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Least Prime Factor function , lpf(i)

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jesko

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Feb 11, 2012, 2:34:17 PM2/11/12
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If we divide that values of lpf(i) in three blocks such that:

A = values < p
B = values of p
C = values > p

then it happens that B/C = 1/(p-1) for certain intervals.
Further if x are the values of the argument of lpf(i) that satisfied
this condition then this values are periodic.

Example

If p =7 then

Argument Ratio 7's > than 7 Diff
30 0.166667 1 6 1
53 0.166667 2 12 23
54 0.166667 2 12 1
55 0.166667 2 12 1
56 0.166667 2 12 1
57 0.166667 2 12 1
58 0.166667 2 12 1
79 0.166667 3 18 21
80 0.166667 3 18 1
81 0.166667 3 18 1
82 0.166667 3 18 1
107 0.166667 4 24 25
108 0.166667 4 24 1
187 0.166667 7 42 79
188 0.166667 7 42 1
189 0.166667 7 42 1
190 0.166667 7 42 1
211 0.166667 8 48 21
212 0.166667 8 48 1
213 0.166667 8 48 1
214 0.166667 8 48 1
215 0.166667 8 48 1
216 0.166667 8 48 1
239 0.166667 9 54 23
240 0.166667 9 54 1
263 0.166667 10 60 23
264 0.166667 10 60 1
265 0.166667 10 60 1
266 0.166667 10 60 1
267 0.166667 10 60 1
268 0.166667 10 60 1
289 0.166667 11 66 21
290 0.166667 11 66 1
291 0.166667 11 66 1
292 0.166667 11 66 1
317 0.166667 12 72 25
318 0.166667 12 72 1
397 0.166667 15 90 79
398 0.166667 15 90 1
399 0.166667 15 90 1
400 0.166667 15 90 1
421 0.166667 16 96 21
422 0.166667 16 96 1
423 0.166667 16 96 1
424 0.166667 16 96 1
425 0.166667 16 96 1
426 0.166667 16 96 1

What does it mean?

Thanks
Francesco

hagman

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Feb 11, 2012, 3:47:11 PM2/11/12
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On 11 Feb., 20:34, jesko <frans...@gmail.com> wrote:
> If we divide that values of lpf(i) in three blocks such that:
>
> A = values < p
> B = values of p
> C = values > p
>
> then it happens that B/C = 1/(p-1) for certain intervals.
> Further if x are the values of the argument of lpf(i) that satisfied
> this condition then this values are periodic.

Consider the primorial function p# = 2*3*5*...*p
Clearly, lpf(n + p#) = lpf(n) if n>1 and either side is <= p.
Over a full period of length p#, there is a proportion of
C = (1-1/2)*(1-1/3)*...*(1-1/p) numbers of the third kind
and B = C / (1-1/p) - C = 1/(p-1) * C of the second kind.
Hence the proportion B/C equals 1/(p-1) at least for intervals
of length p#.

Hagman

quasi

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Feb 11, 2012, 3:51:02 PM2/11/12
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On Sat, 11 Feb 2012 11:34:17 -0800 (PST), jesko <fran...@gmail.com>
wrote:
Firstly, you missed 29.

Secondly, what do you mean when you say the values are
periodic? Periodic in what sense?

quasi

jesko

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Feb 11, 2012, 4:22:59 PM2/11/12
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On Feb 11, 8:51 pm, quasi <qu...@null.set> wrote:
> On Sat, 11 Feb 2012 11:34:17 -0800 (PST), jesko <frans...@gmail.com>
> quasi- Hide quoted text -
>
> - Show quoted text -

In the sense that for example for 53 + k210,54 + k210,...... the ratio
1/6 holds.
But is this useful to evaluate prime distribution or prime counting?


Thanks

hagman

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Feb 12, 2012, 11:59:17 AM2/12/12
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I don't think it is useful, as (see my other post) 210 is simply 7# =
2*3*5*7
so much for the significance of the period.
I count 24 cases per period of 210, that's about 11 percent.
Note that only 6 of these are "primitive" in that they come from a
change
of B or C count (shown in your table with a difference >1). The other
18 come
from "A" numbers following a primitive case (shown in your table with
difference=1).
I don't know if this deviates essentially from what we'd observe with
a random
sequence of 48 "C", 8 "B" and 154 "A" numbers.

Hagman

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