A346730 Irregular triangle read by rows: T(n,k) is the number of n-bit numbers with k divisors.

1, 0, 2, 0, 2, 1, 1, 0, 2, 1, 4, 0, 1, 0, 5, 1, 4, 1, 3, 0, 2, 0, 7, 1, 11, 0, 6, 0, 4, 1, 1, 0, 1, 0, 13, 1, 20, 1, 9, 1, 9, 1, 2, 0, 6, 0, 0, 0, 1, 0, 23, 1, 39, 0, 15, 0, 25, 2, 3, 0, 12, 0, 1, 1, 3, 0, 2, 0, 1, 0, 43, 2, 74, 0, 27, 0, 48, 3, 6, 0, 25, 0, 2, 2, 13, 0, 5, 0, 2, 0, 0, 0, 4

Offset: 1

Author

Jon E. Schoenfield, Jul 30 2021

Keywords

• nonn
• base
• tabf

Comments

The number of terms in row n is A346729(n).

Examples

			There are four 3-bit numbers: 4 = 100_2, 5 = 101_2 = 5, 6 = 110_2, 7 = 111_2. 5 and 7 are both prime, so each has 2 divisors; 4 = 2^2 has 3 divisors (1, 2, and 4), and 6 = 2*3 has 4 divisors (1, 2, 3, and 6). Thus, among the 3-bit numbers, the counts of those having 1, 2, 3, and 4 divisors are 0, 2, 1, and 1, respectively, so the 3rd row of the table is 0, 2, 1, 1.
Triangle begins:
  1;
  0,  2;
  0,  2, 1,  1;
  0,  2, 1,  4, 0, 1;
  0,  5, 1,  4, 1, 3, 0, 2;
  0,  7, 1, 11, 0, 6, 0, 4, 1, 1, 0, 1;
  0, 13, 1, 20, 1, 9, 1, 9, 1, 2, 0, 6, 0, 0, 0, 1;
		

Links

• Alois P. Heinz, Rows n = 1..26, flattened

Crossrefs

Row sums give A011782.

Columns k=1-2 give: A063524, A162145 (for n>=2).

Cf. A000005, A153876, A300509, A346729.

Programs

Maple

T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(add
        (x^numtheory[tau](i), i=2^(n-1)..2^n-1)):
seq(T(n), n=1..10);  # Alois P. Heinz, Jul 31 2021

Mathematica

Map[BinCounts[#, {0, Max[#] + 1, 1}] &, Table[DivisorSigma[0, 2^n + k], {n, 0, 8}, {k, 0, 2^n - 1}]] // Flatten (* Michael De Vlieger, Aug 29 2021 *)

Formula

T(n,2) = A162145(n) for n > 1.

T(n,n) = A300509(n).

Sum_{k>=1} k * T(n,k) = A153876(n). - Alois P. Heinz, Aug 01 2021