OFFSET
1,3
COMMENTS
Is there a formula that is easy to compute?
EXAMPLE
a(6) = 3, because there are two 6-bit squares 36 = 100100_2 and 49 110001_2 with 4 and 3 zeros, respectively.
a(2) = -1, because the first two perfect squares 1 = 1_2 and 4 = 100_2 have 1 and 3 bits, respectively.
PROG
(Python)
from gmpy2 import is_square, popcount
for n in range(1, 33):
m=n+1
for k in range(2**(n-1), 2**n):
if is_square(k):
m=min(m, n-popcount(k))
print(n, -1 if m>n else m)
(Python 3.10+)
def A352631(n): return -1 if n == 2 else min(n-(k**2).bit_count() for k in range(1+isqrt(2**(n-1)-1), 1+isqrt(2**n))) # Chai Wah Wu, Mar 28 2022
CROSSREFS
KEYWORD
sign,base
AUTHOR
Martin Ehrenstein, Mar 25 2022
EXTENSIONS
a(43)-a(71) from Pontus von Brömssen, Mar 26 2022
a(72)-a(80) from Chai Wah Wu, Apr 01 2022
STATUS
approved