mersenneforum.org > Math Smallest prime of the form a^2^m + b^2^m, m>=14
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2018-03-16, 20:18   #34
paulunderwood

Sep 2002
Database er0rr

D6C16 Posts

Quote:
 Originally Posted by Batalov Another droid crossed my path. 260^32768+179^32768 Not proven minimal yet, though.
Firming up this result:

Code:
time ./pfgw64 -k -f0 -od -q"260^32768+179^32768" | ../../coding/gwnum/hybrid -

Testing (x + 2)^(n + 1) == 7 (mod n, x^2 - x + 1)...
Likely prime!

real	3m34.502s
user	5m52.580s
sys	1m13.776s

2018-03-16, 20:28   #35
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by CRGreathouse Yes, but 216 and 109 aren't (though the former is a cube).
m=0, 2^1 + 1^1
m=1, 2^2 + 1^2
m=2, 2^4 + 1^4
m=3, 2^8 + 1^8
m=4, 2^16 + 1^16
m=5, 9^32 + 8^32
m=6, 11^64 + 8^64
m=7, 27^128 + 20^128
m=8, 14^256 + 5^256
m=9, 13^512 + 2^512
m=10, 47^1024 + 26^1024
m=11, 22^2048 + 3^2048
m=12, 53^4096 + 2^4096
m=13, 72^8192 + 43^8192

Aka

(72^2)^4096+(43^2)^4096
(53^2)^2048+(2^2)^2048
(22^2)^1024+(3^2)^1024
(47^2)^512+(26^2)^512
(13^2)^256+(2^2)^256
(14^2)^128+(5^2)^128
(27^2)^64+(20^2)^64
(11^2)^32+(8^2)^32
(9^2)^16+(8^2)^16
(2^2)^8+(1^2)^8
(2^2)^4+(1^2)^4
(2^2)^2+(1^2)^2
(2^2)^1+(1^2)^1

Minimal next level= minimal square based current level. Aka Serges previous result has a>14 or b>10 or both for the next level.

Last fiddled with by science_man_88 on 2018-03-16 at 21:05

2018-03-16, 23:00   #36
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3×3,041 Posts

Quote:
 Originally Posted by Batalov Another droid crossed my path. 260^32768+179^32768 Not proven minimal yet, though.
Now proven minimal.

m=16, anyone?

2018-03-17, 00:20   #37
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by Batalov Now proven minimal. m=16, anyone?
Not unless we find or can easily apply more because with opposite parity and gcd(x,y)=1, my calculations had under 9000 pairs up to (209,208) but more than 18200 up to (300,299)

2018-03-17, 01:06   #38
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3×3,041 Posts

Quote:
 Originally Posted by science_man_88 my calculations had under 9000 pairs up to (209,208) but more than 18200 up to (300,299)
Yeah, my calculations also show that 209/300 ~= 0.7 is a damn good approximation of the 1/ sqrt2. Consequently, but of course, there are almost exactly 2 times more pairs for a<=300 than for a<=209. So what? The number of pairs with a<=n will be O(n2) (in reality, slightly less: the gcd criterion will help).

And under a<=600 there are ~4 times more pairs than under a<=300. So what, just keep sieving and testing! What seems to be the problem?

It is excellent that the number of tries goes quadratic -- you go almost flat in size (and in test runtime); and one likely need to test γ*2^m pairs, i.e. ~30,000 to 40,000. Just keep sieving and testing!

2018-03-17, 01:59   #39
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20B116 Posts

Quote:
 Originally Posted by Batalov Yeah, my calculations also show that 209/300 ~= 0.7 is a damn good approximation of the 1/ sqrt2. Consequently, but of course, there are almost exactly 2 times more pairs for a<=300 than for a<=209. So what? The number of pairs with a<=n will be O(n2) (in reality, slightly less: the gcd criterion will help). And under a<=600 there are ~4 times more pairs than under a<=300. So what, just keep sieving and testing! What seems to be the problem? It is excellent that the number of tries goes quadratic -- you go almost flat in size (and in test runtime); and one likely need to test γ*2^m pairs, i.e. ~30,000 to 40,000. Just keep sieving and testing!
I might see if I can figure out my additive inverse property to help. It may even be relatable to the arithmetic mean of the pair.

 2018-03-17, 02:39 #40 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3×3,041 Posts Aw well, m=16 was too easy. Even a pair for the good measure: 124^65536+57^65536 143^65536+106^65536
2018-03-17, 02:58   #41
axn

Jun 2003

2·13·181 Posts

Quote:
 Originally Posted by Batalov Aw well, m=16 was too easy. Even a pair for the good measure: 124^65536+57^65536 143^65536+106^65536
Are all of these being reported to TOP PRP site? I know you're not reporting them to factordb.

2018-03-17, 03:08   #42
paulunderwood

Sep 2002
Database er0rr

22·859 Posts

Quote:
 Originally Posted by Batalov Aw well, m=16 was too easy. Even a pair for the good measure: 124^65536+57^65536 143^65536+106^65536
Tests for Serge's latest finds:
Code:
time ./pfgw64 -k -f0 -od -q"124^65536+57^65536" | ../../coding/gwnum/hybrid -

Testing (x + 1)^(n + 1) == 2 + 3 (mod n, x^2 - 3*x + 1)...
Likely prime!

real	8m40.139s
user	15m28.568s
sys	1m48.060s
Code:
time ./pfgw64 -k -f0 -od -q"143^65536+106^65536" | ../../coding/gwnum/hybrid -

Testing (x + 2)^(n + 1) == 7 (mod n, x^2 - x + 1)...
Likely prime!

real	8m29.679s
user	15m1.912s
sys	1m47.804s

2018-03-17, 03:45   #43
Dr Sardonicus

Feb 2017
Nowhere

3·7·132 Posts

Quote:
 Originally Posted by Batalov Yeah, my calculations also show that 209/300 ~= 0.7 is a damn good approximation of the 1/ sqrt2. Consequently, but of course, there are almost exactly 2 times more pairs for a<=300 than for a<=209. So what? The number of pairs with a<=n will be O(n2) (in reality, slightly less: the gcd criterion will help).
The number of pairs (a, b) with 1 <= a <= b <= N, and gcd(a, b) = 1, is SUM, b = 1 to N, eulerphi(b). This is the number of fractions in the Farey sequence of order N.

Asymptotically, this is 6/pi^2 * N^2. Long known.

The additional condition that a + b is odd, knocks out the pairs (a, b) with a and b both odd. In this case, b - a is even and gcd(b - a, b) = 1, so for odd b > 1, the condition knocks out exactly 1/2*eulerphi(b) pairs. For b = 1 it knocks out the pair (1, 1).

I'm too lazy to work out the modified asymptotic. I'm sure it is long known, but alas I'm also too lazy to look it up
;-)

Last fiddled with by Dr Sardonicus on 2018-03-17 at 03:53

2018-03-17, 04:28   #44
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

912310 Posts

Quote:
 Originally Posted by axn Are all of these being reported to TOP PRP site? I know you're not reporting them to factordb.
Sure did.

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