

A174682


Positive integers which cannot be represented as halfsums (averages) of two primes with prime subscripts.


4



1, 2, 6, 9, 12, 13, 15, 16, 19, 20, 25, 27, 28, 30, 33, 34, 37, 40, 46, 48, 51, 52, 53, 55, 58, 61, 64, 68, 73, 74, 76, 77, 78, 82, 85, 86, 89, 90, 100, 102, 103, 106, 113, 115, 117, 124, 128, 130, 132, 134, 138, 145, 146, 148, 149, 151, 152, 155, 156, 158, 161, 163, 164
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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 PIPGoldbach Conjecture is: every sufficiently large even number can be represented as the sum of two primes with prime subscripts.


LINKS

Jason Kimberley, Table of n, a(n) for n = 1..733


FORMULA

Complement of A174681. Complement of {(A006450(i) + A006450(j))/2} = Complement of {(A000040(A000040(i)) + A000040(A000040(j)))/2}.


EXAMPLE

a(1) = 1 and a(2) = 2 are in the sequence because they are smaller than the first halfsum (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 halfsum between (5 + 5)/2 = 5 and (3 + 11)/2 = 7.


CROSSREFS

Cf. A000040, A006450, A174681.
Sequence in context: A032713 A257475 A086939 * A287126 A270095 A153043
Adjacent sequences: A174679 A174680 A174681 * A174683 A174684 A174685


KEYWORD

easy,nonn


AUTHOR

Jonathan Vos Post, Mar 26 2010


STATUS

approved



