A293975 If n is even, divide by 2; otherwise, add the next larger prime.

0, 3, 1, 8, 2, 12, 3, 18, 4, 20, 5, 24, 6, 30, 7, 32, 8, 36, 9, 42, 10, 44, 11, 52, 12, 54, 13, 56, 14, 60, 15, 68, 16, 70, 17, 72, 18, 78, 19, 80, 20, 84, 21, 90, 22, 92, 23, 100, 24, 102, 25, 104, 26, 112, 27, 114, 28, 116, 29, 120, 30, 128, 31, 130, 32, 132, 33, 138, 34

Offset: 0

Author

M. F. Hasler, Nov 04 2017

Keywords

• nonn

Comments

Inspired by the "PrimeLatz" map A174221 (where the next three primes are added).

The trajectory under iterations of this map seems to end in the cycle 1 -> 3 -> 8 -> 4 -> 2 -> 1, for any starting value n. Can this be proved?

In order to develop a proof, one can consider the "condensed" version of the map which is: h(x) = odd_part(x+nextprime(x)); i.e., add the next prime, then remove all factors of 2. It is easy to see that this map verifies, for all x > 2, h(x) <= x + g(x)/2 where g(x) is the gap between the x and the next larger prime. Often, h(x) will be close to x/2 or even to x/4 or smaller. Nonetheless, for any power (iteration) of h, there are numbers for which h^m is increasing, e.g., h(h(h(x))) > x for x = 1, 525, 891, 1071, 1135, ..., and h^4(x) > x for x = 2, 1329, 5591, 8469, 9555, ...

From Robert Israel, Nov 08 2017: (Start)

It suffices to prove that if n > 1 is odd, the trajectory {x(i)} starting at x(0)=n contains some number < n. Let p = nextprime(n). As long as x(2k) is odd we have x(2k+1) = x(2k)+p and x(2k+2)=(x(2k)+p)/2 with

n <= x(2k) < x(2k+2) < p. But this can only continue finitely many times: eventually x(2k) must be even, and then x(2k+1) < p/2 < n (by Bertrand's postulate). (End)

Links

• Robert Israel, Table of n, a(n) for n = 0..10001

Crossrefs

Cf. A174221 (the "PrimeLatz" map), A006370 (the "3x+1" map).

Programs

Maple

seq(op([k,2*k+1+nextprime(2*k+1)]),k=0..100); # Robert Israel, Nov 08 2017

Mathematica

Array[If[EvenQ@ #, #/2, NextPrime@ # + # &@ #] &, 69, 0] (* Michael De Vlieger, Nov 08 2017 *)

PARI

A293975(n)=if(bittest(n,0),n+nextprime(n+1),n\2)