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.

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.

Links

Table of n, a(n) for n=0..71.

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

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

Sequence in context: A118747 A129779 A068215 * A096775 A331665 A171989

Adjacent sequences: A305397 A305398 A305399 * A305401 A305402 A305403

Keyword

nonn

Author

M. F. Hasler, May 31 2018

Status

approved