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A103981 Number of prime factors (with multiplicity) of octahedral numbers (A005900).

%I A103981 #19 Feb 16 2025 08:32:56
%S A103981 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 A103981 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 A103981 4,3,8,2,5,4,4,3,4,4,4,4,7,5,3,3,5,3,3
%N A103981 Number of prime factors (with multiplicity) of octahedral numbers (A005900).
%C A103981 When a(n) = 2, n is a term of A103982: indices of octahedral numbers (A005900) which are semiprimes.
%D A103981 J. H. Conway and R. K. Guy, The Book of Numbers, New York, Springer-Verlag, p. 50, 1996.
%D A103981 L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
%H A103981 Robert Israel, <a href="/A103981/b103981.txt">Table of n, a(n) for n = 0..10000</a>
%H A103981 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 A103981 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/OctahedralNumber.html">Octahedral Number</a>.
%F A103981 a(n) = A001222(A005900(n)), n>0. a(n) = Bigomega((2*n^3 + n)/3), n>0.
%e A103981 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 A103981 a(4) = 3 because A005900(4) = (2*4^3 + 4)/3 = 44 = 2 * 2 * 11, which has (with multiplicity) three prime factors.
%p A103981 seq(numtheory:-bigomega((2*n^3+n)/3),n=0..100); # _Robert Israel_, Aug 10 2014
%t A103981 a[n_] := PrimeOmega[n*(2*n^2 + 1)/3]; a[0] = 0; Array[a, 100, 0] (* _Amiram Eldar_, Oct 11 2024 *)
%Y A103981 Cf. A001222, A005900, A103946, A103982.
%K A103981 easy,nonn
%O A103981 0,3
%A A103981 _Jonathan Vos Post_, Feb 24 2005
%E A103981 More terms from _Wesley Ivan Hurt_, Aug 11 2014