Binary Numbers Arranged

[Leroy Quet]

I thought that since a(25) = 16 is so very small when compared to 25!, 
then finding an acceptable permutation for the first 25 positive integers 
would be relatively difficult.

But it did not actually require much effort on my part, and so definitely 
would not require much effort on the part of most of you, to find an 
acceptable permutation (just for fun).

The example I got of an allowable permutation of (1,2,3,...,25) is below, 
with the replied-to post used as spoiler space.

thanks,
Leroy Quet


>Max wrote
>>Max wrote:
>>> Max wrote:
>>>> Leroy Quet wrote:
>>>>> What is the sequence where the nth term, a(n), is the number of 
>>>>> permutations of (1,2,3,...,n) written in binary so that no adjacent 
>>>>> elements share a common 1-bit?
>>>>>
>>>>> In other words, if b(m) and b(m+1) are adjacent elements written in 
>>>>> binary,
>>>>> then (b(m) AND b(m+1)) = 0 for 1 <= m <= n-1.
>>>>> (If a logical AND is applied to each pair of adjacent terms, the 
>>>>> result is zero.)
>>
>>...
>>
>>> With 3 more terms:
>>> 1 2 0 4 2 0 0 0 8 32 0 8 0 0 0 0 0 64 0 1968 508 0 0
>>
>>And 2 more terms:
>>1 2 0 4 2 0 0 0 8 32 0 8 0 0 0 0 0 64 0 1968 508 0 0 0 16
>>
>>Max
>
>
>Hmmm.. I wonder if there are any more 2's in the sequence. (2 being the 
>lowest possible positive value for the terms of the sequence past the 
>first.)
>It might make a fun puzzle to arrange (1 through n) so that no binary 
>one's are next to each other, for some n where a(n) is low and positive.
>
>thanks,
>Leroy Quet


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15, 16, 7, 24, 3, 20, 11, 4, 19, 12, 17, 14,
1, 22, 9, 6, 25, 2, 13, 18, 5, 10, 21, 8, 23