A364608 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.

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

Offset

1,1

Comments

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

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

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.

Links

Table of n, a(n) for n=1..75.

Example

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.

Prog

(Python)
def A364608(n):
    k = 0
    p = 1
    while 2*p.bit_count() != p.bit_length():
        k += 1
        p *= 2*n+1
    return k

Crossrefs

Cf. A031443.

Sequence in context: A021417 A368517 A355346 * A105791 A116515 A303118

Adjacent sequences: A364605 A364606 A364607 * A364609 A364610 A364611

Keyword

nonn,base

Author

Pontus von Brömssen, Jul 30 2023

Status

approved