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.
And Instead of
" 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
Answer to Q2:
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)^((p1)/2)) / (p1).
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.
***