OFFSET
1,2
COMMENTS
Because the cubic factors into n time a quadratic, the icosahedral numbers can never be prime, but can be semiprime (only if n is prime and also n*(5*n^2 - 5*n + 2)/2 is prime, as with n = 31, 61, ...
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
IcosahedralNumber(13) = 5083 = 13 * 17 * 23 so Omega(IcosahedralNumber(13)) = 3.
IcosahedralNumber(37) = 123247 = 37 * 3331 so Omega(IcosahedralNumber(37)) = 2, hence the 37th icosahedral number is the smallest to be semiprime.
MATHEMATICA
Table[PrimeOmega[n*(5*n^2-5*n+2)/2], {n, 120}] (* Harvey P. Dale, Jun 06 2015 *)
PROG
(PARI) a(n)=bigomega(n)+bigomega(5*binomial(n, 2)+1) \\ Charles R Greathouse IV, Mar 09, 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Feb 19 2005
STATUS
approved