Problems & Puzzles: Conjectures

Conjecture 58. Primes m^2+1

Luis Rodríguez sent the following conjecture:

Conjecture:

The number of primes of the form m^2 + 1 and less than m, is asymptotic to 4m / 3Log(m).

Experimental evidence:

 m Number of primes 4m / 3Log(m) 1000 110 108 10000 839 814 50000 3610 3465 100000 6654 6514 200000 12389 12288

Q1.- Can you find an heuristic argument to justify that formula?
Q2.- can you extend the table

Contributrions came from Jan van Delden, Farideh Firoozbakht & J. K. Andersen.

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Jan wrote:

Did a small test until m=10^7 (or actually for all k<10^7, since it stated less than m).

I found:

I) The column with the number of primes is 2 off: (There should be 2 more then specified).
II) The column with the expected number of primes is always rounded downwards.

If one rounds in the usual sense:
m=10^3 should be 109
m=2*10^5 should be 12289
III) There is a misprint with regard to the asymptotic behaviour! The 3 and 4 should be switched!

Q2:

Next three powers of 10 (first three expected numbers rounded downwards, rounding normally would make them 1 bigger):
I extended the table by adding number of primes/ (m/ln(m))

10^6 54110 54286 0.748
10^7 456362 465315 0.736
10^8 3954181 4071510 0.728

[Calculated the first two rows myself, can be extended simply by using Sloane’s A083844!

Q1:

It looks like the fraction [number of primes/ (m/ln(m))] is steadily decreasing below 0.75. So the conjectured asymptotic behaviour is probably false.
In fact more is known on the conjectured asymptotic behaviour.

On the same page: A083844 there is a link to an article of C.K. Caldwell: http://www.utm.edu/~caldwell/preprints/Heuristics.pdf . In paragraph 3.7 he conjectures that the limiting constant should be C/2 with C=1.3728134628 and C represented by an specific integral (like the twins constant).

***

Farideh wrote:

I think he made some mistakes in definition, formula and
Experimental evidences.

" The number of primes of the form m^2 + 1 and less than m, is asymptotic to 4m / 3Log(m). " must be
" The number of k's less than m such that k^2 + 1 is prime, is
asymptotic to 3m / 4Log(m). ".

Also the table must change in this way:

m number of primes 3m/4Log(m)

1000 112 108
10000 841 814
50000 3613 3465
100000 6656 6514
200000 12391 12288

m number of primes 3m/4Log(m)

500000 28563 28577
1000000 54110 54286
2000000 102205 103386
5000000 239185 243112
10000000 456362 465315

***

Andersen wrote:

The puzzle is full of errors. It's apparently about the number of primes of form x^2+1 with x < m. All the listed prime counts are 2 too small. The table column "4m / 3Log(m)" does not display the claimed value (which is far from the prime count) for any of the listed m.

Marek Wolf's paper "Search for primes of the form m^2+1" at http://arxiv.org/abs/0803.1456 includes the number of primes of form x^2+1 with x <= 10^n for n = 3 to 10:
112, 841, 6656, 54110, 456362, 3954181, 34900213, 312357934.

By common conjectures, the asymptotically expected number of primes of form x^2+1 with x < m is C/2 * m/log m, where C = 1.372813... is the product over all odd primes p of 1 - ((-1)^((p-1)/2)) / (p-1).
The puzzle formula 4m / 3Log(m) corresponds to setting C = 8/3 = 2.666... which seems far too high.

A better (but asymptotically identical) estimate than C/2 * m/log m is
C/2 * (integral from x = 2 to m^2 of 1/(sqrt(x)*log x))

Wolf's paper shows this gives 312353428 instead of 298102838 for m = 10^10, where the real prime count is 312357934.

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