OFFSET
1,4
COMMENTS
For each prime, the end of the trajectory is reached when one cannot generate another prime number from it.
For example, p(3) = 5 -> 2 (1 iteration), so a(3)=1. Also p(5) = 11 -> 5 -> 2 (2 iterations), 23 -> 11 -> 5 -> 2 (3 iterations) and 47 -> 23 -> 11 -> 5 -> 2 (4 iterations). Hence a(3) = 1, a(5) = 2, a(9) = 3 and a(15) = 4.
a(n) = 0 for n = 1, 7, 8, 10, 11, 13, 14, 16, 19, 20, 22, 24, 25, ... The corresponding primes are A176902(n) = 2, 17, 19, 29, 31, 41, 43, ... .
The sequence of the last terms of the trajectories begins with 2, 2, 2, 2, 2, 2, 17, 19, 2, 29, 31, 19, 41, 43, 2, 53, 29, 31, 67, ...
The following table gives the trajectories of the smallest prime requiring 0, 1, 2, 3, 4, 5, 6, iterations:
+------------+----------+------------------------------------------+
| Number of | smallest | trajectory |
| iterations | prime | |
+------------+----------+------------------------------------------+
| 0 | 2 | 2 |
| 1 | 3 | 3 -> 2 |
| 2 | 7 | 7 -> 3 -> 2 |
| 3 | 13 | 13 -> 7 -> 3 -> 2 |
| 4 | 47 | 47 -> 23 -> 11 -> 5 -> 2 |
| 5 | 2879 | 2879 -> 1439 -> 719 -> 359 -> 179 -> 89 |
| 6 | 1065601 | 1065601 -> 532801 -> 266401 -> 133201 -> |
| | | 66601 -> 33301 -> 16651 |
+------------+----------+------------------------------------------+
EXAMPLE
a(15) = 4 because prime(15) = 47 and 47 -> 23 -> 11 -> 5 -> 2 with 4 iterations.
MAPLE
for n from 1 to 100 do:
ii:=0:it:=0:p:=ithprime(n):
for i from 1 to 100 while(ii=0) :
p1:=(p-1)/2:p2:=(p+1)/2:
if type(p1, prime)=false and type(p2, prime)=false
then
ii:=1:printf(`%d, `, it):
else
it:=it+1:
if isprime(p1)
then
p:=p1:
else
p:=p2:
fi:
fi:
od:
od:
MATHEMATICA
f[p_] := If[PrimeQ[(q = (p-1)/2)], q, If[PrimeQ[(r = (p+1)/2)], r, 0]]; g[n_] := -2 + Length @ NestWhileList[f, n, #>0 &]; g /@ Select[Range[457], PrimeQ] (* Amiram Eldar, Nov 16 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 25 2019
STATUS
approved