[seqfan] Re: Binary Palindrome Subsequences

[Benoît Jubin]

It seems that you are counting "factors" or "subwords" starting and
ending with one, rather than subsequences.
Indeed, 111, 1111 and 11111 are also subsequences, for 203.

Benoit


On Mon, Dec 15, 2008 at 9:02 AM, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
> Let a(n) = the number of distinct palindromic subsequences, each subsequence starting and ending with 1, that are contained in the binary representation of n.
>
> For most n, it seems, a(n) = A000120(n).
>
> (A000120(n) is the number of 1's digits in the binary representation of n.)
>
> But there are exceptions.
>
> For instance, 203 = 11001011 in binary.
>
> A000120(203) = 5, obviously.
>
> But I count 4 palindromic subsequences, each ending and starting with 1: 1, 11, 101, 1001.
>
> What is the sequence of positive integers n where a(n) does NOT = A000120(n).
>
> ....203, 211,...
>
> (I guess this sequence starts with 203, but I may have made an error.)
>
> Thanks,
> Leroy Quet
>
>
>
>
>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>