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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.
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%I #14 Jan 10 2020 06:16:30

%S 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,

%T 44,25,6,35,32,21,26,11,36,41,46,51,8,37,66,47,28,9,38,43,48,121,2,39,

%U 68,49,54,7,88,69,50,55,12,65,70,99,64

%N 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.

%C 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.

%H Amiram Eldar, <a href="/A116912/b116912.txt">Table of n, a(n) for n = 1..10000</a>

%F 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), ...

%e 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.

%e 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.

%e 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.

%e 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.

%e 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.

%t 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 *)

%Y Cf. A002476, A003627, A072010.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, Mar 18 2006

%E Data corrected by _Amiram Eldar_, Jan 10 2020