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, ...]
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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On 8/19/07, David Wilson <davidwwilson@comcast.net> wrote:
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.
Maybe so, but (although they're easily seen to be equivalent) this formulation is a worthwhile improvement on my own solution. The sequence posed was associated with p = 2, which behaves atypically: instead of the explicit formula depending on the residue n mod 2, for general p it depends on n mod(p-1). These sequences arise as the residual between the exact value of ord_p s(m, n) and the sharp monotonic lower bound for 0 <= n <= m SMLB(m, n) = \sum_{j > 0} \max(0, -[n - m/p^j]) --- which I earlier tried (and failed) hard (I really did) to announce without any misprints. In the case that m = p^k it's possible to conjecture the explicit form of this residual for all n, thus giving the exact order. Proving these formulae, or extending them to general m, still looks a fairly distant prospect. It's nice that these sequences of residuals are independent of k. And David's expression unexpectedly depends on n, n+1; whereas the basic Stirling number recursion depends on n, n-1. I've now recast my solution for general p along the lines suggested --- thanks for the input. Fred Lunnon [19/08/07]
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Fred lunnon