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

%I A346730 #32 Aug 30 2021 01:43:11
%S A346730 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,
%T A346730 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,
%U A346730 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
%N A346730 Irregular triangle read by rows: T(n,k) is the number of n-bit numbers with k divisors.
%C A346730 The number of terms in row n is A346729(n).
%H A346730 Alois P. Heinz, <a href="/A346730/b346730.txt">Rows n = 1..26, flattened</a>
%F A346730 T(n,2) = A162145(n) for n > 1.
%F A346730 T(n,n) = A300509(n).
%F A346730 Sum_{k>=1} k * T(n,k) = A153876(n). - _Alois P. Heinz_, Aug 01 2021
%e A346730 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.
%e A346730 Triangle begins:
%e A346730   1;
%e A346730   0,  2;
%e A346730   0,  2, 1,  1;
%e A346730   0,  2, 1,  4, 0, 1;
%e A346730   0,  5, 1,  4, 1, 3, 0, 2;
%e A346730   0,  7, 1, 11, 0, 6, 0, 4, 1, 1, 0, 1;
%e A346730   0, 13, 1, 20, 1, 9, 1, 9, 1, 2, 0, 6, 0, 0, 0, 1;
%p A346730 T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(add
%p A346730         (x^numtheory[tau](i), i=2^(n-1)..2^n-1)):
%p A346730 seq(T(n), n=1..10);  # _Alois P. Heinz_, Jul 31 2021
%t A346730 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 *)
%Y A346730 Row sums give A011782.
%Y A346730 Columns k=1-2 give: A063524, A162145 (for n>=2).
%Y A346730 Cf. A000005, A153876, A300509, A346729.
%K A346730 nonn,base,tabf
%O A346730 1,3
%A A346730 _Jon E. Schoenfield_, Jul 30 2021