- View SourceAndrey Kulsha wrote:

> reaches 1 for p(p(n)) near exp(25) (so n is about 145e6),

...

> and converges to the value of 1.04...

>

> n p(n) p(p(n)) sum(1/p(p(n)))

> 125791 1666589 26717767 0.978334479698888415

> 6037968 105097561 2147483563 0.992403845704372124

My program agrees with Andrey's table.

An extension shows his estimate for sum>1 is good (it's n=148189304):

n p(n) p(p(n)) sum(1/p(p(n)))

10000000 179424673 3767321791 0.993782079779048

20000000 373587883 8132396761 0.995553008182270

30000000 573259391 12736743289 0.996528886890023

40000000 776531401 17500350067 0.997196394520154

50000000 982451653 22383430879 0.997700609931158

60000000 1190494759 27363009697 0.998104122267683

70000000 1400305337 32423573141 0.998439516689021

80000000 1611623773 37553793989 0.998725881592459

90000000 1824261409 42745397623 0.998975327892169

100000000 2038074743 47991893393 0.999196010613537

110000000 2252945251 53287844179 0.999393678625278

120000000 2468776129 58629206579 0.999572530777542

130000000 2685457421 64011376427 0.999735723538573

140000000 2902958801 69432251137 0.999885688879310

148189304 3081648379 73898684653 1.000000000008883

150000000 3121238909 74889755137 1.000024339564021

160000000 3340200037 80380156427 1.000153205905392

170000000 3559788179 85901366807 1.000273531253509

180000000 3780008329 91452739153 1.000386339616163

190000000 4000846301 97033105289 1.000492482024935

200000000 4222234741 102639953363 1.000592674128579

--

Jens Kruse Andersen - View SourceMilton wrote:

> As I indicated, this sum does not converge.

Based on the above (the entire message quoted to please Milton :-),

>

> The comparable sum to compare it to is the reciprocal of the squares.

>

> Consider to the denominators compared to the squares.

>

> 3 5 11 17 31 41 59 67 83 10- 127 157 179 191 211 241 277 283 331 ...

>

> 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...

>

> After 191<196 the squares remain larger so 1/196 < 1/191

>

> Any series like this, where you select from an infinite number of primes,

> will diverge.

>

> Milton L. Brown

I guess he assumes the sum of the reciprocals of squares diverges,

and then he thinks the discussed primeth sum must also diverge

because the terms soon become larger.

One minor problem with that argument:

The sum of reciprocal of squares converges to pi^2/6.

I guess Milton was thinking of the harmonic series

1 + 1/2 + 1/3 + 1/4 + ... which is famous for diverging.

(The pi^2/6 sum is also a little famous.)

I wonder whether I'm better at mind reading than Milton is at maths ;-)

--

Jens Kruse Andersen