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.

2, 5, 3, 13, 17, 7, 11, 29, 37, 19, 41, 23, 31, 53, 61, 43, 73, 47, 89, 97, 59, 101, 109, 67, 71, 79, 113, 137, 83, 103, 149, 157, 107, 173, 127, 181, 131, 193, 197, 139, 229, 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

Offset

1,1

Comments

After 2, for each n >= 1, swap the places of primes A002144(n) and A002145(n) in A000040.

Links

Antti Karttunen, Table of n, a(n) for n = 1..10001

Formula

a(1) = 2; after which, if prime(n) modulo 4 = 1, a(n) = A002145(A267097(n)), otherwise a(n) = A002144(A267098(n)).

a(n) = A000040(A267100(n)).

a(n) = A267099(A000040(n)).

Example

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.
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.
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.
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.

Prog

(Scheme)
(define (A267101 n) (cond ((= 1 n) 2) ((= 1 (modulo (A000040 n) 4)) (A002145 (A267097 n))) (else (A002144 (A267098 n)))))

Crossrefs

Cf. A000040, A002144, A002145, A267097, A267098, A267099, A267100.

Cf. also A108546.

Sequence in context: A262373 A028415 A211306 * A035334 A243506 A341351

Adjacent sequences: A267098 A267099 A267100 * A267102 A267103 A267104

Keyword

nonn

Author

Antti Karttunen, Feb 01 2016

Status

approved