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%I #6 Oct 13 2012 00:24:05
%S 1,2,6,9,12,13,15,16,19,20,25,27,28,30,33,34,37,40,46,48,51,52,53,55,
%T 58,61,64,68,73,74,76,77,78,82,85,86,89,90,100,102,103,106,113,115,
%U 117,124,128,130,132,134,138,145,146,148,149,151,152,155,156,158,161,163,164
%N Positive integers which cannot be represented as half-sums (averages) of two primes with prime subscripts.
%C _Jason Kimberley_ 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.
%H Jason Kimberley, <a href="/A174682/b174682.txt">Table of n, a(n) for n = 1..733</a>
%F Complement of A174681. Complement of {(A006450(i) + A006450(j))/2} = Complement of {(A000040(A000040(i)) + A000040(A000040(j)))/2}.
%e a(1) = 1 and a(2) = 2 are in the sequence because they are smaller than the first half-sum (average) of two primes with prime subscripts 3 = (3 + 3)/2 because 3 is the first prime with prime subscript, p(2). a(3) = 6 because there is no such half-sum between (5 + 5)/2 = 5 and (3 + 11)/2 = 7.
%Y Cf. A000040, A006450, A174681.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Mar 26 2010
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