

A352631


Minimum number of zeros in a binary ndigit perfect square, or 1 if there are no such numbers.


1



0, 1, 2, 2, 2, 3, 2, 4, 3, 4, 3, 4, 4, 5, 2, 5, 4, 6, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 6, 8, 8, 6, 7, 7, 8, 8, 9, 8, 9, 9, 8, 9, 10, 9, 9, 10, 9, 9, 9, 9, 10, 10, 10, 10, 11, 10, 11, 11, 11, 9, 9, 11, 11, 11, 12, 11, 12, 11, 12
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

Is there a formula that is easy to compute?


LINKS



EXAMPLE

a(6) = 3, because there are two 6bit 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**(n1), 2**n):
if is_square(k):
m=min(m, npopcount(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**(n1)1), 1+isqrt(2**n))) # Chai Wah Wu, Mar 28 2022


CROSSREFS



KEYWORD

sign,base


AUTHOR



EXTENSIONS



STATUS

approved



