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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: A237130 A058773 A122805 * A029270 A090350 A199133

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 December 3 21:14 EST 2016. Contains 278745 sequences.