OFFSET
1,2
COMMENTS
This is a variation of the Enots Wolley sequence A336957 and A360519, with an additional restriction that no term a(n) can have a common factor with n. For the sequence to be infinite a(n) must always have a prime factor that is not a factor of a(n-1)*(n+1). See the examples below.
Other than no term being a prime or prime power, see A336957, no term can be an even number with only two distinct prime factors. Clearly no term a(2*k) can be even, so if we assume that a(2*k+1) = 2^n*p^m, with n and m>=1, then a(2*k) must have p as a factor. But as a(2*k+2) must share a factor with a(2*k+1) and cannot have 2 as a factor, it must also have p as a factor. However that is not allowed as a(n) cannot share a factor with a(n-2), so no term can be even with only two distinct prime factors. Therefore the smallest even number is a(7) = 30.
LINKS
Scott R. Shannon, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Image of the first 100000 terms. The green line is a(n) = n.
EXAMPLE
a(2) = 15 as 15 is the smallest number that is not a prime power and does not have 2 as a factor.
a(3) = 35 as a(3) is chosen so it shares a factor with a(2) = 3*5 while not having 3 as a factor; it therefore must be a multiple of 5 while not being a power of 5. The smallest number meeting those criteria is 10, but a(2)*(3+1) = 15*4 = 60, and 10 has no prime factor not in 60, so choosing 10 would mean a(4) would not exist. The next smallest available number is 35.
a(4) = 77 as a(4) must be a multiple of 7 but not a power of 7, not a multiple of 2, 3 or 5, while having a prime factor not in 35*(4+1) = 165. The smallest number satisfying these criteria is 77.
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Dec 18 2023
STATUS
approved