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

3, 4, 5, 7, 8, 10, 11, 14, 17, 18, 21, 22, 23, 24, 26, 29, 31, 32, 35, 36, 38, 39, 41, 42, 43, 44, 45, 47, 49, 50, 54, 56, 57, 59, 60, 62, 63, 65, 66, 67, 69, 70, 71, 72, 75, 79, 80, 81, 83, 84, 87, 88, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 104, 105, 107, 108, 109, 110, 111

Offset: 1

Author

Jonathan Vos Post, Mar 26 2010

Keywords

• easy
• nonn

Comments

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.

Examples

			a(1) = 6/2 = 3 = (3 + 3)/2 because 3 is the first prime with prime subscript, p(2).
a(2) = 8/2 = 4 = (3 + 5)/2 because 5 is the second prime with prime subscript, p(3).
a(3) = 10/2 = 5 = (5 + 5)/2.
a(4) = 14/2 = 7 = (3 + 11)/2 because 11 is the second prime with prime subscript, p(5).
		

Crossrefs

Cf. A000040, A006450, A174682.

Programs

Maple

hfs := {} ;
for i from 1 to 100 do
for j from 1 to i do
ithprime(ithprime(i))+ithprime(ithprime(j)) ;
hfs := hfs union {%/2}
end do:
end do: sort(hfs) ;

Formula

{(A006450(i) + A006450(j))/2} = {(A000040(A000040(i)) + A000040(A000040(j)))/2}.