 1 2 3 4 5 6 7 8 9

%e =================================================================

%e 0 : 1

%e 1 : 2 1

%e 2 : 4 3 1

%e 3 : 10 9 4 1

%e 4 : 34 33 16 5 1

%e 5 : 154 153 76 25 6 1

%e 6 : 874 873 436 145 36 7 1

%e 7 : 5914 5913 2956 985 246 49 8 1

%e 8 : 46234 46233 23116 7705 1926 385 64 9 1

%e 9 : 409114 409113 204556 68185 17046 3409 568 81 10 1

%e etc.

%t T[n_, k_] := Sum[i!, {i, k, n}]/k!; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Oct 20 2021 *)

%Y Cf. A109398, A094587, A002104 (row sums), A173184 (alt. row sums), A000012 (main diagonal), A000027(1st subdiagonal), A000290 (2nd subdiagonal), A081437 (3rd subdiagonal), A192398 (4th subdiagonal), A003422 (column 0), A007489 (column 1), A345889 (column 2), A143122.

%K nonn,easy,tabl

%O 0,2

%A _Werner Schulte_, Oct 20 2021

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Last modified August 20 18:17 EDT 2024. Contains 375338 sequences. (Running on oeis4.)