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A340349 a(n) is the smallest k such that A292849(k) = 2n-1.

%I A340349 #30 Jan 10 2021 11:10:17
%S A340349 1,3,13,5,57,35,21,9,241,219,49,45,169,83,73,17,993,59,941,53,3197,51,
%T A340349 185,93,209,81,349,85,41,89,105,33,4033,491,4749,247,449,227,429,363,
%U A340349 3249,401,193,259,233,107,117,189,697249,1355,173,517,473,1091,101,231,725,305
%N A340349 a(n) is the smallest k such that A292849(k) = 2n-1.
%C A340349 This implies that a(n)=k is a solution to A000120(2*n-1) = A000120((2*n-1)*k), where A000120 is the Hamming weight. If no such number exists we define a(n) = 2.
%C A340349 Does this sequence consist only of odd numbers? It appears so.
%C A340349 This will be the case if A292849 contains all odd numbers, because a(n) will then never become 2.
%C A340349 A292849 must contain all odd numbers if two conditions are satisfied:
%C A340349 First: For every odd number 2n-1 there must be an odd k > 1 that satisfies A000120(2n-1) = A000120((2n-1)*k). To prove that this condition is satisfied, it may be helpful that if k*m = 2^j+r, we know that A000120(k*m) = A063787(r) and for each Hamming weight there exists an r such that A063787(r) = A063787(r+1). This allows us to choose r such that the Hamming weight becomes A000120(m) = A063787(r). For a given r, k*m = 2^j+r may have no solutions if k or m are divisors of r, but there may still exist a solution for k*m = 2^j+r+1. Of course this is not a complete proof.
%C A340349 Second: For every n there needs to be a number k such that 2n-1 is the smallest solution to A000120(2n-1) = A000120((2n-1)*k). This is satisfied if no row in A340441 is a subset of a previous row.
%C A340349 No odd number can appear more than once in this sequence, but not all odd numbers will appear, so this sequence is not a permutation of the odd numbers.
%C A340349 This sequence can be constructed from A340441. Start with a(1) = 1, then a(n) is the least number in row n-1 of A340441 that has not already appeared in A340441.
%o A340349 (MATLAB)
%o A340349 function a = A340349(maxA292849)
%o A340349 c = A340351(maxA292849,1);
%o A340349 n = 1; run = 1;
%o A340349 while run == 1
%o A340349      i = find(c==(n*2)-1);
%o A340349      if ~isempty(i);
%o A340349          a(n) = i(1);
%o A340349          n = n+1;
%o A340349      else
%o A340349          run = 0;
%o A340349      end
%o A340349 end
%o A340349 end
%o A340349 function a = A340351(max_n,max_m)
%o A340349     for n = 1:max_n
%o A340349         m = 1; k = 1;
%o A340349         while m < max_m+1
%o A340349             c = length(find(bitget(k,1:32)== 1));
%o A340349             if c == length(find(bitget(n*k,1:32)== 1))
%o A340349                 a(n,m) = k;
%o A340349                 m = m+1;
%o A340349             end
%o A340349             k = k +1;
%o A340349         end
%o A340349     end
%o A340349 end
%o A340349 (PARI) f(n) = my(k=1); while ((hammingweight(k)) != hammingweight(k*n), k++); k; \\ A292849
%o A340349 a(n) = my(k=1); while(f(k) != 2*n-1, k++); k; \\ _Michel Marcus_, Jan 09 2021
%Y A340349 Cf. A000120, A292849, A340069, A263132, A077459, A295827, A340441, A340351 table of multiplications.
%K A340349 nonn,base
%O A340349 1,2
%A A340349 _Thomas Scheuerle_, Jan 05 2021