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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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified June 26 20:19 EDT 2019. Contains 324380 sequences. (Running on oeis4.)