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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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS When a(n) = 2, n is an element of A103982: indices of octahedral numbers (A005900) which are semiprimes. REFERENCES Conway, J. H. and Guy, R. K. The Book of Numbers. New York, Springer-Verlag, p. 50, 1996 Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952. LINKS Robert Israel, Table of n, a(n) for n = 0..10000 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 CROSSREFS Cf. A001222, A005900, A103946, A103982. Sequence in context: A336037 A058773 A122805 * A029270 A364503 A090350 Adjacent sequences: A103978 A103979 A103980 * A103982 A103983 A103984 KEYWORD easy,nonn AUTHOR Jonathan Vos Post, Feb 24 2005 EXTENSIONS More terms from Wesley Ivan Hurt, Aug 11 2014 STATUS approved

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Last modified February 21 06:16 EST 2024. Contains 370219 sequences. (Running on oeis4.)