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A267101 2 followed by permutation of odd primes, where each n-th prime of the form 4k+1 (A002144) has been replaced with the n-th prime of the form 4k+3 (A002145) and vice versa.

%I A267101 #12 Feb 03 2016 17:02:14
%S A267101 2,5,3,13,17,7,11,29,37,19,41,23,31,53,61,43,73,47,89,97,59,101,109,
%T A267101 67,71,79,113,137,83,103,149,157,107,173,127,181,131,193,197,139,229,
%U A267101 151,233,163,167,241,257,269,277,179,191,281,199,293,211,313,223,317,227,239,337,251,349,353,263,271,373,283,389
%N A267101 2 followed by permutation of odd primes, where each n-th prime of the form 4k+1 (A002144) has been replaced with the n-th prime of the form 4k+3 (A002145) and vice versa.
%C A267101 After 2, for each n >= 1, swap the places of primes A002144(n) and A002145(n) in A000040.
%H A267101 Antti Karttunen, <a href="/A267101/b267101.txt">Table of n, a(n) for n = 1..10001</a>
%F A267101 a(1) = 2; after which, if prime(n) modulo 4 = 1, a(n) = A002145(A267097(n)), otherwise a(n) = A002144(A267098(n)).
%F A267101 a(n) = A000040(A267100(n)).
%F A267101 a(n) = A267099(A000040(n)).
%e A267101 For n=2, for which A000040(2) = 3, the first prime of the form 4k+3, we select the first prime of the form 4k+1, which is 5, thus a(2) = 5.
%e A267101 For n=3, for which A000040(3) = 5, the first prime of the form 4k+1, we select the first prime of the form 4k+3, which is 3, thus a(3) = 3.
%e A267101 For n=4, for which A000040(4) = 7, the second prime of the form 4k+3, we select the second prime of the form 4k+1, which is 13, thus a(4) = 13.
%e A267101 For n=5, for which A000040(5) = 11, the third prime of the form 4k+3, we select the third prime of the form 4k+1, which is 17, thus a(5) = 17.
%o A267101 (Scheme)
%o A267101 (define (A267101 n) (cond ((= 1 n) 2) ((= 1 (modulo (A000040 n) 4)) (A002145 (A267097 n))) (else (A002144 (A267098 n)))))
%Y A267101 Cf. A000040, A002144, A002145, A267097, A267098, A267099, A267100.
%Y A267101 Cf. also A108546.
%K A267101 nonn
%O A267101 1,1
%A A267101 _Antti Karttunen_, Feb 01 2016