

A234649


Difference between the first members of the widest and the narrowest prime pair having an arithmetic mean of n.


1



2, 2, 4, 2, 6, 4, 6, 6, 10, 8, 12, 0, 14, 14, 10, 14, 14, 16, 18, 16, 16, 12, 22, 16, 20, 24, 24, 26, 26, 28, 26, 32, 30, 26, 36, 16, 36, 36, 28, 36, 36, 18, 44, 38, 40, 44, 42, 40, 50, 48, 40, 42, 52, 30, 42, 46, 42, 56, 56, 58, 48, 60, 64, 56, 66, 60, 48, 60, 70, 68, 68, 54, 68, 74, 60, 56
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

8,1


COMMENTS

Existence of a(n) for all n depends on A061357(n) > 0.
Even numbers missing in the subsequence with n<10^5 are 34,62,82,88,112,116,118,122,130,140,152...
a(n) = 0 for n=4,5,6,7,19 because A061357(n) = 1.


LINKS



FORMULA



EXAMPLE

The prime pairs with an arithmetic mean of 18 are (17,19), (13,23), (7,29), and (5,31), so a(18) = 175 = 3119 = 12. The only pair with mean of 19 is (7,31) so a(19) = 0.


PROG

(PARI) a(n)=mi=0; ma=0; forprime(p=3, n1, if(isprime(2*np), if(!mi, mi=2*np); ma=2*np)); if(!ma, 1, mima)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



