tea.mathoverflow.net - Discussion Feed (Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1)

Andy Putman comments on "Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1" (16885)

Andy Putman — Sat, 05 Nov 2011 20:29:43 -0700

nb to Scott : the thread has not been closed...

Scott Morrison comments on "Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1" (16882)

Scott Morrison — Sat, 05 Nov 2011 18:00:42 -0700

For help with MarkDown syntax (which you can use here on meta, and also on the main site), please read http://daringfireball.net/projects/markdown/syntax.

(I'm closing this thread now, per the comment two above.)

Vassilis Parassidis comments on "Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1" (16873)

Vassilis Parassidis — Sat, 05 Nov 2011 13:57:44 -0700

Is there a way to include a Table in the text?

Vassilis Parassidis comments on "Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1" (16872)

Vassilis Parassidis — Sat, 05 Nov 2011 13:55:04 -0700

Please withdraw this. The formatting has been lost in the post.

Vassilis Parassidis comments on "Range of ascending primes for diameters D_1=5^2^{n-1} and D_2=p_1∙p_2∙…p_n with ratio ≈ 1" (16871)

Vassilis Parassidis — Sat, 05 Nov 2011 13:44:04 -0700

Let’s have a circle with diameter D_1=5^2^{n-1} where n any non-zero natural number. The number of primitive right angle triangles which can be inscribed in this circle is equal to 2^{n-1}. Additionally, the number of primitive right angle triangles inscribed in a circle with diameter equal to a product of primes of the form p=a^2+b^2 is obtained from the diameter D_2=p_1∙p_2∙…p_n=2^{n-1} .

n Length of Diameter Length of Diameter Number of
5^2^{n-1} p_1∙p_2∙…p_n Primes

1 5^1
2 5^2 5∙13 2
3 5^4 5∙13∙17 3
4 5^8 5∙13∙17∙29 4
5 5^16 5∙13∙17∙29∙37 5

We can increase the numbers indefinitely; the ratio of the two diameters increases rapidly. Now we have to find primes whose product gives a ratio 5^2^{n-1}/{ p_1∙p_2∙…p_n}≈1. At this point I want to remind you that the primes are of the form p=a^2+b^2 and from both diameters we obtain equal numbers of primitive right angle triangles inscribed in the circles. In order to find primes in ascending order that give such ratios, we have to find the vicinity in which they lie as follows:
Let’s take the diameter 5^16. Five primes are required and 5^16=5^15∙5^1 and (5^{1/15}∙5)^3≈172. So we have found the vicinity in which these primes lie. Below are the diameters equal to the new products of primes and their vicinities..

Length of Diameter Length of Diameter Vicinity in which Number of Primes
D_1=5^2^{n-1} equal to the new product of primes these primes lie

5^4 5∙13∙17 same 3
5^8 13∙17∙29∙37 25 4
5^16 149∙157∙173∙181∙193 172 5
5^32 5317∙5323∙5327∙5339∙5353∙5363 5343 6
∙ ∙ ∙
∙ ∙ ∙
5^32768 5^2048 15

We can increase the exponents indefinitely. We know the vicinity we have to find the range in which the primes lie. Conjecturally I say, this range can be found as follows:
Vicinity /(1+1/Number of Primes) Vicinity ∙(1+1/ Number of Primes)
Let’s take the Vicinity 172:
172/(1+1/5)=143+1/3 172∙(1+1/5)=206+4/10
or 172^2=(143+1/3)( 206+4/10) and 206/2ln206-143/2ln143≈5
Is this conjecture correct?