OFFSET
0,1
COMMENTS
It appears that a product P of distinct primes rarely has the property that P +- 1 has a square factor, and this is even more rare when P has all of the first n primes as factor. This sequence is one possible way to quantify this observation. (One could also display the gap between a(n) and prime(n), or consider b(n) the least product of distinct primes > prime(n) that yields a product with the desired property.)
See also Zumkeller's contested comment in A007018 and the discussion in the linked MathOverflow page.
LINKS
Dan Asimov, Interesting sequence on MathOverflow, math-fun mailing list, Mar 28 2023.
Fredrick M. Nelson, Does a(0)=6, a(n+1)=a(n)^3-a(n), define a square-free sequence?, MathOverflow, Mar 24 2023.
EXAMPLE
a(0) = 3 because for P = (product of the first 0 primes) = 1, p = 3 is the least prime such that p*P + 1 = 4 = 2^2 is a square; for p = 2 neither p*P - 1 = 1 nor p*P + 1 = 3 has a nontrivial square factor.
a(1) = 5 because for P = (product of the first prime) = 2, p = 5 is the least prime such that p*P - 1 = 9 = 3^2 is a square; for p = 3 none of p*P - 1 = 5 nor p*P + 1 = 7 has a nontrivial square factor.
a(2) = 29 because for P = (product of the first two primes) = 6, p = 29 is the least prime such that p*P + 1 = 5^2*7 has a square factor; for all primes 3 < p < 29 both of p*P +- 1 are squarefree.
MATHEMATICA
Map[(k = 1; While[AllTrue[Prime[k] # + {-1, 1}, SquareFreeQ], k++]; Prime[k]) &, FoldList[Times, 1, Prime@ Range[24] ] ] (* Michael De Vlieger, Mar 28 2023 *)
PROG
(PARI) A361085(n, P=vecprod(primes(n)))=forprime(p=prime(n)+1, , (issquarefree(p*P-1)&&issquarefree(p*P+1))||return(p))
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
M. F. Hasler, Mar 28 2023
EXTENSIONS
a(30) from Michael S. Branicky, Mar 29 2023
a(31)-a(40) from Jinyuan Wang, Mar 30 2023
STATUS
approved