OFFSET
1,1
COMMENTS
This list is an incomplete run of the Script pplus2ton(1000,20000,8) which has been running for 2 days. These numbers are "provable" defiant primes (PVDP). Except for 47 and 79, the primes in this list are of the form 7m-1. So p + 8^k = 7m-1 + (7+1)^k. Expanding this, we get 7m-1 + 7H+1 == 0 mod(7) for some H.
For 47 we construct the sums to get congruences as follows.
39 + 8 + 8^4m = 39 + 8(8^(4m-1)+1) == 0 mod(3)
45 + 2 + 8^(4m+1) = 45 + 8^(4m+1)+2 == 0 mod(5)
39 + 8 + 8^(4m+2) = 39 + 8(8^(4m+1)+1) == 0 mod(3)
39 + 8 + 8^(4m+3) = 39 + 8(8^(4m+2)+1) = 39 + 8(64^(2m+1)+1) == 0 mod(13)
Similarly for 79,
75 + 4 + 8^4m = 75 + 8^4m+4 == 0 mod(5)
78 + 1 + 8^(4m+1) = 78 + 8^(4m+1)+1 == 0 mod(3)
78 + 1 + 8^(4m+2) = 78 + 8^(4m+2)+1 = 78 + 64^(2m+1)+1 == 0 mod(13)
78 + 1 + 8^(4m+3) = 78 + 8^(4m+3)+1 = 78 + 8^(4m+3)+1 == 0 mod(3)
For fewer iterations of k < 5000 in the script, the next number is 271. However, 271, while provable for k=4m+1,2,3, I could not prove it to be defiant for k = 4m. I call 271 a "probable" defiant prime (PDP). In fact, fewer iterations quickly produce provable defiant primes many of which are of the form 7m-1. k <= 1000 produces 113 as a probable defiant prime (PDP) but further checking shows 113 + 8^2265 is a PRP eliminating 113 from the table. Maybe someone can come up with general method to prove a PDP candidate is PVDP prime or not.
PROG
(PARI) pplus2ton(n, m, b) = /* defiant primes base b */ { local(k, s, p, y, flag); s=0; forprime(p=2, n, flag=1; for(k=0, m, y=p+b^k; if(ispseudoprime(y), /* print1(k, ", "); */ /* k such that p + b^k is not prime */ s++; flag=0; break) ); if(flag, print1(p", ")); /* search for defiant primes. */ ); print(); print(s); }
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Cino Hilliard, Jun 28 2005
STATUS
approved