[seqfan] Re: New sequence?

[Richard Guy]

Well, that went over like a lead balloon!  Here's what I guess to be the 
next three members of A000048:  954437120, 1857283155, 3616814565.

This gives  4096, 2, 4  for the next three members of the suggested
sequence.  Conjecture: in this sequence, if  n  is 3-smooth. then
a(n)  is a power of 2.  [probably easy to prove.]

A000048  is a sort of pseudo-divisibility sequence.  E.g.

2 has ranks of apparition  4, 9 and 25 (and 49, ...???) in the
sense that  A000048(n) is even just if  n  is a multiple of
4 or 9 or 25 (or 49 or ???)

5 has ranks  6, 13, 14, 17, 22, ???

7 has ranks  9, 13, 19, ??

11 has ranks 21, 25, 33, ??

13 has ranks 14, 18, 37, 38, ??

17 has ranks 10, 12, 21, 26, ??

31 has ranks 11, 25, ??

However,  3  does not divide  A000048(15), so maybe I'm barking
up the wrong tree.  Check?  Mod 2^15 + 1, there is a 2-cycle
{10923,-10923}, a 6-cycle {3641,7282,14564,-3641,-7282,-14564},
three 10-cycles of which one is
{2979,5958,11916,-8937,14895,-2979,-5958,...}, and 1091
30-cycles.    R.

On Thu, 19 Apr 2012, Richard Guy wrote:

> Would an editor more competent than I like to enter the following
> sequence into OEIS, if it's not there already (I'm not a good looker) ?
>
> [check & extend.  These are only hand calculations.  A000048 could
> also easily be extended]  For  n = (0) 1 2 3 ...
>
> (0),0,0,2,0,2,4,2,0,8,4,2,16,2,4,38,0,2,64,2,16,134,4,2,256,32,4,
> 512,16,2,1084,2,0,2054,4,159,
>
> It's the total length of all cycles which are strictly less than
> the full length of  2n.
>
>    2^n  -  2 * n * A000048(n)
>
> a(2^k) = 0,  a(prime) = 2,  a(2p) = 4.
>
> There's a simple formula using the Moebius function (v. A000048).
>
> Let me know if I've made errors.   Thanks!   R.
>