Another Sequence Recursive & Palindromic

Leroy Quet qqquet at
Sat Mar 29 02:46:28 CET 2003

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:



The average of the terms approaches:

1 + sum{k=1 to oo} a(k)/2^(k+1)

Again, a(k) is the k_th term of the original sequence, the sequence
whose terms we are calculating the average of the terms of...

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;


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
odd integer, instead of 0.)

And, in this case, the average of the terms will by 1 less, but so
will the a()'s. So the average, with a'(k) = a(k)-1, a(k) = original

1/2 + sum{k=1 to oo} a'(k)/2^(k+1)


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 sequence-name??)

Leroy Quet 

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