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%I #14 Aug 11 2014 05:08:14
%S 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,
%T 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,
%U 4,3,8,2,5,4,4,3,4,4,4,4,7,5,3,3,5,3,3
%N Number of prime factors (with multiplicity) of octahedral numbers (A005900).
%C When a(n) = 2, n is an element of A103982: indices of octahedral numbers (A005900) which are semiprimes.
%D Conway, J. H. and Guy, R. K. The Book of Numbers. New York, Springer-Verlag, p. 50, 1996
%D Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
%H Robert Israel, <a href="/A103981/b103981.txt">Table of n, a(n) for n = 0..10000</a>
%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2002), 65-75.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/OctahedralNumber.html">Octahedral Number</a>
%F a(n) = A001222(A005900(n)), n>0. a(n) = Bigomega((2*n^3 + n)/3), n>0.
%e 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.
%e a(4) = 3 because A005900(4) = (2*4^3 + 4)/3 = 44 = 2 * 2 * 11, which has (with multiplicity) three prime factors.
%p seq(numtheory:-bigomega((2*n^3+n)/3),n=0..100); # _Robert Israel_, Aug 10 2014
%Y Cf. A001222, A005900, A103946, A103982.
%K easy,nonn
%O 0,3
%A _Jonathan Vos Post_, Feb 24 2005
%E More terms from _Wesley Ivan Hurt_, Aug 11 2014
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