A225851 For the Collatz (3x+1) iterations starting with a prime number p, a(n) is the smallest p such that the trajectory contains n successive prime numbers.

3, 3, 7, 7, 7, 19, 59, 59, 59, 59, 157, 13397, 2312267, 97760291, 1042776437

Offset

1,1

Comments

a(13)-a(15) were found by Farideh Firoozbakht. See Carlos Rivera's website. - T. D. Noe, May 17 2013

Links

Table of n, a(n) for n=1..15.

Carlos Rivera, Puzzle 476: A question related to Collatz sequence

Example

a(11) = 157 because the Collatz sequence of odd numbers is 157 -> 59 -> 89 -> 67 -> 101 -> 19 -> 29 -> 11 -> 17 -> 13 - > 5 -> 1 with 11 consecutive prime numbers.

Maple

nn:=300:T:=array(1..nn):
for n from 1 to 15 do:jj:=0:
for m from 2 to 10^5 while(jj=0) do:p:=ithprime(m):
for i from 1 to nn while(jj=0) do:
T[i]:=0:od:a:=1:T[1]:=p:x:=p:
for it from 1 to nn while (x>1) do:
if irem(x, 2)=0 then
x := x/2:
else
a:=a+1:T[a]:=x:
x := 3*x+1: fi:
od:
jj:=0:aa:=a:itr:=0:
for j from 2 to n+1 do:
if type(T[j], prime)=true then
itr :=itr+1 :
else fi:
od:
if itr=n then
jj:=1: printf ( "%d %d \n", n, p):
else
fi:
od:
od:

Mathematica

RemoveEven[n_] := n/2^IntegerExponent[n, 2]; Collatz2[n_] := NestWhileList[RemoveEven[3 # + 1] &, n, # > 1 &]; PrimeCnt[lst_] := Module[{i = 1}, While[PrimeQ[lst[[i]]], i++]; i - 1]; nn = 12; t = Table[0, {nn}]; found = 0; n = 2; While[found < nn, n = NextPrime[n]; ps = PrimeCnt[Collatz2[n]]; If[ps > nn, ps = nn]; While[ps > 0 && t[[ps]] == 0, t[[ps]] = n; found++; ps--]]; t (* T. D. Noe, May 17 2013 *)

Crossrefs

Cf. A006577.

Sequence in context: A374187 A003817 A092474 * A107470 A228325 A327122

Adjacent sequences: A225848 A225849 A225850 * A225852 A225853 A225854

Keyword

nonn,hard

Author

Michel Lagneau, May 17 2013

Status

approved