OFFSET
1,1
COMMENTS
It's clear that a(2^i)=-1 if i>0, because 2^i*n always has more 0's than n does. I do not know if every number other than a power
of 2 has such a k. Local maxima seem to be near odd powers of 2. It is not hard to show that a(2^n +- 1)!=-1 for n>=1.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program
FORMULA
a(n) = A358858(n)/n unless a(n) = -1. - Pontus von Brömssen, Dec 03 2022
EXAMPLE
For n = 3 both 49 = [110001] and 49*3 = [10010011] have the same number of 0's as 1's, and this is the least such.
PROG
(Python)
from itertools import count
from sympy.utilities.iterables import multiset_permutations
def isbalanced(n): b = bin(n)[2:]; return b.count("0") == b.count("1")
def A031443gen(): yield from (int("1"+"".join(p), 2) for n in count(1) for p in multiset_permutations("0"*n+"1"*(n-1)))
def a(n):
if n > 1 and bin(n)[2:].strip("0") == "1": return -1
return next(k for k in A031443gen() if isbalanced(k*n))
print([a(n) for n in range(1, 61)]) # Michael S. Branicky, Dec 03 2022
(PARI) See Links section.
CROSSREFS
KEYWORD
sign,base
AUTHOR
Jeffrey Shallit, Dec 03 2022
STATUS
approved