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%I #27 Jul 17 2017 00:39:44
%S 40,88,184,424,808,1048,1384,1528,1864,2104,2184,3080,4504,4744,5224,
%T 5928,6440,6568,7224,8104,8360,8840,9784,10264,10472,11480,11544,
%U 11848,12808,12904,14136,14840,14968,16280,16648,18664,19608,20344,21080,22040,23240,23704,24440,24648,24920,26008,26584,27384,27608,27688,28264,28952,29240
%N For positive n with prime decomposition n = Product_{j=1..m} (p_j^k_j) define A_n = Sum_{j=1..m} (p_j*k_j) and B_n = Sum_{j=1..m} (p_j^k_j). This sequence gives those n for which A_n and B_n are both prime and B_n = A_n + 2 (i.e., form a twin prime pair).
%C Assuming the twin prime conjecture, my advisor and I are able to prove there are infinitely many of these pairs. In other words, there are infinitely many n such that A_n and B_n are prime and B_n = A_n + 2.
%C From _Bobby Browning_, Feb 14 2011: (Start)
%C 8*A046138 is a subsequence of A185718 for the following reasons:
%C i) the n in A185718 for which A_n and B_n form a twin prime pair are of the form n=2^3*p_1*p_2*...p_k.
%C ii) the A046138 sequence consists of primes p such that p+6 and p+8 form a twin prime pair.
%C iii) so if p is a prime such that p+6 and p+8 form a twin prime pair and n = 2^3*p then A_n = p+6 and B_n = p+8. Thus, the integers such that n = 2^3*p are a subsequence of A185718. (End)
%H T. D. Noe, <a href="/A185718/b185718.txt">Table of n, a(n) for n = 1..1000</a>
%e a(1) = 40 = 2^3*5^1, with a = 11 and b = 13.
%e a(2) = 88 = 2^3*11^1 with a = 17 and b = 19.
%e a(3) = 184 = 2^3*23^1 with a = 29 and b = 31.
%t okQ[n_] := Module[{p, e, a, b}, {p, e} = Transpose[FactorInteger[n]]; a = Plus @@ (p*e); b = Plus @@ (p^e); b == a + 2 && PrimeQ[a] && PrimeQ[b]]; Select[Range[30000], okQ]
%o (PARI) sopfr(n)=my(f=factor(n));sum(i=1,#f[,1],f[i,1]*f[i,2]);
%o forstep(n=8,1e5,16,if(issquarefree(n/8)&&isprime(k=sopfr(n))&isprime(k+2), print1(n", ")))
%Y Cf. A001414 (A_n), A008475 (B_n).
%K nonn
%O 1,1
%A _Bobby Browning_, Feb 10 2011
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