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.
***
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)^((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.
***
