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A103981
Number of prime factors (with multiplicity) of octahedral numbers (A005900).
2
0, 0, 2, 1, 3, 2, 2, 3, 4, 2, 3, 5, 4, 2, 3, 3, 7, 2, 4, 2, 5, 2, 4, 2, 4, 4, 4, 3, 4, 4, 3, 2, 6, 2, 4, 4, 4, 3, 5, 3, 6, 3, 3, 4, 4, 3, 4, 3, 6, 3, 4, 4, 5, 2, 5, 3, 7, 3, 3, 3, 5, 3, 4, 4, 7, 5, 3, 3, 4, 3, 8, 2, 5, 4, 4, 3, 4, 4, 4, 4, 7, 5, 3, 3, 5, 3, 3
OFFSET
0,3
COMMENTS
When a(n) = 2, n is a term of A103982: indices of octahedral numbers (A005900) which are semiprimes.
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, New York, Springer-Verlag, p. 50, 1996.
L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
LINKS
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, Octahedral Number.
FORMULA
a(n) = A001222(A005900(n)), n>0. a(n) = Bigomega((2*n^3 + n)/3), n>0.
EXAMPLE
a(3) = 1 because OctahedralNumber(3) = A005900(3) = 19, which is prime and thus has only one prime factor. Because the cubic polynomial for octahedral numbers factors into n time a quadratic, the octahedral numbers can never be prime after a(3) = 19.
a(4) = 3 because A005900(4) = (2*4^3 + 4)/3 = 44 = 2 * 2 * 11, which has (with multiplicity) three prime factors.
MAPLE
seq(numtheory:-bigomega((2*n^3+n)/3), n=0..100); # Robert Israel, Aug 10 2014
MATHEMATICA
a[n_] := PrimeOmega[n*(2*n^2 + 1)/3]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Oct 11 2024 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Feb 24 2005
EXTENSIONS
More terms from Wesley Ivan Hurt, Aug 11 2014
STATUS
approved