Prime 5673250207638707288871898411075037017304674344103088757834723767

[enzocreti]

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!

[enzocreti]

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!

[CRGreathouse]

Originally Posted by enzocreti

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.

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;
}

[CRGreathouse]

Originally Posted by enzocreti

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

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.

[Dr Sardonicus]

Originally Posted by enzocreti

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.