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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