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%I #35 Aug 30 2021 22:22:37
%S 1,1,1,1,1,2,2,9,9,2,1,5,0,0,5,2,8,4,11,0,0,5,4,11,2,11,19,3,11,3,20,
%T 7,6,6,7,3,4,16,20,4,4,4,14,17,0,26,4,2,17
%N a(n) is 1 plus the number of iterations of the map prime(n)# + k -> prime(n)# + lpf(prime(n)# + k) required to reach a prime, starting at k=1, where prime(n)# is the n-th primorial and lpf() is the least prime factor, or 0 if no prime is ever reached.
%C The algorithmic process goes on like this.
%C Given a primorial prime(n)#.
%C 1) Add 1 to this primorial.
%C 2) If this number is prime, we are done.
%C 3) If not, we look at the prime factors of the number prime(n)# + 1.
%C 4) Now we add the smallest prime factor p1 of prime(n)# + 1 to prime(n)#.
%C 5) If prime(n)# + p1 is prime, then we are done.
%C 6) If not we add the smallest prime factor p2 of prime(n)# + p1 to prime(n)#.
%C 7) If prime(n)# + p2 is prime, then we are done. If not we repeat this process until we find a prime.
%C 8) In the case that we will never find a prime through this proces, we set the value to 0.
%C There are numbers n such that a(n)=0. Some known values where this happens are n = 13, 14, 20, 21, 45.
%F Open question: Are there infinitely many n such that a(n)=0?
%e The 6th primorial is 2*3*5*7*11*13. Adding 1 gives us prime(6)# + 1 = 59*509, prime(6)# + 59 is prime. So a(6)=2.
%e a(14)=0 because the least prime factors that occur in the process form an infinite loop:
%e prime(14)# + 1 = 167 * 78339888213593
%e prime(14)# + 167 = 89 * 487 * 301842542779
%e prime(14)# + 89 = 1049537 * 12465269287
%e prime(14)# + 1049537 = 53 * 246844553447539
%e prime(14)# + 53 = 12911 * 1013303487853
%e prime(14)# + 12911 = 3393197 * 3855585553
%e prime(14)# + 3393197 = 53 * 27031 * 9131906089
%e ...
%t a[n_] := Module[{primorial = Product[Prime[i], {i, 1, n}], ps = {}, sum, p, count = 1}, sum = primorial + 1; While[! PrimeQ[sum], p = FactorInteger[sum][[1, 1]]; If[MemberQ[ps, p], count = 0; Break[]]; AppendTo[ps, p]; sum = primorial + p; count++]; count]; Array[a, 30] (* _Amiram Eldar_, Aug 20 2021 *)
%o (PARI)
%o primorial=1 ; for(n=1, 35, primorial=primorial*prime(n) ; iterations=1 ; addtoprimorial=1 ; addtoprimorialvector=[]; done=0 ; while(!done, if(isprime(primorial + addtoprimorial) , print1(iterations, ", ") ; done=1 , iterations++ ; addtoprimorial=factor(primorial + addtoprimorial)[1,1] ; if(vecsearch(addtoprimorialvector,addtoprimorial), print1(0, ", "); done=1) ; addtoprimorialvector=vecsort(concat(addtoprimorialvector,addtoprimorial)) )))
%Y Cf. A002110, A020639.
%K nonn,more,hard
%O 1,6
%A _Kim Hollesen_, Aug 07 2021
%E a(13)-a(49) from _Jinyuan Wang_, Aug 20 2021
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