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A174681 Half-sums (averages) of two primes with prime subscripts.

%I A174681 #8 Feb 28 2020 22:52:41
%S A174681 3,4,5,7,8,10,11,14,17,18,21,22,23,24,26,29,31,32,35,36,38,39,41,42,
%T A174681 43,44,45,47,49,50,54,56,57,59,60,62,63,65,66,67,69,70,71,72,75,79,80,
%U A174681 81,83,84,87,88,91,92,93,94,95,96,97,98,99,101,104,105,107,108,109,110,111
%N A174681 Half-sums (averages) of two primes with prime subscripts.
%C A174681 11 is the smallest value generated in two ways, as 22 is the smallest sum of two primes with prime subscripts in two ways: 11 + 11 = 17 + 5. Corrected and extended by _R. J. Mathar_, who also supplied the Maple code. _Jason Kimberley_ also computed the first 733 positive integers which cannot be represented as the half sum of two primes with prime subscripts (A174682) as found using the first 998 values of A006450. From computing the first 20 thousand terms of A006450 (the 20000th term is 3118459), he shows the next value in the sequence of complements must be greater than 2907940. The PIP-Goldbach Conjecture is: every sufficiently large even number can be represented as the sum of two primes with prime subscripts.
%F A174681 {(A006450(i) + A006450(j))/2} = {(A000040(A000040(i)) + A000040(A000040(j)))/2}.
%e A174681 a(1) = 6/2 = 3 = (3 + 3)/2 because 3 is the first prime with prime subscript, p(2).
%e A174681 a(2) = 8/2 = 4 = (3 + 5)/2 because 5 is the second prime with prime subscript, p(3).
%e A174681 a(3) = 10/2 = 5 = (5 + 5)/2.
%e A174681 a(4) = 14/2 = 7 = (3 + 11)/2 because 11 is the second prime with prime subscript, p(5).
%p A174681 hfs := {} ;
%p A174681 for i from 1 to 100 do
%p A174681 for j from 1 to i do
%p A174681 ithprime(ithprime(i))+ithprime(ithprime(j)) ;
%p A174681 hfs := hfs union {%/2}
%p A174681 end do:
%p A174681 end do: sort(hfs) ;
%Y A174681 Cf. A000040, A006450, A174682.
%K A174681 easy,nonn
%O A174681 1,1
%A A174681 _Jonathan Vos Post_, Mar 26 2010