A364608 - OEIS

%I #42 Aug 02 2023 13:45:21

%S 2,4,2,1,4,2,2,21,8,5,10,2,4,2,2,61,1,1,34,1,18,12,6,1,63,11,49,2,496,

%T 2,2,58,24,12,11,40,7,43,59,3,3,53,31,54,15,7,59,30,5,185,2,5,97,28,

%U 10,2,6,4,42,2,27,2,2,15,3,72,1,7,1,1,15,37,1,1,129

%N Smallest k such that there are as many 0's as 1's in the binary representation of (2*n+1)^k, or -1 if no such k exists.

%C a(n) is the smallest k such that (2*n+1)^k is in A031443.

%C Conjecture: a(n) != -1 for all n.

%C For even m, the proportion of 1's in the binary representation of m^k conjecturally converges to less than 1/2, so unless the numbers of 0's and 1's are equal for some small k, it is likely that no such k exists. For m = 2, 6, 10, 12, 14, 38, 42, 44, 46, ... there do exist such small k = 1, 3, 1, 1, 3, 1, 1, 1, 5, ..., respectively.

%e For n = 2, (2*n+1)^k = 5^k in binary is 101, 11001, 1111101, 1001110001 for k = 1, 2, 3, 4. Only the last of these has as many 0's as 1's, so a(2) = 4.

%o (Python)

%o def A364608(n):

%o     k = 0

%o     p = 1

%o     while 2*p.bit_count() != p.bit_length():

%o         k += 1

%o         p *= 2*n+1

%o     return k

%Y Cf. A031443.

%K nonn,base

%O 1,1

%A _Pontus von Brömssen_, Jul 30 2023