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A110502
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Numbers n such that n in binary representation has a block of exactly a nontrivial square number of zeros.
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2
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16, 33, 48, 66, 67, 80, 97, 112, 132, 133, 134, 135, 144, 161, 176, 194, 195, 208, 225, 240, 264, 265, 266, 267, 268, 269, 270, 271, 272, 289, 304, 322, 323, 336, 353, 368, 388, 389, 390, 391, 400, 417, 432, 450, 451, 464, 481, 496, 512, 528, 529, 530, 531
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OFFSET
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1,1
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COMMENTS
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a(n) is the index of zeros in the complement of the square analog of the Baum-Sweet sequence, which is b(n) = 1 if the binary representation of n contains no block of consecutive zeros of exactly a nontrivial square number length; otherwise b(n) = 0.
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REFERENCES
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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 157.
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LINKS
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FORMULA
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a(n) is in this sequence iff n (base 2) has a block (not a sub-block) of k^2 = A000290(k) consecutive zeros for k>1.
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EXAMPLE
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a(1) = 16 because 16 (base 2) = 10000, which has a block of 4 = 2^2 zeros.
a(2) = 33 because 33 (base 2) = 100001, which has a block of 4 zeros.
a(3) = 48 because 48 (base 2) = 110000, which has a block of 4 zeros.
a(49) = 512 because 512 (base 2) = 1000000000, with a block of 9 = 3^2 zeros.
Similarly, there are blocks of exactly 9 zeros in 1025, 1536, 2050, 2051, 3073, 3584, 7149, 8196, 8197, 8198, 8199.
65536, 131073, 196608, 262146 and 262147 are in this sequence because (base 2) they each have a block of 16 = 4^2 zeros.
33554432 has a block of 25 = 5^2 zeros.
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MAPLE
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filter:= proc(n) local L, nL, A, B;
L:= convert(n, base, 2);
nL:= nops(L);
A:= select(t -> L[t]=0 and (t=1 or L[t-1]=1), [$1..nL]);
B:= select(t -> L[t]=1 and L[t-1]=0, [$2..nL]);
ormap(t -> t>3 and issqr(t), B-A)
end proc:select(filter, [$1..1000]); # Robert Israel, Sep 01 2021
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MATHEMATICA
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Select[Range[531], Or @@ (First[ # ] == 0 && Length[ # ] > 1 && IntegerQ[Length[ # ]^(1/2)] &) /@ Split[IntegerDigits[ #, 2]] &] (* Ray Chandler, Sep 12 2005 *)
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PROG
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(Python)
from math import isqrt
from itertools import groupby
def is_nt_sqr(n): # is nontrivial square
return n > 1 and isqrt(n)**2 == n
def ok(n):
b = bin(n)[2:]
return any(k == '0' and is_nt_sqr(len(list(g))) for k, g in groupby(b))
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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