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A352631
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Minimum number of zeros in a binary n-digit perfect square, or -1 if there are no such numbers.
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1
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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
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OFFSET
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1,3
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COMMENTS
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Is there a formula that is easy to compute?
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LINKS
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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sign,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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