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A305400
a(n) = round(1/(A073918(n)/prime(n)# - 1)), where A073918(n) = min { prime p | omega(p-1) = n } and p# = product of primes <= p.
0
1, 2, 6, 30, 210, 2310, 2, 1, 3, 3, 14, 200560490130, 2, 4, 2, 8, 7, 2, 2, 2, 4, 9, 7, 3, 2, 5, 7, 4, 13, 27, 2, 3, 3, 10, 3, 8, 9, 4, 41, 7, 4, 5, 7, 32, 5, 32, 6, 5, 7, 11, 7, 4, 5, 13, 5, 21, 10, 19, 27, 8, 7, 3, 6, 51, 15, 10, 10, 15, 8, 21, 17, 29
OFFSET
0,2
COMMENTS
We conjecture that lim inf A073918(n)/A002110(n) = 1 but the value of the lim sup is unknown. Therefore we consider x defined as A073918(n)/A002110(n) = 1 + 1/x, and a(n) = round(x).
We have lim sup a(n) = oo <=> lim inf A073918(n)/A002110(n) = 1, and lim inf a(n) = m <=> (2m + 1)/(2m - 1) >= lim sup A073918(n)/A002110(n) >= (2m + 3)/(2m + 1), where the first inequality only holds for m >= 1.
FORMULA
a(n) = round(A002110(n)/(A073918(n) - A002110(n))).
a(n) = A002110(n) <=> n in A014545 <=> primorial(n) + 1 is prime.
EXAMPLE
For 0 <= n <= 5, A073918(n) = prime(n)# + 1, therefore a(n) = prime(n)#.
For n = 6, the smallest prime p such that p - 1 has 6 distinct prime factors is prime(5)#*prime(8) + 1, therefore a(n) = round(prime(6)/(prime(8) + 1/prime(5)# - prime(6))) = 2.
PROG
(PARI) apply( a(n)=1\/(A073918(n)/factorback(primes(n))-1), [0..99])
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, May 31 2018
STATUS
approved