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A346918
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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.
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0
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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, 7, 6, 6, 7, 3, 4, 16, 20, 4, 4, 4, 14, 17, 0, 26, 4, 2, 17
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OFFSET
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1,6
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COMMENTS
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The algorithmic process goes on like this.
Given a primorial prime(n)#.
1) Add 1 to this primorial.
2) If this number is prime, we are done.
3) If not, we look at the prime factors of the number prime(n)# + 1.
4) Now we add the smallest prime factor p1 of prime(n)# + 1 to prime(n)#.
5) If prime(n)# + p1 is prime, then we are done.
6) If not we add the smallest prime factor p2 of prime(n)# + p1 to prime(n)#.
7) If prime(n)# + p2 is prime, then we are done. If not we repeat this process until we find a prime.
8) In the case that we will never find a prime through this proces, we set the value to 0.
There are numbers n such that a(n)=0. Some known values where this happens are n = 13, 14, 20, 21, 45.
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LINKS
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FORMULA
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Open question: Are there infinitely many n such that a(n)=0?
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EXAMPLE
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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.
a(14)=0 because the least prime factors that occur in the process form an infinite loop:
prime(14)# + 1 = 167 * 78339888213593
prime(14)# + 167 = 89 * 487 * 301842542779
prime(14)# + 89 = 1049537 * 12465269287
prime(14)# + 1049537 = 53 * 246844553447539
prime(14)# + 53 = 12911 * 1013303487853
prime(14)# + 12911 = 3393197 * 3855585553
prime(14)# + 3393197 = 53 * 27031 * 9131906089
...
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MATHEMATICA
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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 *)
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PROG
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(PARI)
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)) )))
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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