OFFSET
1,2
COMMENTS
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.
Does this sequence consist only of odd numbers? It appears so.
This will be the case if A292849 contains all odd numbers, because a(n) will then never become 2.
A292849 must contain all odd numbers if two conditions are satisfied:
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.
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.
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.
PROG
(MATLAB)
function a = A340349(maxA292849)
c = A340351(maxA292849, 1);
n = 1; run = 1;
while run == 1
i = find(c==(n*2)-1);
if ~isempty(i);
a(n) = i(1);
n = n+1;
else
run = 0;
end
end
end
function a = A340351(max_n, max_m)
for n = 1:max_n
m = 1; k = 1;
while m < max_m+1
c = length(find(bitget(k, 1:32)== 1));
if c == length(find(bitget(n*k, 1:32)== 1))
a(n, m) = k;
m = m+1;
end
k = k +1;
end
end
end
(PARI) f(n) = my(k=1); while ((hammingweight(k)) != hammingweight(k*n), k++); k; \\ A292849
a(n) = my(k=1); while(f(k) != 2*n-1, k++); k; \\ Michel Marcus, Jan 09 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Thomas Scheuerle, Jan 05 2021
STATUS
approved