%I #15 Mar 15 2022 20:58:39
%S 1,2,1,3,2,3,4,3,6,1,5,4,7,2,7,6,5,12,3,14,3,7,6,14,4,15,6,7,8,7,15,5,
%T 27,7,14,1,9,8,24,6,28,12,15,2,15,10,9,28,7,30,14,19,3,30,7,11,10,30,
%U 8,31,15,28,4,31,14,3,12,11,31,9,39,24,30,5,43,15,6,3,13,12
%N Square array, read by descending antidiagonals, where row n gives all solutions k > 0 to A000120(k)=A000120(k*n), A000120 is the Hamming weight.
%C Square array is read by descending antidiagonals, as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.
%C Rows at positions 2^k are 1, 2, 3, ..., (A000027). Row 2n is equal to row n.
%C Values are different from those in A115872, because we use multiplication with carry here.
%e Eight initial terms of rows 1 - 8 are listed below:
%e 1: 1, 2, 3, 4, 5, 6, 7, 8, ...
%e 2: 1, 2, 3, 4, 5, 6, 7, 8, ...
%e 3: 3, 6, 7, 12, 14, 15, 24, 28, ...
%e 4: 1, 2, 3, 4, 5, 6, 7, 8, ...
%e 5: 7, 14, 15, 27, 28, 30, 31, 39, ...
%e 6: 3, 6, 7, 12, 14, 15, 24, 28, ...
%e 7: 7, 14, 15, 19, 28, 30, 31, 37, ...
%e 8: 1, 2, 3, 4, 5, 6, 7, 8, ...
%e a(6,3) = 7 because: 7 in binary is 111 and 6*7 = 42 in binary is 101001, both have 3 bits set to 1.
%o (MATLAB)
%o function [a] = A340351(max_n)
%o for n = 1:max_n
%o m = 1;
%o k = 1;
%o while m < max_n
%o c = length(find(bitget(k,1:32)== 1));
%o if c == length(find(bitget(n*k,1:32)== 1))
%o a(n,m) = k;
%o m = m+1;
%o end
%o k = k +1;
%o end
%o end
%o end
%Y Cf. A000120, A292849 (1st column), A340069, A077459 (3rd row).
%K nonn,base,tabl
%O 1,2
%A _Thomas Scheuerle_, Jan 05 2021