Problems & Puzzles: Conjectures

Conjecture 68. Based on an Euler's Conjecture

Luis Rodríguez sent the following conjecture:

In Polya's "Mathematics and Plausible Reasoning . Vol II.",  it appear the following  Euler's conjecture:

"Any integer of the form 8n + 3 can be decomposed in the sum of a square and double of a prime"

That's equivalent to: "For each number N there is a triangular ( T = x(x + 1)/2 ) such that 4(N - T) + 1  = prime

This means that given x = 0, 1, 2 , 3.... , before  T  attains N  there will be a prime = 4(N - T) + 1

My conjecture is : "The ratio  x /  (Log N)^2  is bounded by a constant < 2"

Here are some experimental results:

N                         X                   T                        x / (Log(N)^2
6101                   88                   3872               1.1583
108977               154              11935               1.1447
275141               226              25651               1.4406
295832               220              24310               1.3863
324041               196              19012               1.2174
552116               232              27028               1.3272

Questions.

1. - Can someone find a relation > 1.4406 ?
2.- There is some heuristic argument to fund the conjecture ?

Contribution came from Hakan Summakoğlu & Emmanuel Vantieghem

***

Hakan wrote:

Q1:

For N=2 , X=1 , produce T=1 , 4*(N-T)+1=5
X /(Log(N)^2= 2,0814 (counter value)

For N=11890877 , X=400 , produce T=80200 , 4*(N-T)+1=47242709
X /(Log(N)^2=1.5071

For N=49592066  , X=478 , produce T=114481 , 4*(N-T)+1=197910341
X /(Log(N)^2=1.5224

For N=93975566 , X=532 , produce T=141778 , 4*(N-T)+1=375335153
X /(Log(N)^2=1.5785

***

Emmanuel wrote:

This is a small table with 'record' values of  u = x / ( Log N )^2  ( p  = 4N+1 - 2x(x+1), the corresponding prime) :

N                     x                   p                   u
11890877           400            47242709            1.50713
49592066           478           197910341           1.52241
93975566           532           375335153           1.57847

No bigger values than  1.57847  are found for  N <= 10^9.

I believe the conjecture is true because the average number of positive primes of the form  4N+1 - 2x(x+1)  is increasing...

To be more precise : the average number of positive primes of the form  4N+1 - 2x(x+1)  is about  2Sqrt[2N]/Log[4N]. This can easily be deduced from the prime number theorem and also coincides with the results of my computations.

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