Hofstadteriana with primes

franktaw at netscape.net franktaw at netscape.net
Tue Jun 26 22:20:58 CEST 2007


This is almost certainly wrong.  The limit is almost certain to be 0, 
on probabilistic grounds.  One would expect it to decline gradually 
(roughly on the order of 1/log(n)), so you need to look at an 
exponential sequence of values to see it, not (as Jacques did) a 
linearly increasing sequence.

E.g., using his data,

3000 38857028 194 6.4666666667
6000 169071260 354 5.9000000000
12000 730829204 642 5.3500000000
24000 3141343994 1193 4.9708333333

Franklin T. Adams-Watters

-----Original Message-----
From: Jacques Tramu <jacques.tramu at echolalie.com>

From: "Eric Angelini" <Eric.Angelini at kntv.be>
> S = 1   3   8   15  26  39  56  75  98  127  158  195  236  279  
326...
> d =   2   5   7   11  13  17  19  23  29   31   37   41   43   47 ...
>
> - start S with 1
> - add the smallest prime not yet added and not already present in S
>
> Question:
> - What could be the ratio primes/composites of S?

...

I have a proof that the limit of the ratio is 5%, unfortunately,
the left margin of this sheet of paper is too small ....





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>It really shouldn't.
>
>It is common, although incorrect, to regard the prime factorization of
>1 as 1.  See, for example, A027746.  But really, the prime
>factorization of 1 is the empty multiset, which sums to 0.
>
>The description of A096461 should be changed to "Starting with a(1)=1
>and a(2)=2, ..."
>
>Franklin T. Adams-Watters
>
>-----Original Message-----
>From: Neville Holmes <nholmes at leven.comp.utas.edu.au>
>
>A096461, starting 1 2 4 8 14 23 ..., is described as
>"Starting with a(1)=1, increment a(n) by the sum of
>its prime factors."  I can see how it would work
>with a(1)=2, but why and how does it start with 1 ?


A096460 has the same problem.

Tony






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