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%I #32 Nov 21 2013 07:38:02
%S -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
%N 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.
%C 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.
%C Conjecture: Apart from n=1, 2 and 4, no other a(n) is -1.
%C See A214560 for a related conjecture.
%e Here is a table whose columns give:
%e N, N^2, number of bits in N^2, number of 1's in N^2, number of 0's in N^2:
%e 0 0 1 0 1
%e 1 1 1 1 0
%e 2 4 3 1 2
%e 3 9 4 2 2
%e 4 16 5 1 4
%e 5 25 5 3 2
%e 6 36 6 2 4
%e 7 49 6 3 3
%e 8 64 7 1 6
%e 9 81 7 3 4
%e 10 100 7 3 4
%e 11 121 7 5 2
%e 12 144 8 2 6
%e 13 169 8 4 4
%e 14 196 8 3 5
%e 15 225 8 4 4
%e 16 256 9 1 8
%e 17 289 9 3 6
%e 18 324 9 3 6
%e 19 361 9 5 4
%e ...
%e 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.
%e 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.
%Y Cf. A000120, A023416, A159918, A214560, A230097, A231897.
%K sign,more
%O 1,3
%A _N. J. A. Sloane_, Nov 19 2013
%E Missing word in definition supplied by _Jon Perry_, Nov 20 2013.
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