A108763 If n-th prime is 8m+1, then a(n) = 8m+3. If n-th prime is 8m+3, then a(n) = 8m+5. If n-th prime is 8m+5, then a(n) = 8m+7. If n-th prime is 8m+7, then a(n) = 8m+1.

5, 7, 1, 13, 15, 19, 21, 17, 31, 25, 39, 43, 45, 41, 55, 61, 63, 69, 65, 75, 73, 85, 91, 99, 103, 97, 109, 111, 115, 121, 133, 139, 141, 151, 145, 159, 165, 161, 175, 181, 183, 185, 195, 199, 193, 213, 217, 229, 231, 235, 233, 243, 253, 259, 257, 271, 265, 279, 283

Offset

2,1

Comments

a(n) is asymptotically equal to prime(n). There are infinitely many a(n) > a(n) and infinitely many a(n+1) < a(n). There are arbitrarily large gaps between a(n) and a(n+1) for sufficiently large n. If Goldbach's conjecture is true, this sequence contains infinitely many primes. Conjecture: for any whole number k, there exists at least one a(n) with exactly k prime factors (for k = 3, for instance, 8*20 + 3 = 163 which is prime and the sequence includes 8*20 + 5 = 3 x 5 x 11).

Links

Harvey P. Dale, Table of n, a(n) for n = 2..1000

Formula

a(n) = 8*quotient(n-th prime, 8) + mod(2 + n-th prime, 8) = 2 + n-th prime unless mod(n-th prime, 8) = 7 then -6 + n-th prime.

Example

Prime(1) = 2, which is not of the form 8m+1, 8m+3, 8m+5, nor 8m+7, hence this sequence starts with a(2) = 5 because prime(2) = 8*0 + 3, so 8*0+5 = 5.
Prime(11) = 31 = 8*3 + 7, so a(11) = 8*3 + 1 = 25 = 5^2, which happens not to be a prime.

Mathematica

Table[p = Prime[n]; If[ Mod[p, 8] == 7, p - 6, p + 2], {n, 2, 61}] (* Robert G. Wilson v, Jun 24 2005 *)
Table[If[Mod[n, 8]==7, n-6, n+2], {n, Prime[Range[2, 100]]}](* Harvey P. Dale, Dec 04 2017 *)

Crossrefs

Cf. A107323, A108245.

Sequence in context: A316132 A261159 A145737 * A061415 A196847 A087455

Adjacent sequences: A108760 A108761 A108762 * A108764 A108765 A108766

Keyword

easy,nonn

Author

Jonathan Vos Post, Jun 19 2005

Extensions

Edited, corrected and extended by Robert G. Wilson v, Jun 24 2005

Status

approved