A309280 T(n,k) is (1/k) times the sum of the elements of all subsets of [n] whose sum is divisible by k; triangle T(n,k), n >= 1, 1 <= k <= n*(n+1)/2, read by rows.

1, 6, 1, 1, 24, 6, 4, 1, 1, 1, 80, 20, 9, 4, 4, 2, 2, 1, 1, 1, 240, 60, 30, 14, 12, 7, 5, 3, 3, 3, 2, 2, 1, 1, 1, 672, 168, 84, 42, 29, 20, 15, 10, 9, 7, 5, 5, 4, 4, 4, 3, 2, 2, 1, 1, 1, 1792, 448, 202, 112, 71, 49, 40, 27, 23, 17, 15, 12, 10, 10, 8, 8, 7, 7, 6, 5, 5, 4, 3, 2, 2, 1, 1, 1

Offset: 1

Author

Alois P. Heinz, Jul 20 2019

Keywords

• nonn
• look
• tabf

Comments

T(n,k) is defined for all n >= 0, k >= 1. The triangle contains only the positive terms. T(n,k) = 0 if k > n*(n+1)/2.

The sequence of column k satisfies a linear recurrence with constant coefficients of order 3*A000593(k).

Examples

			The subsets of [4] whose sum is divisible by 3 are: {}, {3}, {1,2}, {2,4}, {1,2,3}, {2,3,4}.  The sum of their elements is 0 + 3 + 3 + 6 + 6 + 9 = 27.  So T(4,3) = 27/3 = 9.
Triangle T(n,k) begins:
    1;
    6,  1,  1;
   24,  6,  4,  1,  1, 1;
   80, 20,  9,  4,  4, 2, 2, 1, 1, 1;
  240, 60, 30, 14, 12, 7, 5, 3, 3, 3, 2, 2, 1, 1, 1;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..50, flattened

Crossrefs

Columns k=1..10 give: A001788, A309294, A309295, A309296, A309297, A309298, A309299, A309300, A309301, A309302.

Row sums give A309281.

Row lengths give A000217.

T(n,n) gives A309128.

Rows reversed converge to A000009.

Cf. A000593, A309402.

Programs

Maple

b:= proc(n, m, s) option remember; `if`(n=0, [`if`(s=0, 1, 0), 0],
      b(n-1, m, s) +(g-> g+[0, g[1]*n])(b(n-1, m, irem(s+n, m))))
    end:
T:= (n, k)-> b(n, k, 0)[2]/k:
seq(seq(T(n, k), k=1..n*(n+1)/2), n=1..10);
# second Maple program:
b:= proc(n, s) option remember; `if`(n=0, add(s/d *x^d,
      d=numtheory[divisors](s)), b(n-1, s)+b(n-1, s+n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n, 0)):
seq(T(n), n=1..10);

Mathematica

b[n_, m_, s_] := b[n, m, s] = If[n == 0, {If[s == 0, 1, 0], 0}, b[n-1, m, s] + Function[g, g + {0, g[[1]] n}][b[n-1, m, Mod[s+n, m]]]];
T[n_, k_] := b[n, k, 0][[2]]/k;
Table[T[n, k], {n, 1, 10}, {k, 1, n(n+1)/2}] // Flatten (* Jean-François Alcover, Oct 04 2019, after Alois P. Heinz *)

Formula

T(n+1,n*(n+1)/2+1) = A000009(n) for n >= 0.