Template talk:Sequence of the Day for October 3

On Template:Sequence_of_the_Day_for_October_3 you asked,

How can this sequence be computed most efficiently? (...)

but then give the answer which is the same as the FORMULA given there since the day it was created,

a(n) = 2*A018900(n)+1.

Going a bit further, one could mention that the sequence could be arranged as triangle where a row contains all numbers having the same binary length:

111
1011 1101
10011 10101 11001
100011 100101 101001 110001
etc

It is seen that the r-th row contains r elements of binary length r+2, the first term being 2^(r+1)+3, and the last term being 3*2^r+1, and in between them the middle 1 "wanders" from the right to the left. This yields an easy way to find which is the position of a given term of the sequence, and conversely, get a(n) explicitely as function of n. — M. F. Hasler 19:02, 3 October 2011 (UTC)

That sounds very clever. It might be worth editing both the entry for A084468 and editing this Sequence of the Day write-up. Alonso del Arte 20:51, 3 October 2011 (UTC)