A259560 Primes p such that p = 2kq + 1 for k a positive integer, q an odd prime and 2k <= q - 3.

11, 23, 29, 47, 53, 59, 67, 79, 83, 89, 103, 107, 131, 137, 139, 149, 167, 173, 179, 191, 223, 227, 229, 233, 239, 263, 269, 277, 283, 293, 311, 317, 347, 349, 359, 367, 373, 383, 389, 431, 439, 461, 467, 479, 499, 503, 509, 523, 557, 563, 569, 587, 593, 607

Offset: 1

Author

Christopher Hunt Gribble, Jun 30 2015

Keywords

• nonn

Comments

This sequence is associated with the conjecture in A245664 that p + q is prime-partitionable.

There are 138438 values of p in the first 216816 primes, i.e., 63.85%, all of which are distinct.

Examples

			The table lists values of n, q, 2k and p for 1 <= n <= 20.
.n      q     2k      p (a(n))
.1      5      2     11
.2     11      2     23
.3      7      4     29
.4     23      2     47
.5     13      4     53
.6     29      2     59
.7     11      6     67
.8     13      6     79
.9     41      2     83
10     11      8     89
11     17      6    103
12     53      2    107
13     13     10    131
14     17      8    137
15     23      6    139
16     37      4    149
17     83      2    167
18     43      4    173
19     89      2    179
20     19     10    191
		

Links

• Christopher Hunt Gribble, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A059756, A245664.

Programs

Maple

ppgen := proc (n)
  local i, j, k, nprimes, p1a, p1b, p1b_ind, pless, pless_idx, p1b_ind_num_0, p1b_ind_num_1;
  pless := {};
  for i from 3 to n do
    if isprime(i) then
      pless := `union`(pless, {i})
    end if
  end do;
  nprimes := numelems(pless);
  p1b_ind := Vector(nprimes);
  for j to nprimes do
    p1a := pless[j];
    if (1/2)*pless[-1]+1/2 < p1a then
      break
    end if;
    for k to (1/2)*p1a-3/2 do
      p1b := 2*k*p1a+1;
      if member(p1b, pless, 'pless_idx') then
        p1b_ind[pless_idx] := 1
      elif pless[-1] < p1b then
        break
      end if
    end do
  end do;
  p1b_ind_num_0 := 1;
  p1b_ind_num_1 := 0;
  for i to nprimes do
    if p1b_ind[i] = 0 then
      p1b_ind_num_0 := p1b_ind_num_0+1
    else
      p1b_ind_num_1 := p1b_ind_num_1+1;
      fprintf(fop, "%d %d\n", p1b_ind_num_1, pless[i])
    end if
  end do
end proc;
n := 376200;
ppgen(n);

PARI

is(n)=my(f=factor(n\2)[,1]); for(i=1,#f, if(n\2/f[i]*2<=f[i]-3, return(isprime(n)))); 0 \\ Charles R Greathouse IV, Jul 15 2015