

A305400


a(n) = round(1/(A073918(n)/prime(n)#  1)), where A073918(n) = min { prime p  omega(p1) = 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
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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 A171989 A233438
Adjacent sequences: A305397 A305398 A305399 * A305401 A305402 A305403


KEYWORD

nonn


AUTHOR

M. F. Hasler, May 31 2018


STATUS

approved



