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A259301 Taken over all those prime-partitionable numbers m for which there exists a 2-partition of the set of primes < m that has one subset containing two primes only, a(n) is the frequency with which the smaller prime occurs, where n is the prime index.

%I A259301 #16 Aug 27 2015 10:58:55
%S A259301 0,0,1,1,3,3,3,2,4,4,3,4,5,7,8,5,8,7,8,9,10,10,11,12,12,14,13,13,12,
%T A259301 15,14,14,17,14,19,17,12,18,13,19,20,22,20,23,21,15,21,21,23,25,26,23,
%U A259301 26,26,19,23,27,24,29,27,26,28,31,29,30,25,30,29,34,30
%N A259301 Taken over all those prime-partitionable numbers m for which there exists a 2-partition of the set of primes < m that has one subset containing two primes only, a(n) is the frequency with which the smaller prime occurs, where n is the prime index.
%C A259301 A number n is called a prime partitionable number if there is a partition {P1,P2} of the primes less than n such that for any composition n1+n2=n, either there is a prime p in P1 such that p | n1 or there is a prime p in P2 such that p | n2.
%C A259301 To demonstrate that a positive integer m is prime-partitionable, a suitable 2-partition {P1, P2} of the set of primes < m must be found. In this sequence we are interested in prime-partitionable numbers such that P1 contains 2 odd primes.
%C A259301 Conjecture: If P1 = {p1a, p1b} with p1a and p1b odd primes, p1a < p1b and p1b = 2*k*p1a + 1 for some positive integer k such that 2*k <= p1a - 3 and if m = p1a + p1b then m is prime-partitionable.
%H A259301 Christopher Hunt Gribble, <a href="/A259301/b259301.txt">Table of n, a(n) for n = 1..9592</a>
%e A259301 The table below shows all p1a and p1b pairs for p1a <= 29 that demonstrate that m is prime-partitionable.
%e A259301 . n    p1a    p1b     2k      m
%e A259301 . 3      5     11      2     16
%e A259301 . 4      7     29      4     36
%e A259301 . 5     11     23      2     34
%e A259301 .       11     67      6     78
%e A259301 .       11     89      8    100
%e A259301 . 6     13     53      4     66
%e A259301 .       13     79      6     92
%e A259301 .       13    131     10    144
%e A259301 . 7     17    103      6    120
%e A259301 .       17    137      8    154
%e A259301 .       17    239     14    256
%e A259301 . 8     19    191     10    210
%e A259301 .       19    229     12    248
%e A259301 . 9     23     47      2     70
%e A259301 .       23    139      6    162
%e A259301 .       23    277     12    300
%e A259301 .       23    461     20    484
%e A259301 .10     29     59      2     88
%e A259301 .       29    233      8    262
%e A259301 .       29    349     12    378
%e A259301 .       29    523     18    552
%e A259301 By examining the p1a column it can be seen that
%e A259301 a(1) = 0, a(2) = 0, a(3) = 1, a(4) = 1, a(5) = 3, a(6) = 3,
%e A259301 a(7) = 3, a(8) = 2, a(9) = 4, a(10) = 4.
%p A259301 # Makes use of conjecture in COMMENTS section.
%p A259301 ppgen := proc (ub)
%p A259301   local freq_p1a, i, j, k, nprimes, p1a, p1b, pless;
%p A259301   # Construct set of primes < ub in pless.
%p A259301   pless := {};
%p A259301   for i from 3 to ub do
%p A259301     if isprime(i) then
%p A259301       pless := `union`(pless, {i});
%p A259301     end if
%p A259301   end do;
%p A259301   nprimes := numelems(pless);
%p A259301   # Determine frequency of each p1a.
%p A259301   printf("0, ");    # For prime 2.
%p A259301   for j to nprimes do
%p A259301     p1a := pless[j];
%p A259301     freq_p1a := 0;
%p A259301     for k to (p1a-3)/2 do
%p A259301       p1b := 2*k*p1a+1;
%p A259301       if isprime(p1b) then
%p A259301         freq_p1a := freq_p1a+1;
%p A259301       end if;
%p A259301     end do;
%p A259301     printf("%d, ", freq_p1a);
%p A259301   end do;
%p A259301 end proc:
%p A259301 ub := 1000:
%p A259301 ppgen(ub):
%Y A259301 Cf. A059756, A245664.
%K A259301 nonn
%O A259301 1,5
%A A259301 _Christopher Hunt Gribble_, Jun 23 2015