A285268 -id:A285268 - cp's OEIS frontend

A121662 Triangle read by rows: T(i,j) for the recurrence T(i,j) = (T(i-1,j) + 1)*i.

1, 4, 2, 15, 9, 3, 64, 40, 16, 4, 325, 205, 85, 25, 5, 1956, 1236, 516, 156, 36, 6, 13699, 8659, 3619, 1099, 259, 49, 7, 109600, 69280, 28960, 8800, 2080, 400, 64, 8, 986409, 623529, 260649, 79209, 18729, 3609, 585, 81, 9, 9864100, 6235300, 2606500, 792100, 187300, 36100, 5860, 820, 100, 10

Offset: 1

Thomas Wieder, Aug 15 2006

The first column is A007526 = "the number of (nonnull) "variations" of n distinct objects, namely the number of permutations of nonempty subsets of {1,...,n}." E.g. for n=3 there are 15 subsets: {a}, {b}, {c}, {ab}, {ba}, {ac}, {ca}, {bc}, {cb}, {abc}, {acb}, {bac}, {bca}, {cab}, {cba}. These are subsets with a number of elements l=1,...,n. The second column excludes all subsets with l=n elements. For n=3 one has therefore only the 9 subsets {a}, {b}, {c}, {ab}, {ba}, {ac}, {ca}, {bc}, {cb}. The third column excludes all subsets with l>=n-1 elements. For n=3 one has therefore only the 3 subsets {a}, {b},{c}. See also A121684. The second column is A038156 = n!*Sum(1/k!, k=1..n-1). The first lower diagonal are the squares A000290 = n^2. The second lower diagonal (15, 40, 85...) is A053698 = n^3 + n^2 + n + 1. The row sum is A030297 = a(n) = n*(n+a(n-1)).

T(i, j) is the total number of ordered sets of size 1 to i-j+1 that can be created from i distinct items. - Manfred Boergens, Jun 22 2022

			Triangle T(i,j) begins:
       1
       4     2
      15     9     3
      64    40    16     4
     325   205    85    25    5
    1956  1236   516   156   36   6
   13699  8659  3619  1099  259  49  7
   ...

Cf. A007526, A030297, A053698, A038156, A000290, A121682.

Mirror of triangle A285268.

T:= proc(i, j) option remember;
      `if`(j<1 or j>i, 0, T(i-1, j)*i+i)
    end:
seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Jun 22 2022

Table[Sum[m!/(m - i)!, {i, n}], {m, 9}, {n, m, 1, -1}] // Flatten (* Michael De Vlieger, Apr 22 2017 *)
(* Sum-free code *)
b[j_] = If[j == 0, 0, Floor[j! E - 1]];
T[i_, j_] = b[i] - i! b[j - 1]/(j - 1)!;
Table[T[i, j], {i, 24}, {j, i}] // Flatten
(* Manfred Boergens, Jun 22 2022 *)

From Manfred Boergens, Jun 22 2022: (Start)
T(i, j) = Sum_{k=1..i-j+1} i!/(i-k)! = Sum_{k=j-1..i-1} i!/k!.
Sum-free formula: T(i, j) = b(i) - i!*b(j-1)/(j-1)! where b(0)=0, b(j)=floor(j!*e-1) for j>0.
(End)

Edited by N. J. A. Sloane, Sep 15 2006

Formula in name corrected by Alois P. Heinz, Jun 22 2022

A347667 Triangle read by rows: T(n,k) = Sum_{j=0..k} binomial(n,j) * j! (0 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 3, 5, 1, 4, 10, 16, 1, 5, 17, 41, 65, 1, 6, 26, 86, 206, 326, 1, 7, 37, 157, 517, 1237, 1957, 1, 8, 50, 260, 1100, 3620, 8660, 13700, 1, 9, 65, 401, 2081, 8801, 28961, 69281, 109601, 1, 10, 82, 586, 3610, 18730, 79210, 260650, 623530, 986410

Offset: 0

Ilya Gutkovskiy, Sep 10 2021

			Triangle begins:
  1;
  1, 2;
  1, 3,  5;
  1, 4, 10,  16;
  1, 5, 17,  41,   65;
  1, 6, 26,  86,  206,  326;
  1, 7, 37, 157,  517, 1237, 1957;
  1, 8, 50, 260, 1100, 3620, 8660, 13700;
  ...

T(n,n) = A000522, T(2*n,n) = A066211.

Cf. A008279, A008949, A111063, A121662, A285268.

T[n_, k_] := Sum[Binomial[n, j] j!, {j, 0, k}]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten

cp's OEIS Frontend

A121662 Triangle read by rows: T(i,j) for the recurrence T(i,j) = (T(i-1,j) + 1)*i.

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