A305400 - OEIS

%I #10 Jun 19 2018 05:57:38

%S 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,

%T 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,

%U 21,10,19,27,8,7,3,6,51,15,10,10,15,8,21,17,29

%N 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.

%C 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).

%C 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.

%F a(n) = round(A002110(n)/(A073918(n) - A002110(n))).

%F a(n) = A002110(n) <=> n in A014545 <=> primorial(n) + 1 is prime.

%e For 0 <= n <= 5,  A073918(n) = prime(n)# + 1, therefore a(n) = prime(n)#.

%e 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.

%o (PARI) apply( a(n)=1\/(A073918(n)/factorback(primes(n))-1), [0..99])

%Y Cf. A073918, A002110, A014545, A305398, A305399.

%K nonn

%O 0,2

%A _M. F. Hasler_, May 31 2018