A272893 Number of iterations of the map p -> (p + q)/2 needed to reach the end of the cycle where the initial odd p = prime(n) and q is the smallest prime such that (p+q)/2 is prime, q different from p.

3, 2, 2, 2, 2, 3, 3, 3, 4, 4, 1, 4, 4, 5, 5, 5, 2, 1, 2, 5, 5, 5, 6, 2, 6, 6, 6, 3, 6, 2, 2, 3, 3, 6, 6, 3, 6, 7, 7, 7, 3, 3, 7, 7, 7, 7, 7, 4, 7, 3, 3, 3, 3, 3, 4, 4, 4, 4, 7, 7, 7, 4, 4, 7, 8, 8, 8, 8, 4, 8, 4, 8, 8, 4, 8, 8, 8, 5, 8, 8, 8, 5, 8, 8, 8, 5, 5

Offset

2,1

Comments

The growth of a(n) is very slow. It seems that 3 < log_10(n)/log_10(a(n)) < 4 when n tends into infinity.

a(n) = 1 for n = 12 and 19 (initial p = 37 and 67) for n < 10^5. Are there other values of n for which a(n) = 1?

We observe long subsequences of consecutive identical numbers. For example, a(115) = a(116) = ... = a(132) = 9.

Links

Table of n, a(n) for n=2..88.

Example

a(15) = 5 because prime(15) = 47, and we obtain the following trajectory: 47 -> 29 -> 17 -> 11 -> 7 -> 5. The prime 5 is the last term because 5 -> 11 with 11 = (5+17)/2 is already present in the trajectory. The elements of the trajectory are: 29 = (47+11)/2; 17 = (29+5)/2; 11 = (17+5)/2; 7 = (11+3)/2; 5 = (7+3)/2.

Maple

for n from 2 to 50 do:
p0:=ithprime(n) :lst:={p0}:
for it from 1 to 50 do :
  ii:=0:
   for m from 2 to 1000 while(ii=0) do:
     p:=ithprime(m):q:=(p0+p)/2:
      if type(q, prime)=true and q<>p
       then
       ii:=1:lst:=lst union {q}:
       else
      fi:
    od:
     p0:=q:
   od:
    n0:=nops(lst):printf(`%d, `, n0-1):
  od:

Mathematica

Table[Length@ Union@ NestList[Function[p, (p + SelectFirst[Prime@ Range[10^4], PrimeQ[(p + #)/2] && p != # &])/2], Prime@ n, 10] - 1, {n, 2, 88}] (* Michael De Vlieger, May 09 2016, Version 10 *)

Crossrefs

Cf. A000040.

Sequence in context: A057934 A058758 A122396 * A037199 A145376 A084402

Adjacent sequences: A272890 A272891 A272892 * A272894 A272895 A272896

Keyword

nonn

Author

Michel Lagneau, May 09 2016

Status

approved