A210882 a(1)=1, a(n)=a(n-1)-1 if a(n-1)-1 > 0 is not in the set {a(k)| 1<=k<n}, else a(n)=p, where p is the least prime number not yet in the sequence.

1, 2, 3, 5, 4, 7, 6, 11, 10, 9, 8, 13, 12, 17, 16, 15, 14, 19, 18, 23, 22, 21, 20, 29, 28, 27, 26, 25, 24, 31, 30, 37, 36, 35, 34, 33, 32, 41, 40, 39, 38, 43, 42, 47, 46, 45, 44, 53, 52, 51, 50, 49, 48, 59, 58, 57, 56, 55, 54, 61, 60, 67, 66, 65, 64, 63, 62, 71

Offset

1,2

Comments

A reordering of the natural numbers.

The sequence is self-inverse in that a(a(n))=n.

If n is a prime, then a(n+1) is the next prime > n. Hence, the subsequence 2, a(2+1), a(a(2+1)+1), a(a(a(2+1)+1)+1), a(a(a(a(2+1)+1)+1)+1), ... generates the sequence of primes A000040.

Links

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Formula

a(1)=1, a(n)=p (where p is the least prime number > a(k) for 1<=k<n) if the minimal natural number not yet in the sequence is greater than a(n-1), else a(n)=a(n-1)-1.

a(n)<>n for all n>3.

p(n+1)=a(p(n)+1), where p(n) is the n-th prime.

a(n+1)=p(m+2), if a(n)-1 is the m-th prime, else a(n+1)=a(n)-1, for n>2.

a(n)=p(m)+p(m-1)-n+1, where m is the least index such that p(m)>n-1 (valid for n>2).

Example

a(4)=5, since 5 is the least prime > a(1), a(2), a(3), and the minimal number not yet in the sequence (=4) is greater than 3=a(3).
a(5)=4, since 4 is not in the set {1,2,3,5}={a(k)| 1<=k<n}.
7=p(4)=a(p(3)+1)=a(a(p(2)+1)+1)= a(a(a(p(1)+1)+1)+1)= a(a(a(2+1)+1)+1).

Crossrefs

Cf. A020703, A000040, A038722, A132666, A132664, A132665, A132674, A182194.

Sequence in context: A232640 A085790 A257339 * A117120 A181095 A276345

Adjacent sequences: A210879 A210880 A210881 * A210883 A210884 A210885

Keyword

nonn

Author

Hieronymus Fischer, Apr 30 2012

Status

approved