A102294 Number of prime divisors (with multiplicity) of icosahedral numbers.

0, 3, 5, 3, 3, 5, 3, 5, 3, 4, 5, 4, 3, 7, 4, 5, 3, 5, 5, 5, 3, 6, 4, 5, 4, 5, 6, 5, 3, 11, 3, 7, 4, 5, 9, 6, 2, 6, 5, 6, 3, 5, 4, 6, 4, 6, 5, 6, 3, 6, 6, 5, 3, 7, 5, 7, 4, 4, 6, 6, 2, 8, 6, 8, 4, 6, 6, 5, 3, 6, 5, 6, 3, 5, 6, 4, 4, 7, 3, 8, 6, 6, 6, 5, 3, 6, 5, 5, 4, 8, 5, 5, 3, 8, 6, 8, 3, 7, 10, 6

Offset: 1

Author

Jonathan Vos Post, Feb 19 2005

Keywords

• easy
• nonn

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, ...

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001222, A006564.

Programs

Mathematica

Table[PrimeOmega[n*(5*n^2-5*n+2)/2],{n,120}] (* Harvey P. Dale, Jun 06 2015 *)

PARI

a(n)=bigomega(n)+bigomega(5*binomial(n,2)+1) \\ Charles R Greathouse IV, Mar 09 2012

Formula

a(n) = A001222(A006564(n)) = Bigomega(n*(5*n^2 - 5*n + 2)/2).