[seqfan] Re: Powers of two in permutation of digits

David Corneth davidacorneth at gmail.com
Fri Apr 22 22:27:59 CEST 2016


Thank you. Forgot the word 'least', so we would have
'For a number n, consider all permutation of digits. Select the least
number divisible by the highest power of 2, formed by one of the
permutations.'

I have a method to do this, which leans on the sequence A261370. Writing a
good explanation for that code is something I'm not to good at.

However, maybe that sequence isn't needed. Say we're investigating n =
546338. Candidates for last 1 digit are
3, 4, 5, 6, 8. At least one of those is divisible by 2^1 so a sequence of
last digits must be 4, 6 or 8, leaving those as candidates.

Proceding for the last 2 digits,
They are eighter 34, 54, 64, 84, 36, 46, 56, 86, 38, 48, 58, 68. At least
one of those is divisible by 2^2 so a sequence of last digits must be 64,
84, 36, 56, 48 or 68.

Proceed similarily for the last 3, 4 etc digits. However, the list of
candidates may well get long.
We should get 354368.

For large numbers n, constructing a(n) like this at least for a part may be
an advantage. Whether solely constructing like this or lateron just trying
permutations is a good way, I'd like to see some insight on that if
possible.

I submitted A272215 as the sequence described above and A272216 as the
exponent of the highest power of 2 an element of A272216 is divisible by.
Any more thoughts on these sequences?





On Fri, Apr 22, 2016 at 9:24 PM, Frank Adams-Watters <franktaw at netscape.net>
wrote:

> You do need to note that in the case of a tie, the smallest tied
> permutation is selected.
>
> I would say go ahead and submit it.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: David Corneth <davidacorneth at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Fri, Apr 22, 2016 1:52 pm
> Subject: [seqfan] Powers of two in permutation of digits
>
> For a number n, consider all permutation of digits. Select the number
> divisible by the highest power of 2, formed by one of the permutations.
> The sequence is, with offset 1 as follows:
>
> 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,12,22,32,24,52,26,72,28,92,30,13,32,33,34,35...
> Would this be nice to add to the OEIS?
> --Seqfan Mailing list - http://list.seqfan.eu/
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>


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