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Old 2019-11-06, 13:46   #1
enzocreti
 
Mar 2018

10178 Posts
Default binary form of the exponents 69660, 92020, 541456

pg(69660), pg(92020) and pg(541456) are probable primes with 69660, 92020 and 541456 multiple of 86




69660 in binary is 10001000000011100
92020 in binary is 10110011101110100
541456 in binary is 10000100001100010000


you can see that the number of the 1's is always a multiple of 5




a chance?
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Old 2019-12-18, 08:26   #2
enzocreti
 
Mar 2018

17×31 Posts
Default 69660, 92020, 541456

69660, 92020 and 541456 are 6 mod 13 (and 10^m mod 41)




69660, 92020 and 541456 are multiple of 43


is there a reason why




(69660-6)/26 is congruent to 13 mod 43
(92020-6)/26 is congruent to 13 mod 43
(541456-6)/26 is congruent to 13 mod 43?


215, 69660, 92020, 541456 are multiple of 43

let be log the log base 10

int(x) let be the integer part of x so for example int(5.43)=5

A=10^2*log(215)-215/41

int(A)=227=B

215 (which is odd) is congruent to B+1 mod 13
69660 which is even is congruent to B mod 13
92020 is congruent to B mod 13
and 541456 is also congruent to B mod 13






215 (odd) is congruent to -1215 mod 13
69660 (even) is congruent to 1215 mod 13
92020 (even) is congruent to 1215 mod 13
541456 (even) is congruent to 1215 mod 13










215, 69660, 92020, 541456 can be written as 13x+1763y+769


(541456-769-13*93)/1763=306


(69660-769-13*824)/1763=33


(92020-769-13*781)/1763=46


as you can see 306,33 and 46 are all 7 mod 13

769+13*824
769+13*781
769+13*93 are multiples of 43


(69660-(10^3+215))/13=5265 which is 19 mod 43
(92020-(10^3+215))/13=6985 which is 19 mod 43
(541456-(10^3+215))/13=41557 which is 19 mod 43


10^3+215 is 6 mod 13


now
69660/13=5358,4615384...
92020/13=7078,4615384...
541456/13=41650,4615384...


the repeating term 4615384 is the same...so that numbers must have some form 13s+k?


215 (odd) is congruent to 307*2^2-10^3 or equivalently to - (19*2^6-1) mod 13
69660 (even) is congruent to 307*2^2-1001 or equivalenly to (19*2^6-1) mod 13
92020 (even) the same
541456 (even) the same




(19*2^6*(541456-92020)/(13*43)+1)/13+10=75215 is a multiple of 307=(215*10-1)/7=(5414560-1)/17637




pg(51456) is another probable prime with 51456 congruent to 10^n mod 41


75215=(51456*19+1)/13+10=(19*2^6*(541456-92020)/(13*43)+1)/13+10




215 is congruent to -1215=5*3^5 mod 13

69660 is congruent to 1215 mod 13 and so also 92020 and 541456


1215=(51456/2-2*13*10^2-43)/19




...so summing up...


215 (odd) is congruent to 3*19*2^2 mod((41*43-307)/(7*2^4)=13) where 41*43-307 is congruent to 10^3+3*19*2^2 mod (3*19*2^2=228)
69660 (even) is congruent to (3*19*2^2-1) mod 13 ...
the same for 92020 and 541456


another way is 215 (odd) is congruent to -15*81 mod ((41*43-307)/(307-15*13))
69660 (even) is congruent to 15*81 mod ((41*43-307)/(307-15*13))
the same for 92020 and 541456




215 is congruent to (41*43-307-10^3)/2 mod ((41*43-307)/(7*2^4))
69660 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))
92020 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))
541456 is congruent to (41*43-307-10^3)/2-1 mod ((41*43-307)/(7*2^4))




215 is congruent to 3*19*2^2 mod ((41*43-307)/(2*(19*3-1)))
69660 is congruent to 3*19*2^2-1 mod ((41*43-307)/(2*(19*3-1)))
and so 92020 and 541456




215 is congruent to 3*19*2^2 mod ((3^6-1)/(3*19-1))
69660 is congruent to 3*19*2^2-1 mod((3^6-1)/(3*19-1)) and so 92020 and 541456

51456 (pg(51456) is probable prime and 51456 is 10^n mod 41) is congruent to 19*3*2^4 mod 13

Pg(2131) is probable prime
2131 is prime
227=2131-307*4-26^2
So 215 is congruent also to 2131-307*4-26^2 mod 13
And 69660 is congruent to 2131-307*4-26^2+1 mod 13 and so also 92020 and 541456


69660 is congruent to 1763-307*5-1 mod ((1763-307)/112)=13)

And so 92020 and 541456

215 which is odd is congruent to - 1763+307*5+1 mod 13
215+1763-307*5-1 is divisible by 17 and by 13
And also 541456-1763+307*5+1 is divisible by 13 and 17
215 and 541456 have the same residue 10 mod 41

((541456-1763+307*5+1)/(13*17)+1) *200+51456=541456


69660 92020 541456 are congruent to 7*2^6 mod 26
7*2^6-(1763-307*5-1)=221=13*17
215+7*2^6 is a multiple of 221
541456-7*2^6 is a multiple of 221

Last fiddled with by enzocreti on 2020-01-26 at 17:58
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Old 2020-01-24, 15:31   #3
enzocreti
 
Mar 2018

17×31 Posts
Default 215 , 51456, 69660, 92020, 541456

51456, 69660, 92020, 541456 are even and congruent to 10^n mod 41


pg(51456), pg(69660), pg(92020) and pg(541456) are prp


51456, 69660, 92020, 541456 are all congruent to 7*2^q-1 mod 13 with q a nonnegative integer




215 is odd and pg(215) is prp


215 is congruent to 7*2^q mod 13
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Old 2020-03-30, 11:52   #4
enzocreti
 
Mar 2018

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Default 69660 and 92020

69660 and 92020 are multiple of 215 and congruent to 344 mod 559
92020=lcm(215,344,559)+69660


| denotes concatenation in base 10


2^69660-1 | 2^69559-1 is prime
2^92020-1 | 2^92019-1 is prime!!!

Last fiddled with by enzocreti on 2020-03-30 at 11:53
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Old 2020-03-30, 12:19   #5
LaurV
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Jun 2011
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Default

please show us a proof that they are prime
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Old 2020-03-30, 12:50   #6
enzocreti
 
Mar 2018

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Default ...

Well...
actually they are only probable primes... maybe in future they will be proven primes

Last fiddled with by enzocreti on 2020-03-30 at 12:52
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Old 2020-03-30, 17:39   #7
enzocreti
 
Mar 2018

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Default

Quote:
Originally Posted by LaurV View Post
please show us a proof that they are prime


http://factordb.com/index.php?id=1100000001110801143

http://factordb.com/index.php?query=...%2B2%5E92019-1

Last fiddled with by enzocreti on 2020-03-30 at 17:55
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Old 2020-03-30, 23:06   #8
enzocreti
 
Mar 2018

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Default ... I note also...

69660 I note also that


(lcm(215,344,559))^2=4999*10^5+69660-60


I notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)


I note that the polynomial

X^2-X*429^2+7967780460=0 has the solution x=69660


If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43

I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344.



I note that 92020*2+1=429^2

The discriminant of the polynomial is 429^4-4*7967780460 which is a perfect square and lcm(215,344,559) divides 429^4-4*7967780460-1


Pg(331259) is prime and pg(92020) is prime.
92020+(92020/215-1)*559+546=331259

Again magic numbers 559 and 546 strike!

So 331259 is a number of the form
215*(13s-1)+(13*b-2)*559+546

For some s, b positive integers

So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g.

Pg(69660) is prime. 69660=(3067*8-546)*3-11# where # is the primorial and 3067 is a prime of the form 787+456s

notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

So we have pg(215) is prime pg(69660) is prime pg(92020) is prime

With 215 69660 and 92020 multiple of 43

69660=215+(18*18-1)*215

92020=69660+18*18*69+4


-215 is congruent to 108 mod (18*18-1=323)
541456 is congruent to 108 mod (18*18-1)
92020 is congruent to (17*17-1) mod (18*18-1)
69660 is congruent to 215 mod (18*18-1)

108=6^3-18^2

17*17-1+36-216=17*17-1+6^2-6^3=108


215=6^3-1

To make it easier

215, 69660, 541456 are congruent to plus or minus 215 mod 323

92020 is congruent to (17*17-1) mod (18*18-1)

curious that 289/215 is about 1.(344)...

and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559


215, 69660, 541456 are congruent to plus or minus 6^3-1 mod 323

92020 is congruent to (12/9)*6^3 mod 323


92020*9/12 is congruent to 6^3 mod 323

92020 is congruent to (17^2-1) mod 323 and to - (6^2-1) mod 323

92020 is a number of the form 8686+13889s


13889=(6^3+1)*64


215 69660 92020 541456 are + or - 344 mod 559

lcm(215,344,559)-86*(10^2+1)+6^3-1=(6^3+1)*2^6+1

92020=69660+lcm(215,344,559) so you can substitute

92020=69660+86*(10^2+1)-6^3+1+(6^3+1)*2^6+1

86*(10^2+1) mod 323 is 17^2-1



215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41

They are congruent to plus or minus 215 mod 323

92020 is congruent to 2^4 (not a power of 10) mod 41

92020 is congruent to (2^4+1)^2-1 mod 323

288 is 17^2-1

288 in base 16 is 120

120=11^2-1

also 323=18^2-1 in base 16 is 143=12^2-1


344*((1444456-1763*2^9) /344-1)=541456

1444456=lcm(13,323,344)


541456=lcm(13,323,344)-344*(41*2^6+1)




215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323

215 is congruent to - 108 mod 323
541456 is congruent to 108 mod 323
69660 is congruent to - 108 mod 323
92020 is congruent to 288 mod 323


108 and 288 are numbers of the form 3^a*2^b

So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3-smooth number mod 323

108 and 288 are both divisible by 36

Last fiddled with by enzocreti on 2020-08-18 at 21:00 Reason: notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is - 21
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Old 2020-08-19, 05:48   #9
enzocreti
 
Mar 2018

17×31 Posts
Default

Q77I
Quote:
Originally Posted by enzocreti View Post
69660 I note also that


(lcm(215,344,559))^2=4999*10^5+69660-60


I notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)


I note that the polynomial

X^2-X*429^2+7967780460=0 has the solution x=69660


If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43

I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344.



I note that 92020*2+1=429^2

The discriminant of the polynomial is 429^4-4*7967780460 which is a perfect square and lcm(215,344,559) divides 429^4-4*7967780460-1


Pg(331259) is prime and pg(92020) is prime.
92020+(92020/215-1)*559+546=331259

Again magic numbers 559 and 546 strike!

So 331259 is a number of the form
215*(13s-1)+(13*b-2)*559+546

For some s, b positive integers

So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g.

Pg(69660) is prime. 69660=(3067*8-546)*3-11# where # is the primorial and 3067 is a prime of the form 787+456s

notice that lcm(215,344,559)=22360

22360/(18*18)=69.01234567...

curious
I notice also that 541456 (multiple of 43),is - 215 mod (18*18-1)
and 69660 (multiple of 43) is 215 mod (18*18-1)

So we have pg(215) is prime pg(69660) is prime pg(92020) is prime

With 215 69660 and 92020 multiple of 43

69660=215+(18*18-1)*215

92020=69660+18*18*69+4


-215 is congruent to 108 mod (18*18-1=323)
541456 is congruent to 108 mod (18*18-1)
92020 is congruent to (17*17-1) mod (18*18-1)
69660 is congruent to 215 mod (18*18-1)

108=6^3-18^2

17*17-1+36-216=17*17-1+6^2-6^3=108


215=6^3-1

To make it easier

215, 69660, 541456 are congruent to plus or minus 215 mod 323

92020 is congruent to (17*17-1) mod (18*18-1)

curious that 289/215 is about 1.(344)...

and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559


215, 69660, 541456 are congruent to plus or minus 6^3-1 mod 323

92020 is congruent to (12/9)*6^3 mod 323


92020*9/12 is congruent to 6^3 mod 323

92020 is congruent to (17^2-1) mod 323 and to - (6^2-1) mod 323

92020 is a number of the form 8686+13889s


13889=(6^3+1)*64


215 69660 92020 541456 are + or - 344 mod 559

lcm(215,344,559)-86*(10^2+1)+6^3-1=(6^3+1)*2^6+1

92020=69660+lcm(215,344,559) so you can substitute

92020=69660+86*(10^2+1)-6^3+1+(6^3+1)*2^6+1

86*(10^2+1) mod 323 is 17^2-1



215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41

They are congruent to plus or minus 215 mod 323

92020 is congruent to 2^4 (not a power of 10) mod 41

92020 is congruent to (2^4+1)^2-1 mod 323

288 is 17^2-1

288 in base 16 is 120

120=11^2-1

also 323=18^2-1 in base 16 is 143=12^2-1


344*((1444456-1763*2^9) /344-1)=541456

1444456=lcm(13,323,344)


541456=lcm(13,323,344)-344*(41*2^6+1)




215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323

215 is congruent to - 108 mod 323
541456 is congruent to 108 mod 323
69660 is congruent to - 108 mod 323
92020 is congruent to 288 mod 323


108 and 288 are numbers of the form 3^a*2^b

So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3-smooth number mod 323

108 and 288 are both divisible by 36

215 69660 92020 541456 are congruent to plus or minus (6^k-1) mod 323 for k=3,2


69660=(2^5*3^7)-(2^3*3^4) so it is the difference of two 3 smooth numbers (2^a*3^b)-(2^(a-3)*3^(b-3))



69660 is multiple of 3 and congruent to 0 mod (6^2-1)

215, 92020, 541456 are not multiple of 3 and multiple of 43 and are congruent to plus or minus 2^k mod 36 for some k


1763*323-(6^3+1)*(2^7+1)=541456 or

(42^2-1)*(18^2-1)-(6^3+1)*(2^7+1)=541456

I also note that

69660=(2^7+1)*(6^3+1)+(2^7+1)*(18^2-1)

And by the way 215=(42^2-1)*4^2-(2^7+1)*(6^3+1)

27993=(217*129)=(2^7+1)*(6^3+1)

I notice that - 541456 mod 27993=9202*2

92020=10*9202

And 9202*2*10+1=429^2

27993=3/5*(6^6-1)


27993 in base 6 is 333333

27993 has also the representation:

(42^2-1)*4^2-(6^3-1)=27993=2*43*(18^2-1)+(6^3-1)


x/5+(42^2-1)*(18^2-1)-3/5*(6^6-1)=20*3/5*(6^6-1)

The solution of this equation x=92020


this identity:

(42^2-1)*(18^2-1)=(10^3+18^2)*430+43*3

Maybe it is not a chance that pg(10^3+18^2-1=1323) is prime


pg(1323), pg(215), pg(69660), pg(92020), pg(541456) are primes

1323, 215, 69660, 92020, 541456 are congruent to plus or minus (2^a*3^b-1) mod 323 where 2^a*3^b is a 3 smooth number with 2^a*3^b<323


or 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (p-1) mod 323 where p is a perfect power




92020 has the factorization 2*(6^3-1)*(6^3-2)




lcm(215,344,559)=(6^3-1)*(2*(6^3-2)-18^2)=(2*(6^3-2)-18^2)*(2*(6^3-2)-18^2+111)




69660=92020-lcm(215,344,559)


215=2*(6^3-2)+2*(6^3+1)-2*18^2+1




92020=(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2




so 215=2*(6^3-2)+2*(6^3+1)-2*18^2+1


69660==(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2-((6^3-1)*(2*(6^3-2)-18^2))


92020=(2*(6^3-2)+2*(6^3+1)-2*18^2+1)*(2*(6^3-2)+2*(6^3+1)-2*18^2)*2




541456=(42^2-1)*(18^2-1)-(7^3-6^3+2)*(6^3+1)




69660=111111-5*11111+8*(42^2-1)




344=7^3+1=2*(111111-5*11111)/(18^2-1)=A




so lcm(215,344,559)=lcm(215, A, A+215)


541456=5*111112-(42^2-1)*8




(12^2-1) divides (111111-5*11111-6^3+1)




559=(111111-5*11111-6^3+1)/(12^2-1)+(111111-5*11111)/(18^2-1)

Quite clear that 215, 69660, 92020,541456 multiple of 43

are congruent to plus or minus (18^2-6^k) mod (18^2-1) where k is 2 or 3 and 3 indeed is the maximum exponent such that 18^2-6^k i 1s positive




(541456-18^2+6^3)/323=41^2-5=1676
(92020-18^2+6^2)/323=17^2-5=284
(69660+18^2-6^3)/323=6^3
(215+18^2-6^3)/323=1


There is clearly a pattern!!!




I notice also that (1+5)=6 is a semiprime
(6^3+5)=13*17 is a semiprime
(1676+5)=41*41 is a semiprime
(284+5)=17*17 is a semiprime




(6,221,1681,289) are either squares of primes or product of two consecutive primes




so when pg(k) is prime and k is a multiple of 43, then k can be expressed in this way:


(p*q-5)*(18^2-1) plus or minus (18^2-6^k) with 6^k<18^2


p and q are primes


when pg(k) is prime and k is a multiple of 546, then


k is congruent to 78 mod (18^2-6^3)


as in the cases k=56238 and k=75894




Curio of the curios:


pg(331259) is probable prime and 331259 is prime


magic:


71*6^6-331*(10^4-3*331)=331259




71*6^6 and 331259 have in common the first five digits 33125=182^2+1


If you consider 71*6^n for n>3

For n even

71*6^n is congruent to (11^2-1) mod (13^2-1)

and for n odd

Is congruent to (7^2-1) mod (13^2-1)

For n=4

71*6^4=92016

for n=6

71*6^6=3312576

As you can see

Pg(92020) is prime and pg(331259) is prime

92020 has the last two digits 20 different from 92016

The same for 3312576 and 331259

Moreover 331259 mod (71*6^3) =9203


Both 331259 and 92020 are 5 mod 239

Is there something connected with the fact

ord (71*6^k)=4

I mean the smallest value k for which 71*6^k is congruent to 1 mod 239 is k=4???




71 and 6 are both quadratic residues mod 239






92020 and 331259 are congruent to 71*3^3 mod (239*13)


I wonder if this concept could be generalized




pg(51456), pg(92020), pg(331259) are probable primes


51456 is congruent to 71 mod (239*(6^3-1))
92020 is congruent to 71*3^3 mod (239*13)
331259 is congruent to 71*3^3 mod (239*13)


92020 and 331259 are congruent to 6 mod 13


51456 and 331259 are congruent to 2^3 mod 109


71 and 71*3^3 are both congruent to 6 mod 13




so


51456, 92020, 331259 are congruent to 13*(5+71*k)+6 either mod (239*215) or mod (239*13)


are these primes infinite?


the odd thing is that


51456, 92020, 331259 are either congruent to 10^m mod 41 or prime (331259 infact is prime)

So I think it is no chance that 51456, 92020, 331259 are either congruent to 2^j mod 71 or to 13*2^i mod 71

No chance at all!

There is a file Rouge!
Pg(541456) is probable prime as pg(51456)

And 541456 mod (239*215)=9202*3, that is 3*( 92020/10 )and pg(92020) is prime




541456 is congruent to (3/20)*(429^2-1) mod (239*215)

541456 is congruent to 14*71*3^3+3*2^8 (mod (239*215))


Maybe there is some connection to the fact that

215, 69660, 92020 and 541456 are plus or minus 344 mod 559

lcm(344,559)=4472=71*(2^6-1)-1




I suspect that something in field F(239) is in action!




331259^(-1) mod (215*239)=49999=(10^5-2)/2

((71*(2^6-1))-1)/2=lcm(344,559)=9202-6966

Multiplying both sides by 10 you have

92020=69660+lcm(215,344,559)


I would suggest to study these exponents in field F(51385=239*215)

(239*10*215+3*(429^2-1)/10-541456) /3=9202

So

10*
(239*10*215+3*(429^2-1)/10-541456) /3=92020

In this equation we have 215, 92020, 541456 multiple of 43 and not of 3

The other multiple of 43 is 69660 which is multiple of 3

And 92020=69660+lcm(344,215,559)

239*10*215+3*(429^2-1)/10 is a multiple of 559

239*10*215+3*(429^2-1)/10=(42^2-1)*(18^2-2)+43*2^5


This identity

(239*10*215+(3/10)*(429^2-1)-541456+92020)/559=92020/(215*2)


One can play around with this expression containing 215 and 559


And substitute for example 92020 with (429^2-1)/2

Pg(331259) is prime and also 331259 is prime

The inverse modulo (215*239) of 331259 is the prime 5*10^4-1=49999


(331259*49999-1)/(215*239)=7*(18^2-1)*(12^2-1)-10^3+1




So 331259=((((18^2-1)*(10^3+1)-10^3+1)*239*215)+1)/(5*10^4-1)




Using Wolphram Alpha i considered this equation:




((323*(10^x+1)-10^x+1)*239*215+1)/(5*10^(x+1)-1)=y


wolphram say that the integer solution is x=3, y=331259


wolphram gives an alternate form:


84898302/(5*(2^(x+1)*5^(x+2)-1))+1654597/5=y


the number 1654597=69660+(30^2-1)*(42^2-1)




so 69660 and 331259 (both 6 mod 13 and pg(69660) pg(331259) primes) are linked by this equation:


84898302/(5*(2^(3+1)*5^(3+2)-1))+(69660+(30^2-1)*(42^2-1))/5=331259


541456 in field F(239*215) and in field F(239*323)

541456 mod (239*215)=9202*3
541456 mod (239*323)=359*3

Pg(359) is prime, pg(9202*10) is prime

9202-359 is a multiple of 239

Pg(92020+239239=331259) is prime




I note also


331259=71*6^6-331*(10^4-3*331)=6^2*9202-13=92020+239239




239239 is congruent to -13 mod 107 and mod 43




22360=(10*(239239+13))/107


92020=69660+22360


92020=331259-239239


92020 is a multiple of 107




lcm(215,344,559)*10=22360




multiple of 86, that is 69660, 92020 and 541456 are congruent to a square mod 428


so the muliple of 86=k for which pg(k) is prime are numbers for which there is a solution to this modular equation:




y is congruent to 36*x^2 mod 428


infact


92020 is congruent to 36*0^2 mod 428
69660 is congruent to 36*3^2 mod 428
541456 is congruent to 36*1^2 mod 428






There are pg(k) primes with k multiple of 215 and k multiple of 546


probably that numbers 215 and 546 are not random at all




look at this equality:


71*6^6-182^2-3^2-331*(10^4-3*331)-10=546^2


546-215=331=546-(6^3-1)



by the way 182^2+3^2+546^2 is prime




546^2 is congruent to 6^3 mod 331



541456-((3*239) ^2-239+(429^2-1)/20)=(429^2-1)/10




541456=(18/5)*(429^2-1)/2+13*((429^2-11)/10-2235)


where 2235=lcm(215,344,559)-1




541456*10=3*331259+239*(136^2+1)



92020 and 331259 are both 5 mod 239

71*6^k is congruent to 1 mod 239 for k=4

But 71*6^4 is 920...
And 71*6^6 is 33125...


92020 mod 71 is 4 and mod 331 is 3


71*6^4 is congruent to 1 mod (385*239)

541456 is congruent to - 239 mod 385

I notice that 92020 and 331259 are congruent to 5 mod 239 but also to - 72 mod (1001)

I notice that 92020 is congruent to 146 mod 71

541456 and 331259 are congruent to 146 mod 703




I notice that 215, 69660, 541456 are plus or minus 215 mod 323
92020 which is congruent to 16 mod 41, is congruent to 288 mod 323 and mod 71
288 and 92020 are both 4 mod 71


so 92020 is congruent to 4 mod 71
and is congruent to 4+284 (mod 323)


where 284 is the residue mod 323 of 71*6^4




215, 69660, 541456 are congruent to plus or minus 215 mod 323


where 215=4+211


211 is the residue of 71*6^4 mod 215




in particular 92020 is congruent to 288 mod 323 and mod 284, infact 92020=71*6^4+4


I think that also 331259 has something to do with 71*6^6


so multiple of 43, that is 215, 69660, 92020, 541456 are either of the form 323k+108, or 323k-108, or 323k-288


I notice that (323-108)=6^3-1 and (323-288)=6^2-1




211 and 284 are also the residues of 71*6^4 and 71*6^6 mod 323




the 18-th pg prime is pg(1323),
the 36-th is pg(360787)


modulo (18^2-1=323), 1323 is 31 and 360787 is 319


the difference between 319 and 31 is again 288


1323 is congruent to 31 mod (6^4-4)
360787 is congruent to 6^2*2^3+31 mod (6^4-4)




319 is congruent to 31 mod 6^2 infact




here:


215, 69660, 541456 are congruent to plus or minus 108 mod 323
92020 is congruent to (6^4-1008=288) mod 323




so multiple of 43 are either congruent to plus or minus (18+90) mod (18^2-1) or to 6^4-(18+90+900) mod (18^2-1)




92020 is congruent to (11*6^2-108) mod (18^2-1)
and to - 11*6^2 mod (304^2)






so great fact 215, 69660, 92020, 541456 are either congruent to plus or minus (108*2-1) mod (108*3-1) or to plus/minus (108/3-1) mod (108*3-1)
















pg(331259) and pg(92020) are probable primes.

331259=92020+239239 as said


331259 and 92020 are congruent to 5 mod 239
331259 and 92020 are congruent to 6 mod 13

using chinese remainder theorem it yields
something like

239x(1)+13y(1)=1

using Euclidean algor you have the solution y(1)=92

92 are the first two digits of 92020

now 331259 can be rewritten as 239*10^3+92*10^3+331-71-1


331259 and 92020 have the property that they are congruent to the last two decimal digits modulo 9200

331259 is congruent to 59 mod 9200
92020 is congruent to 20 mod 9200


(239*77+1)*5=92020

239*77+1 is a multiple of 215

541456 is congruento to 3/2*(239*77+1) mod (239*215)

331259 is congruent to 5 mod (239*77) and also 92020


(541456-9202*3)/215/239=10


(13*331-1)*7*11+5-7*11*13*239=92020


541456 is congruent to (9203+(71*6^4-1)/5) (mod (239*5*43))


9203 is prime

(541456-9203-(71*6^4-1)/5)/215/239=10




331259-13*(9202*2-1)=92020


92020-13*1720=69660


lcm(215,344,559)=13*1720




67*(16683/67+22360/43-1)=51456


where 16683=9202*2-1-1720 and 22360=lcm(215,344,559)




331259 has the representation


331259=36*(7/180+(92020-4)/10)



Let be Floor(x) the floor function floor(5.5)=5 for example

Floor(239*(331259/(71*6^4))/4)=215


So i suspect that there are infinitely many pg(k) primes with k multiple either of 215 or multiple of something transformed by the Floor function in 215 and in these Cases k is alway 6 mod 13


215=(239*6^2-2^2)/40=(239*6^2-2^2)/(6^2+2^2)




331259=(3*71*6^5+7)/5




71*6^6-(3^3*71*6^5-7)/5=331259


(3^3*71*6^5-7)/5=(10^4-3*331)*331




so we have 92020=71*6^6-(3^3*71*6^5-7)/5-13*(71*6^4-1)/5


(69660/3-860)/13=1720


I see where 215 comes from

(331259-23666)/(12^2-1)=9*239-1=215*10


23666 I used CRT x is congruent to 6 mod 13 and to 5 mod (239*7)


As you can see Both 331259 and 92020 are congruent to 23666 mod 143


But this simply means that both 331259 and 92020 are congruent to 71 mod 143

By the way also 215 is congruent to -71 mod 143


so 215, 92020, 331259 are congruent to plus or minus 71 mod 143

215 is 2 mod (71)
92020 is 4 mod (71*6^3)
331259 is 9203 (prime) mod (71*6^3)

there is clearly a pattern

note that also 541456 is congruent to -(9203*3) mod (7*71)


i have no idea how to develop these ideas but i strongly suspect that there is a structure


It should be clear that

541456 215 69660 331259 92020 are congruent to plus or minus (19+13s) mod 143 for some non negative s.


331259 is also congruent to sqrt(239*9-215) mod 71

215*9=44^2-1

So 331259 is congruent to sqrt(215*9+1) mod 71

another way to see the same thing is that 215 92020 and 331259 are congruent to plus or minus 72 (mod 143)

and 92020 and 331259 are even congruent to - 72 (mod 143*7)

infact 92020+72=92092 and 331259+72=331331

It's easy to see that 9203 conguent to 44 mod 71 and 331259 is as well congruent to 44 mod 71

and 9203 has the same residue 259 mod 344 and mod 559

((541456/921)-((41*43*10+7)/30))^(-1)-7=9203


note that 921=3*307

9210=30*307=((541456/921)-((41*43*10+7)/30))^(-1)

2150=307*7+1

92020*(5879)/999-71.107107107=541456

or 92020*(1763*10+7)/2997-71.107107...=541456


541456-(92020*(1763*10+7)/58790)=10*215*239




331259 is congruent to (71*6^3+307*10)/2 (mod (71*7*6^4))




9203=(331259*2-7*71*6^4)/2




331259=(71*6^3*43)/2+307*5




331259=(2^10+1)*(18^2-1)+184






I would conjecture that if pg(k) is prime and k is multiple of 43,


then k is a multiple of (72+143s) for some positive integer s


the multiple of 215=72+143 are 215 itself, 69660, 92020


the other multiple of 43 is 541456 which is a multiple of 787=72+143*5






multiple of 43 are 215, 69660, 92020, 541456


they are either congruent to (plus/minus) 215 mod 323 or to 288 mod 323




215 and 288 are integers for which a integer solution exists for the equation x^2+71*y^2=z (x,y,z positive integers)






multiple of 43 are 215 69660 92020 541456


Now consider the equations:


x^2+71*y^2=215
x^2+71*y^2=69660
x^2+71*y^2=92020
x^2+71*y^2=541456


there are Always non zero integer solutions x and y

541456 is about 271*999*2 i wonder Why???


so multiple of 86, 92020 69660 and 541456 are congruent to (271*999-1) mod 172


this implies (92020 69660 541456 are also 6 mod 13) that 92020 69660 and 541456 are congruent to 344 mod 2236

and I remember that 92020=69660+22360

this could suggest why multiple of 43 are congruent to plus or minus 344 mod 559

infact (271*999-1) is congruent to 172 mod 559

(271*999-1) is congruent to -172 mod 215


pg(51456) and pg(541456) are probable primes

observing that 51456 is congruent to 508 mod 542 and to 507 mod 999 using CRT:

solutions are 51456+541458n

infact lcm(542,999)=541458=541456+2


331259= 11*2^10+39*2^13+507


51456*9 is congruent to 700^2 mod (164^2)

541456=51456+700^2


(700^2-164^2)/51456=3^2

(10*700^2-164^2)/3^2=541456


So i wonder inf there are infinitely many primes pg(k) with k multiple of 86 with the property that k is congruent to 164^2 mod (268) as pg(541456) and pg(92020)
More generally if k is multiple of 86 and k is multiple of 3, 69660 is the example

k/3 is congruent to - 164^2 mod 268

if k is not multiple of 3 and k is multiple of 86, then

Examples are 92020 and 541456 are congruent to 164^2 mod 268


541456=271*9*999*2-18

271*9=2439

Maybe this explain the fact that 541456/41 and 51456/41 have a repenting term 2439 and so they are 10^m mod (700^2+239)*2^8/(271*9)=51456

(2439*111-1)*2-(2439*201-239)=51456

From above reasoning it results that

541456 and 51456 are both congruent to -239 mod 245=7*25

In subastance here the formulas

541456=245*2211-239
51456=245*211-239


Where 211 is prime and 2211 is a multiple of 67 as 51456

(541456-92020) is a multiple of 559 and 67

(541456-92020)/(67*3)=2236


69660+22360=92020

But ed can substitute 67*3 with 2211/11

So (541456-92020)*11/2211=2236

i note also that (541456-210*1001+13)=331259

92020 is so congruent to (245*2211-239) mod (3*67*43*13)

and (245*2211-239=541456) is congruent to (331259+13) mod 559


i think that it shuould be possible to prove that when pg(k) is prime and k is congruent to 6 mod 13, (examples known 215, 69660, 92020, 331259, 541456) then k is congruent to plus or minus (344-13s) mod 559
with s=1 in the case of 331259 which infact is congruent to 331 mod 559 and s=0 in all the other cases where k is a multiple of 43


i think that something interesting could be found examining this formula:


f(x)=((71*6^x+4)/10)*6^2-13 for x integer

x=4 f(4)=331259
331259 congruent to -13 mod 9202

i think that the study of this function could shed light on these numbers

331259 is congruent for example to -7 mod 55211=f(3)

331259=92020+239239


note that also 239239 is congruent to -13 mod 9202


239239 is congruent to 546 mod 559

215*(239239-546)/559+215=92020

I suspect that 546 is not a random residue

infact there are primes pg(k) with k multiple of 546

for example pg(75894) is prime and 75894 is multiple of 546

modulo 559 75894 is 429 (92020=(429^2-1)/2)

(239239-546)/559+2=429


so( ((239239-546)/559+2)^2-1)/2=92020


(239239+(12^2-1)*4)=429


Given the attention this question has received, it is disappointing that it is closed. I have a more concise answer that I cannot post other than as a comment: 245⋅22…11−239=245(2000*(10^m−1)/9+211)−239. Using the congruences: 245≡−1, 2000≡−9, 451≡0mod41 we get (−1)(−9)10m−19+(−1)(211)−239≡10m−1−211−239=10m−451≡10m


245 is congruent to -1 mod 41

modulo 41 we have 245*211=-211

-239-211-1 =451 congruent to 10^m mod 41

pg(451) is prime

((71*6^4+4)-344)/559=164


but we saw before that

51456=(700^2-164^2)/9
541456=(10*700^2-164^2)/9


so we arrived to this:

(700^2-(((71*6^4+4)-344)/559)^2)/9=51456
(10*700^2-(((71*6^4+4)-344)/559)^2)/9=541456


So multiple of 43

215 69660 92020 and 541456 are of the form

s(71*6^4-340)/52 +r with r being the residue 1763s+r such that is congruent to 10^m mod 41


340-52=288 maybe this explain why 92020 is congruent to 288 mod 323???


I think yes because 92020=52*1763+344


(71*6^4+4)
is congruent to 344 both modulo 1763 and modulo 559

(92020-344)/559+(92020-344)/1763=6^3


92020=(71*6^4+4)
331259=(71*6^6+4+10)/10

71*6^6 is congruent to -12^2 mod (71*6^4+4)


2236=(331*1001-81*143)/143

Anyway we have a fact:


92020 and 331259 are congruent to 71*3^3 mod (239*13)..so I wonder if there are infinitely many of such exponents

By the way I note that

541456 is congruent to 71*3^3 mod (1001)


541456 is congruent to 331*1001+71*3^3 mod (1001*13)


Or 541456 is congruent to 71*3^3 mod (11*1001)


so 541456=71*3^3+11011n where 11011 is 3^3 in base 2


71*3^3 is congruent to 344 mod 143

I think that this could be useful

It holds 541456-(331259)+13-210*1001=0

So this could be useful if you have in mind that 331259 is congruent to 71*3^3 mod (239*13)


92020=541456-13*(449*77-1)
331259=92020+239*1001

it is clear that there is a relationship

541456 is congruent to (331259+71*3^3+111) mod (239*13)

331259=92020 mod(239*13)

so 541456 is congruent to 2*71*3^3+111 mod (239*13)


92020 and 331259 are congruent to 71*3^3 mod (239*13) and to 71*3^2+13 mod (1001)


92020 and 331259 are congruent to 929 mod (1001)
541456 is congruent to (929-13) mod 1001


69660*559 is congruent to (331259-559-331) mod (1001)

69660*559 is so also congruent to (92020-559-331) mod (1001)



so 69660*559 is either congruent to (x-890) mod 1001 or to (x-903) mod (1001)

where x=541456, 331259, 92020 numbers congruent to 6 mod 13

331259 and 92020 are -72 (mod 1001) and so 69660*559 is congruent to (x-890) mod (1001) with x=331259, 92020
if x=541456 not congruent to - 72 mod 1001, then

69660*559 is congruent to (x-903) mod 1001

but this is equivalent to say

69660*559 is congruent to either (x +111) or (x+111+13) mod (1001)

x=331259, 541456, 92020


541456, 331259, 92020 are congruent to either (13*2-111) or to (13*3-111) mod 1001

69660 is congruent to 591 mod 1001

591*559 is congruent to 39 mod 1001


69660/3 is congruent to (14^2+1) mod 1001

92020 and 331259 are congruent to -(14^2-111) mod 1001
541456 is congruent to -(14^2-111-13) mod 1001



so


69660 is congruent to 591 mod (1001)
331259 is congruent to 559*591-111 mod (1001)
541456 is congruent to 559*591-111-13 mod (1001)

92020 the same as 331259 if i am not wrong

so 69660 is (14^2+1)*3 mod 1001

331259 is congruent to 559*3*(14^2+1)-111 mod 1001...


-(69660/81) is congruent to 141 mod 1001


so

541456 is congruent to (4*(14^2+1)-860-13) mod (1001)

92020 and 331259 to (4*(14^2+1)-860) mod 1001

69660 is congruent to 3*(14^2+1) mod 1001

14^2+1=197 is a prime

69660/81=860


541456 is congruent to 3*860-41*5*13 mod (1001)
331259 and 92020 are congruent to 3*860-41*5*13+13 mod (1001)
69660 is congruent to -2*41*5 mod 1001


541456 is congruent to 344-13*2^5-13 mod 1001
331259 and 92020 are congruent to 344-13*2^5 mod 1001


541456 is congruent to (344+13*2^5-13*2^6-13) mod 1001
92020 and 331259 to (344+13*2^5-13*2^6) mod 1001
69660 to (344+13*2^5+13*2^6) mod 1001


-(344+3*13*2^5) is congruent to 410 mod 1001
-(344-13*2^5) is congruent to 72 mod 1001
-(344-13*2^5-13) is congruent to 85 mod 1001

69660 is -410 mod 1001
331259 and 92020 are -72 mod 1001
541456 is -85 mod 1001


pg(2131) pg(2131*9=19179) and pg(92020) are probable primes

2131 is congruent to -1^2 mod 164
19179 is congruent to - 3^2 mod 164
92020 is congruent to 4^2 mod 164


92020 is congruent to 43*2^3 mod (2132) and congruent to 43*3^2 mod 2131


19179 is congruent to -81 mod (2131+9)
2131 is congruent to -9 mod (2131+9)
92020 is a multiple of (2131+9)


This implies that

(92020-2131) is congruent to 9 mod 4280
(92020-2131*9) is congruent to 81 mod 4280


92020 is congruent to 2140 mod 4280


92020 s congruent to 2140 mod 8988


2131=p is prime
pg(2131) is prime and pg(2131*9=19179) is prime

pg(92020) is prime

pg(69660) is prime
92020=69660+2236*10

92020 can be written both as (2236*43+344) and (2132*41+344)

so 92020 leaves the same residue 344 mod (41*2236) and mod (2132*43) so 92020 is also congruent to 344 mod 1763

Also 541456 and 69660 multiple of 86 are 344 mod 2236


So the exponents multiple of 86 are 69660 92020 and 541456


They are congruent to 344 mod 2236


But only 92020 is congruent to 344 modulo 2236, mod 1763 and modulo 2132=p+1=2131+1 where this prime p 2131 gives the other two probable prime pg(2131) and pg(19179)


19179 is congruent to -214 mod 1763

92020 is a multiple of 214


-2131*3^2*430 is congruent to 344 mod 1763

92020=214*430

so

modulo 1763

-2131*3^2=214

92020=214*430 congruent to 344 mod 1763

pg(331259) is prime
331259 is congruent to 6 mod 13

331259 is congruent to 344-559*3 mod (2132)


pg(56238) and pg(331259) are probable primes

the difference 331259-56238 is congruent to - 7 mod (2132*43)


331259 is -7 mod 13, 56238 is 0 mod 13

331259+7-56238=43*2132*3

this is equivalent to 92020+239*1001+7-56238=43*2132*3


92020+239239=331259

239239 is congruent to (56238-344-7) mod (2132*43)

92020 is congruent to 344 mod (2132*43)

331259 is congruent to (56238-7) mod (2132*43)


92020 is congruent to 344 mod 2132

331259 is congruent to 344*2+111 mod 2132

56238-7 is congruent to 344*2+111 mod 2132


pg(451=11*41) is prime pg(2131) is prime pg(92020) is prime


92020 is congruent to -2131*9*430 both mod 1763 and mod 451 -2131*9 is 214 mod 451 and mod 1763

92020 is congruent to -215 mod (6149=11*43*41)

541456 is congruent to 344 mod 6149

(215+2131*9-1)/43=451


11*43*41 mod 2131 is 214

92020=214*430

Mod 2131

214*430 is 387

So 92020 is congruent to 387 mod 2131


-(1763*11) mod 2131 is 71*3^3

331259 and 92020 have something tondo with 71*3^3

In fact 92020 and 331259 are congruent to 71*3^3 mod (239*13)


71*3^3=2131*10-41*43*11

This means that 331259 and 92020 are congruent to (2131*10-451*43) mod (239*13)

and to (21310-451*43+13) mod (1001)

541456 is congruent to (21310-451*43) mod (1001)








Consider this set of congruences:


x==-13 mod 214
x==6 mod 13
x+13=344 mod 43


Using CRT calulator I found the solution


x=92007+119626k




92007+13=92020 and pg(92020) is prime


92007+119626*2=331259 and pg(331259) is prime




another curio:


pg(1323) is prime


pg((1323*10+3)*3=39699) is prime


1323 is congruent to -215*3 mod 984

39699 is congruent to -215*3 mod 984


Pg(69660) is prime

69660 is multiple of 215*3=645

1323/3=441


69660 is congruent to -215*3+441 mod 984

Or to -215*3+21^2 mod 984


modulo 984, the number 69660 is a perfect square (42*9)^2
modulo 984

69660 is -1323*108=-(42*9)^2

as you can see 69660 is -1323*108=-(42*9)^2 mod (6^3)

now i consider this modular equation

1323x is congruent to -6^3 mod 984

the soluton wolphram gives me is (40+328n)

for n=-1 you get -288

92020 is congruent to 288 mod 323
and 215, 69660 and 541456 are + or - 108 mod 323


modulo 328:

21^2 is -215


-69660=21^2*42^2 modulo 328

-92020=107*42^2 modulo 328


-215 and -69660 in a certain sense are perfect squares (21^2 and 378^2)

378^2 is congruent to 21^2 mod (42*323)


92020=428*215

428 is congruent to -13 mod 21^2

92020 is congruent to 215*10^2 mod (215*328)

Ah ah this is weird

Also 92020 is a perfect square

-210^2 mod 328

Infact 428 is 10^ 2 mod 328

So mod 328 92020 is -210^2


1323 and 39669 are 11 mod 328

69660 modulo 328 is 11*108

Modulo 328 92020 is 11*4*(108^2-1)

-69660 is congruent to 204 mod 984


204=645-21^2


69660 is a multiple of 645

1323=42^2-21^2

1323*108 mod 984 is 204 as -69660 mod 984


69660 is congruent to 21^2-645 mod 984
92020 is congruent to 13^2-645 mod 984


1323 is 11 mod 328
39699 is 11 mod 328
69660 is (11+21^2) mod 328


1323 is 339 mod 984
39699 is 339 mod 984
69660 is (339+21^2) mod 984


69660 mod 984 is 780

42^2 also is 780 mod 984

this because
1323 is congruent to 339 mod 984

69660 is congruent to (339+21^2) mod 984

so
69660 is congruent to (3*21^2+21^2=42^2) mod 984




239*7*11 is -1 mod 172


69660 92020 and 541456 are congruent to (239*7*11-171) mod 2236


I notice that 92020 and 331259 are both congruent to 5 mod (239*7*11)


331259 is congruent to (239*7*11-171-13) mod 2236




modulo 2236 infact


239*7*11-171 is 344




239*7*11 is 515 mod 2236


so


239*7*11+1 so is 516 mod 2236




69660 is divisible by 516


92020, 541456 are 172 mod 516


331259 is -13 mod 516

-69660 is congruent to 1323 mod (239*11)
69660 is congruent to 6 mod 13
using wolphram alpha solution is
:
69660=1306+34177*2

92020=1306+22360+34177*2
331259=1306+22360+34177*9



what is quite clear is that

there are k such that pg(k) is prime with k congruent to 6 mod 13 and k of the form 215+1001s-287 (case 92020 and 331259) and 541456 which is of the form 215+1001s-287-13

so k is of the form -72+1001s or -85+1001s

215 is congruent to -(929-11*13) mod (7*11*13)
92020 is congruent to (929) mod (7*11*13)
331259 is congruent to (929) mod (7*11*13)
541456 is congruent to (929-13) mod (7*11*13)

929 and 331259 are primes congruent to -72 mod (7*11*13)

so mod 143

215 331259 92020 ...are plus or minus 71 mod 143
541456 is (71-13=58) mod 143..


215 is congruent to -786 mod (11*13*7)

786 is a number of the form 71+143*s

331259 and 92020 are congruent to 929 mod (11*13*7)

929 is of the form 71+143s



so these exponents leading to a prime with k congruent to 6 mod 13 have this property


215 is congruent to (71+143s) mod 1001
541456 is congruent to (58+143s) mod 1001
331259 is congruent to (71+143s) mod 1001
92020 also
I dont know 69660??? maybe not but 69660 is the only one multiple of 3


i note that 215 541456 331259 92020 69660 are congruent to plus or minus (71+r) mod 1001


where r is a 13-smooth number which is not a 11 smooth number


r infact can be one of these numbers: 845, 858 or 520




520=71+18^2+14^2
845=71+18^2+14^2+18^2+1
858=71+18^2+14^2+18^2+14

Last fiddled with by enzocreti on 2021-01-15 at 14:35 Reason: 3
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