Problems & Puzzles: Puzzles

Puzzle 566.- Semiprimes in a given interval

JM Bergot sent the following puzzle:

In the open interval P(n)^2 to P(n+1)^2 find the number of semiprimes, NS. Let AN be all numbers in this interval.

 Q. Does NS/AN stay near 0.3 as the primes interval increase?


Contributions came from Ryan Bailey, Giovanni Resta, Faride Firoozbakht and Jerrold W. Pease.

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Ryan Bailey wrote:

No, although initially when n is low and the interval between P(n)^2 and P(n+1)^2 causes AN to be low, NS/AN can be close to .3. But when the interval increases, so does NS, but AN increases even faster. 

Here's some specific calculations.
 
n = 0 SN//AN = 0.25
n = 1 SN//AN = 0.5
n = 2 SN//AN = 0.4
n = 3 SN//AN = 0.35714285714285715
n = 4 SN//AN = 0.36363636363636365
n = 5 SN//AN = 0.23076923076923078
n = 6 SN//AN = 0.38235294117647056
n = 7 SN//AN = 0.3157894736842105
n = 8 SN//AN = 0.32608695652173914
n = 9 SN//AN = 0.29310344827586204
...
n = 100 SN//AN = 0.07769652650822668
n = 101 SN//AN = 0.07360861759425494
n = 102 SN//AN = 0.06483126110124333
n = 103 SN//AN = 0.0632688927943761
n = 104 SN//AN = 0.07355516637478109
n = 105 SN//AN = 0.07019064124783363
n = 106 SN//AN = 0.07836456558773425
n = 107 SN//AN = 0.05902192242833052
n = 108 SN//AN = 0.07178631051752922
n = 109 SN//AN = 0.064891846921797
Huge Gap from .3
...
n = 400 SN//AN = 0.02000727537286286
n = 401 SN//AN = 0.02015982564475118
n = 402 SN//AN = 0.01969642211781713
n = 403 SN//AN = 0.02106589845156644
n = 404 SN//AN = 0.019541054141269273
n = 405 SN//AN = 0.020601934790397708
n = 406 SN//AN = 0.021094029317125493
n = 407 SN//AN = 0.02070689039628704
n = 408 SN//AN = 0.01962183374955405
n = 409 SN//AN = 0.02075203973040085
n = 410 SN//AN = 0.01906106600776562

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Giovanni Resta wrote:

it seems to me that the ratio decreases with n.
I've attached a plot of the function for n<50000
together with a fit.
I obtained a good fit with this function:
f(x) = (0.13462+0.5384*log(log(x))) / log(x)

In any case the shape of the data seems to suggest
that the ratio goes to 0 as n increases.

I tried to obtain an asymptotic estimation but
it is not trivial.

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Farideh Firoozbakht wrote:

Let f[(n) be NS(n)/ AN(n) where NS(n) is number of semiprimes in the open interval
(prime(n)^2, prime(n+1)^2)  and NS(n) =  prime(n+1)^2-prime(n)^2-1 namely NS(n) is
all integers in this interval. 
 
We have :
 
f(1) = 0.25
f(10) = 0.294118
f(100) = 0.218324
f(1000) = 0.169326
f(2000) = 0.162384
f(3000) = 0.157281
f(4000) = 0.153635
f(5000) = 0.151076 
f(6000) = 0.149134

f(7000) = 0.14768

So the answer of question is no.

I guess the limit goes to zero

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Jerrold W. Pease wrote:

A good estimate for the number of semiprimes less than n is this:
 
         log(log(n)/log(2)) * n / log(n)
 
Also:   P(n) is about (k*log(k)).
 
Therefore:  P(n+1) is about ((k+1)*log(k+1)).

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Emmanuel Vantieghem wrote:

Here is my contribution to Puzzle 566 :
_______________________________________________
 
n = 10 :      SN/AN = 0.294118
n = 100 :    SN/AN = 0.218324
n = 1000 :   SN/AN = 0.169326  
n = 10000 : SN/AN = 0.144421
  ...
n = 16384 (= 2^14) : SN/AN = 0.140126
________________________________________________
 
I send you this because there is probably a mistake in either my results or those of Ryan Bailey.  Besides the case  n = 100, I found for  n = 109 : SN/AN = 0.229262  and for  n = 410 : SN/AN = 0.184615.
 
Could it be that the notion of 'semiprime' is different for us ?

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