A116912 In prime factorization of n replace all primes of form k*6+1 with k*6+5 and primes of form k*6+5 with k*6+1.

1, 2, 3, 4, 1, 6, 11, 8, 9, 2, 7, 12, 17, 22, 3, 16, 13, 18, 23, 4, 33, 14, 19, 24, 1, 34, 27, 44, 25, 6, 35, 32, 21, 26, 11, 36, 41, 46, 51, 8, 37, 66, 47, 28, 9, 38, 43, 48, 121, 2, 39, 68, 49, 54, 7, 88, 69, 50, 55, 12, 65, 70, 99, 64

Offset: 1

Author

Jonathan Vos Post, Mar 18 2006

Keywords

• easy
• nonn

Comments

Primes of form 6n + 1 are also primes of the form 3n+1 and -3 is a quadratic residue mod a prime p iff p is in this sequence. Primes of the form 6n + 5 are the same as A003627 Primes of form 3n-1, except that the latter sequence starts with 2. Every twin prime after (3,5) is of the form (6n+5, 6n+1) hence the current sequence exchanges lesser twin primes with greater twin primes. See also: A072010 In prime factorization of n replace all primes of form k*4+1 by k*4+3 and primes of form k*4+3 by k*4+1. See also: A002476 Primes of form 6n + 1.

Examples

			a(5) = 1 because 5 is a prime of the form 6n + 5 (with n = 0), so is replaced with 6n + 1 (with n = 0), namely 1.
a(7) = 11 because 7 is a prime of the form 6n + 1 (with n = 1), so is replaced with 6n + 5 (with n = 1), namely 11.
a(11) = 7 because 11 is a prime of the form 6n + 5 (with n = 1), so is replaced with 6n + 1 (with n = 1), namely 7.
a(13) = 17 because 13 is a prime of the form 6n + 1 (with n = 2), so is replaced with 6n + 5 (with n = 2), namely 17.
a(14) = 22 because 14 = 2 * 7; but 7 is a prime of the form 6n + 1 (with n = 1), so is replaced with 6n + 5 (with n = 1), namely 11; giving 2 * 11 = 22.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002476, A003627, A072010.

Programs

Mathematica

f[p_, e_] := If[p < 5, p^e, (p + 6 - 2 * Mod[p, 6])^e]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 64] (* Amiram Eldar, Jan 10 2020 *)

Formula

Multiplicative. a(2^n) = 2^n, a(3^n) = 3^n, a(5^n) = 1, a(7^n) = 11^n, a(11^n) = 7^n, a(13^n) = 17^n, a(17^n) = 13^n, a(19^n) = 23^n, a(23^n) = 19^n, a(29^n) = 5^(2n), a(31^n) = (5^n)*(7^n), a(37^n) = 41^n, a(41^n) = 37^n, a(43^n) = 47^n, a(47^n) = 43^n, a(53^n) = 7^(2n), a(59^n) = (5^n)*(11^n), a(61^n) = (5^n)*(13^n), ...

Extensions

Data corrected by Amiram Eldar, Jan 10 2020