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-   -   Smallest prime of the form a^2^m + b^2^m, m>=14 (https://www.mersenneforum.org/showthread.php?t=23160)

 paulunderwood 2018-03-16 20:18

[QUOTE=Batalov;482538]Another droid crossed my path. [SPOILER]260^32768+179^32768[/SPOILER]
Not proven minimal yet, though.[/QUOTE]

Firming up this result:

[CODE]time ./pfgw64 -k -f0 -od -q"[SPOILER]260^32768+179^32768[/SPOILER]" | ../../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
[/CODE]

 science_man_88 2018-03-16 20:28

[QUOTE=CRGreathouse;482537]Yes, but 216 and 109 aren't (though the former is a cube).[/QUOTE]

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.

 Batalov 2018-03-16 23:00

[QUOTE=Batalov;482538]Another droid crossed my path. [SPOILER]260^32768+179^32768[/SPOILER]
[STRIKE]Not proven minimal yet, though[/STRIKE].[/QUOTE]
Now proven minimal.

m=16, anyone?

 science_man_88 2018-03-17 00:20

[QUOTE=Batalov;482562]Now proven minimal.

m=16, anyone?[/QUOTE]

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)

 Batalov 2018-03-17 01:06

[QUOTE=science_man_88;482567]my calculations had under 9000 pairs up to (209,208) but more than 18200 up to (300,299)[/QUOTE]
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(n[SUP]2[/SUP]) (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!

 science_man_88 2018-03-17 01:59

[QUOTE=Batalov;482571]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(n[SUP]2[/SUP]) (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![/QUOTE]

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.

 Batalov 2018-03-17 02:39

Aw well, m=16 was too easy. Even a pair for the good measure:
124^65536+57^65536
143^65536+106^65536

 axn 2018-03-17 02:58

[QUOTE=Batalov;482574]Aw well, m=16 was too easy. Even a pair for the good measure:
124^65536+57^65536
143^65536+106^65536[/QUOTE]

Are all of these being reported to TOP PRP site? I know you're not reporting them to factordb.

 paulunderwood 2018-03-17 03:08

[QUOTE=Batalov;482574]Aw well, m=16 was too easy. Even a pair for the good measure:
124^65536+57^65536
143^65536+106^65536[/QUOTE]

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]

[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[/CODE]

 Dr Sardonicus 2018-03-17 03:45

[QUOTE=Batalov;482571]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(n[SUP]2[/SUP]) (in reality, slightly less: the gcd criterion will help).[/QUOTE]
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
;-)

 Batalov 2018-03-17 04:28

[QUOTE=axn;482577]Are all of these being reported to TOP PRP site? I know you're not reporting them to factordb.[/QUOTE]
Sure did.

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