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

%I #23 Jul 29 2015 05:47:46

%S 11,23,29,47,53,59,67,79,83,89,103,107,131,137,139,149,167,173,179,

%T 191,223,227,229,233,239,263,269,277,283,293,311,317,347,349,359,367,

%U 373,383,389,431,439,461,467,479,499,503,509,523,557,563,569,587,593,607

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

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

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

%H Christopher Hunt Gribble, <a href="/A259560/b259560.txt">Table of n, a(n) for n = 1..20000</a>

%e The table lists values of n, q, 2k and p for 1 <= n <= 20.

%e .n q 2k p (a(n))

%e .1 5 2 11

%e .2 11 2 23

%e .3 7 4 29

%e .4 23 2 47

%e .5 13 4 53

%e .6 29 2 59

%e .7 11 6 67

%e .8 13 6 79

%e .9 41 2 83

%e 10 11 8 89

%e 11 17 6 103

%e 12 53 2 107

%e 13 13 10 131

%e 14 17 8 137

%e 15 23 6 139

%e 16 37 4 149

%e 17 83 2 167

%e 18 43 4 173

%e 19 89 2 179

%e 20 19 10 191

%p ppgen := proc (n)

%p local i, j, k, nprimes, p1a, p1b, p1b_ind, pless, pless_idx, p1b_ind_num_0, p1b_ind_num_1;

%p pless := {};

%p for i from 3 to n do

%p if isprime(i) then

%p pless := `union`(pless, {i})

%p end if

%p end do;

%p nprimes := numelems(pless);

%p p1b_ind := Vector(nprimes);

%p for j to nprimes do

%p p1a := pless[j];

%p if (1/2)*pless[-1]+1/2 < p1a then

%p break

%p end if;

%p for k to (1/2)*p1a-3/2 do

%p p1b := 2*k*p1a+1;

%p if member(p1b, pless, 'pless_idx') then

%p p1b_ind[pless_idx] := 1

%p elif pless[-1] < p1b then

%p break

%p end if

%p end do

%p end do;

%p p1b_ind_num_0 := 1;

%p p1b_ind_num_1 := 0;

%p for i to nprimes do

%p if p1b_ind[i] = 0 then

%p p1b_ind_num_0 := p1b_ind_num_0+1

%p else

%p p1b_ind_num_1 := p1b_ind_num_1+1;

%p fprintf(fop, "%d %d\n", p1b_ind_num_1, pless[i])

%p end if

%p end do

%p end proc;

%p n := 376200;

%p ppgen(n);

%o (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

%Y Cf. A059756, A245664.

%K nonn

%O 1,1

%A _Christopher Hunt Gribble_, Jun 30 2015