# Sequence Pictures

Leroy Quet qq-quet at mindspring.com
Tue Dec 9 23:42:46 CET 2003

```>I did a column about turning sequences into pictures.
>
>http://www.maa.org/editorial/mathgames/mathgames_12_08_03.html
>
>--Ed Pegg Jr.

Interesting.

But there seems to be some problems with where the images appear on the
page.
(For they alternate between 2 locations on the page every time I load it.
Are the images suppose to be grouped together without numerical labels,
as they appear with my browswer?)

I have had my interest resurrected in the following sequence recently.

I would bet that it would look interesting represented as you have the
other sequences, given its recursive definition (a definition which
involves 2 as well).

(Also, a closed-form definition is given below, unless I made a mistake
in 'discovering' it.)

If someone graphically represents the sequence described below, and you
can post it to a website, please send me the link. I am curious now if it
would LOOK interesting...

Relevant parts of my sci.math post describing the sequence(s):

>Let A(1) = 1, a sequence with only a single element.
>
>Let A(m+1) =
>{A(m), a(m)+1, A(m)},
>
>a concatenation, where a(m) is the m_th term of the sequence A(m).
>
>( A(m) has 2^m -1 elements. So a(m) is not the last element of A(m)
>for m >= 2, it should be emphasized.)
>
>The sequence begins:
>
>1,2,1,3,1,2,1,2,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,2,1,2,1,3,1,2,1,...
>
>
>.....

>
>A "closed-form" for this sequence is, I think:
>
>If e(m) is a nonnegative integer such that:
>2^e(m) is the highest power of 2 dividing m;
>
>then:
>
>a(m) is the number of e()'s where:
>
>e(e(e(...e(m)...))) = 0.
>
>--
>It might be more natural, for some purposes, to define A(1) = 0.
>Then we will have the same sequence minus 1.
>
>(In this case, a(m) would be the number of e()'s needed to achieve an
>
>.....

>--
>
>I THINK that the a(k)-sequence, taken mod 2, *might* be likely to be the
>absolute values of the "first differences of the Thue-Morris sequence"
>according to the OEIS. (paraphrased title??)
>

thanks,
Leroy
Quet

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