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Old 2019-04-12, 08:10   #1
enzocreti
 
Mar 2018

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Default Prime 5673250207638707288871898411075037017304674344103088757834723767

I found this prime 5673250207638707288871898411075037017304674344103088757834723767=s


such that (s+2)/67 is again prime. 67 are the last two digits of s. 67 and 5673250207638707288871898411075037017304674344103088757834723767 are both Chen primes.




Can you find other primes p such that (p+2)/c is prime, where c are the last two digits of p and c is a Chen prime? p obvioulsly must be a Chen prime itself


Or better find all Chen primes q such that (q+2) is divisible by a Chen prime!

Last fiddled with by enzocreti on 2019-04-12 at 08:50
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Old 2019-04-12, 09:11   #2
enzocreti
 
Mar 2018

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Default This is the second right version

I found this prime 5673250207638707288871898411075037017304674344103088757834723767=s

such that (s+2)/67 is again prime. 67 are the last two digits of s. 67 and 5673250207638707288871898411075037017304674344103088757834723767 are both Chen primes.



Can you find other primes p such that (p+2)/c is prime, where c are the last two digits of p and c is a Chen prime? p obvioulsly must be a Chen prime itself

Or better find all Chen primes q such that (q+2) is divisible by a Chen prime d and (q+2)/d is again prime!
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Old 2019-04-12, 18:22   #3
CRGreathouse
 
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Aug 2006

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Quote:
Originally Posted by enzocreti View Post
Can you find other primes p such that (p+2)/c is prime, where c are the last two digits of p and c is a Chen prime? p obvioulsly must be a Chen prime itself
That makes c one of A109611(5)-A109611(20) = {11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 67, 71, 83, 89}.

10^1000 + 17817 is the first example with more than a thousand digits. Using PFGW one could extend this to 100,000 digits without much difficulty, I imagine.

Code:
findNextDirect(x)=
{
  my(v=[11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 67, 71, 83, 89]);
  forstep(n=ceil(x)\2*2+1,oo,2,
    my(p=n%100);
    if(setsearch(v,p) && n%p==0 && ispseudoprime(n/p), return(n))
  );
}

findNextResidue(x)=
{
  my(N,mn=(x+10)\11,m=mn%100);
  mn += if(m>1, 101, 1) - m;
  forstep(q=mn,oo,100,
    if(ispseudoprime(q),
      N=11*q;
      break
    )
  );

  forprime(p=13,89,
    if(bigomega(p+2)>2, next);
    mn=(x-1)\p+1;
    m=mn%100;
    mn += if(m>p, 101, 1) - m;
    forstep(q=mn,N\p,100,
      if(ispseudoprime(q),
        N=p*q;
        break
      )
    );
  );
  N;
}

findNextPrimeResidue(x)=
{
  my(N,mn=(x+10)\11,m=mn%100);
  mn += if(m>1, 101, 1) - m;
  forprimestep(q=mn,,100,
    N=11*q;
    break
  );

  forprime(p=13,89,
    if(bigomega(p+2)>2, next);
    mn=(x-1)\p+1;
    m=mn%100;
    mn += if(m>p, 101, 1) - m;
    forprimestep(q=mn,N\p,100,
      N=p*q;
      break
    );
  );
  N;
}
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Old 2019-04-12, 19:08   #4
CRGreathouse
 
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Aug 2006

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Quote:
Originally Posted by enzocreti View Post
Or better find all Chen primes q such that (q+2) is divisible by a Chen prime!
I'll assume that you're not allowing twin primes (q, q+2) which trivially meet the conditions. (Wait, now I see you've posted a second thread with that clarification, let me merge that in.)

The first few are
Code:
7, 13, 19, 23, 31, 37, 47, 53, 67, 83, 89, 109, 113, 127, 131, 139, 157, 167, 181, 199, 211, 233, 251, 257, 263, 293, 307, 317, 337, 353, 359, 379, 389, 401, 409, 443, 449, 467, 479, 487, 491, 499, 503, 509, 541, 557, 563, 571, 577, 587, 631, 647, 653, 677, 683, 701, 719, 743, 751, 761, 769, 787, 797, 811, 829, 839, 863, 877, 887, 911, 919, 937, 941, 947, 953, 971, 977, 983, 991, 1009, 1039, 1097, 1109, 1117, 1163, 1187, 1193, 1201, 1217, 1259, 1283, 1291, 1297, 1327, 1361, 1367, 1381, 1399, 1409, 1439
There are 20162 up to a million, 150230 up to ten million, 1161427 up to a hundred million, and 9222205 up to a billion.
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Old 2019-04-14, 15:22   #5
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by enzocreti View Post
I found this prime 5673250207638707288871898411075037017304674344103088757834723767=s


such that (s+2)/67 is again prime. 67 are the last two digits of s. 67 and 5673250207638707288871898411075037017304674344103088757834723767 are both Chen primes.




Can you find other primes p such that (p+2)/c is prime, where c are the last two digits of p and c is a Chen prime? p obvioulsly must be a Chen prime itself
<snip>
The primes q < 100 for which q+2 is either prime, the product of two primes, or the square of a prime -- and neither q nor q+2 is 5 -- are given in the vector

[7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 67, 71, 83, 89]

There are 223 primes p < 10^6 for which (1) p + 2 is divisible by r = p%100, (2) r is one of the components of the above vector, and (3) (p+2)/r is (pseudo)prime. They are given by the vector

[467, 911, 941, 3407, 4211, 5507, 7607, 10559, 13907, 16007, 16319, 18671, 22307, 24329, 33911, 37517, 39971, 41729, 47111, 50411, 57917, 62207, 63611, 66431, 67619, 67829, 68507, 76907, 78347, 80111, 82571, 83207, 85229, 86711, 86837, 87407, 98717, 104207, 106307, 117389, 118907, 122231, 123941, 126311, 131507, 134417, 141719, 142007, 142811, 147419, 149717, 150407, 160841, 162611, 162947, 168731, 169859, 180317, 185441, 185711, 189011, 192407, 197837, 198719, 200807, 205817, 219707, 221567, 221807, 222011, 227219, 235211, 240659, 241517, 241667, 242807, 250889, 253307, 267929, 274271, 276629, 280607, 282707, 292517, 293207, 294029, 301319, 311111, 312107, 318419, 319937, 320741, 333041, 335519, 340811, 341219, 341507, 347411, 348617, 350711, 352007, 353237, 357689, 360407, 360611, 363911, 363917, 364031, 382241, 382631, 391907, 398219, 403511, 404507, 409817, 410807, 415319, 419141, 421019, 426611, 431807, 433229, 439811, 444671, 459047, 462911, 464237, 465917, 468029, 472319, 473147, 475907, 476111, 503207, 517919, 522017, 530507, 530837, 534707, 553037, 553607, 555029, 557747, 559259, 559907, 562007, 583367, 586319, 591341, 591611, 592019, 593507, 594911, 603467, 603641, 604007, 604811, 613817, 630737, 631307, 635507, 643031, 664211, 665141, 666119, 675407, 676829, 683159, 683807, 685907, 687311, 694619, 698507, 706019, 706907, 707111, 717011, 721571, 725807, 726911, 733511, 736259, 737729, 742607, 755147, 757307, 765707, 773231, 774107, 775037, 778307, 786311, 787217, 791519, 801407, 812711, 812717, 819437, 835811, 859919, 861941, 864989, 866231, 881207, 894119, 894329, 897137, 903029, 916907, 919337, 930437, 933707, 941537, 949931, 950507, 965267, 970817, 974837, 977831, 979907, 985937, 996431, 997037]

The first (pseudo)prime p > 10^100 with properties (1) to (3) is 10^100 + 991417. The first such (pseudo)prime p > 10^200 is 10^200 + 8409211.

EDIT: the first value, p = 467, has the curious property that either r = 7 = p%10 or r = 67 = p%100 will give the desired result.

Last fiddled with by Dr Sardonicus on 2019-04-14 at 15:30
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