

A231898


a(n) = smallest k with property that for all m >= k, there is a square N^2 whose binary expansion contains exactly n 1's and m 0's; or 1 if no such k exists.


5



1, 1, 2, 1, 4, 3, 4, 3, 4, 5, 5, 5, 6, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
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OFFSET

1,3


COMMENTS

a(n) = 1 for n = 1, 2 and 4, because all squares with exactly 1, 2 or 4 1's in their binary expansion must contain an even number of 0's.
Conjecture: Apart from n=1, 2 and 4, no other a(n) is 1.
See A214560 for a related conjecture.


LINKS

Table of n, a(n) for n=1..26.


EXAMPLE

Here is a table whose columns give:
N, N^2, number of bits in N^2, number of 1's in N^2, number of 0's in N^2:
0 0 1 0 1
1 1 1 1 0
2 4 3 1 2
3 9 4 2 2
4 16 5 1 4
5 25 5 3 2
6 36 6 2 4
7 49 6 3 3
8 64 7 1 6
9 81 7 3 4
10 100 7 3 4
11 121 7 5 2
12 144 8 2 6
13 169 8 4 4
14 196 8 3 5
15 225 8 4 4
16 256 9 1 8
17 289 9 3 6
18 324 9 3 6
19 361 9 5 4
...
a(n) is defined by the property that for all m >= a(n), the table contains a row ending n m. For example, there are rows ending 3 2, 3 3, 3 4, 3 5, ..., but not 3 1, so a(3) = 2.
a(5)=4: for t>=0, (11*2^t)^2 contains 5 1's and 2t+2 0's and (25*2^t)^2 contains 5 1's and 2t+5 0's, so for m >= 4 there is a number N such that N^2 contains 5 1's and m 0's. Also 4 is the smallest number with this property, so a(5) = 4.


CROSSREFS

Cf. A000120, A023416, A159918, A214560, A230097, A231897.
Sequence in context: A283166 A130973 A093779 * A276638 A116449 A316433
Adjacent sequences: A231895 A231896 A231897 * A231899 A231900 A231901


KEYWORD

sign,more


AUTHOR

N. J. A. Sloane, Nov 19 2013


EXTENSIONS

Missing word in definition supplied by Jon Perry, Nov 20 2013.


STATUS

approved



