1,1

Each cluster of candidates is about e (2.71828...) times as large as the previous one.

Table of n, a(n) for n=1..27.

The average of p(4) = 7 and p(6) = 13 is 10, which is divisible by 5; so 5 is a term of the sequence.

a = 0; b = 1; c = 2; Do[ c = Prime[ n + 1 ]; If[ IntegerQ[ (a + c)/(2n) ], Print[ n ] ]; a = b; b = c, {n, 1, 2 10^9} ]

Cf. A066706.

Sequence in context: A047374 A241653 A100107 * A163098 A216562 A174009

Adjacent sequences: A066825 A066826 A066827 * A066829 A066830 A066831

nonn

Robert G. Wilson v, Jan 20 2002

approved