1,2

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.

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), ...

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.

Cf. A002476, A003627, A072010.

Sequence in context: A072438 A132739 A060791 this_sequence A030104 A049563 A084453

Adjacent sequences: A116909 A116910 A116911 this_sequence A116913 A116914 A116915

easy,nonn

Jonathan Vos Post (jvospost2(AT)yahoo.com), Mar 18 2006