OFFSET
1,1
COMMENTS
(prime(n) + 2)^2 + 1 is the second minimal possible value of A242719(n) after prime(n)^2 + 1. Indeed, by the definition lpf(A242719(n) - 3) > lpf(A242719(n) - 1) >= prime(n), thus after prime(n)^2 + 1 we should consider prime(n)*(prime(n) + 2) + 1. Then prime(n) should be lesser number of twin primes, but then prime(n) + 1 == 0 (mod 3). So, prime(n)*(prime(n) + 2) - 2 == 0 (mod 3). Analogously one can prove that prime(n)*(prime(n) + 4) - 2 == 0 (mod 3).
LINKS
François Marques, Table of n, a(n) for n = 1..109
FORMULA
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Sep 09 2014
EXTENSIONS
More terms from Peter J. C. Moses, Sep 09 2014
STATUS
approved