Here's the best I could do: Let b(n) be the distance between the first and last 1 in the binary expression of n. For example, b(1) = 0, b(3) = 1, b(5) = 2, b(6) = 1. Then your sequence is f(n) = 0 if n = 0, else b(n) if n even, else b(n)+b(n+1). Yeah, ugly. ----- Original Message ----- From: "Fred lunnon" <fred.lunnon@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, August 16, 2007 1:26 PM Subject: [math-fun] fractal sequence
This sequence arose in connection with divisibility of Stirling numbers of the first kind.
It took me a while to figure out a (slightly clumsy) explicit expression for the n-th term.
Seeing that things are a bit slack on the list just now, I thought I'd share it with everybody.
The terms for n = 0, ..., 257 follow below --- should suffice for a wet afternoon! WFL
[0, 0, 0, 1, 0, 3, 1, 2, 0, 5, 2, 4, 1, 5, 2, 3, 0, 7, 3, 6, 2, 7, 3, 5, 1, 7, 3, 6, 2, 7, 3, 4, 0, 9, 4, 8, 3, 9, 4, 7, 2, 9, 4, 8, 3, 9, 4, 6, 1, 9, 4, 8, 3, 9, 4, 7, 2, 9, 4, 8, 3, 9, 4, 5, 0, 11, 5, 10, 4, 11, 5, 9, 3, 11, 5, 10, 4, 11, 5, 8, 2, 11, 5, 10, 4, 11, 5, 9, 3, 11, 5, 10, 4, 11, 5, 7, 1, 11, 5, 10, 4, 11, 5, 9, 3, 11, 5, 10, 4, 11, 5, 8, 2, 11, 5, 10, 4, 11, 5, 9, 3, 11, 5, 10, 4, 11, 5, 6, 0, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 10, 3, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 9, 2, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 10, 3, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 8, 1, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 10, 3, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 9, 2, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 10, 3, 13, 6, 12, 5, 13, 6, 11, 4, 13, 6, 12, 5, 13, 6, 7, 0, ...]
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