1,1

Because the cubic polynomial for octahedral numbers factors into n time a quadratic, the octahedral numbers can never be prime after a(3) = 19, but can be semiprime (if n is prime and 2*n^2+1 is triple a prime, or if n is triple a prime and 2*n^2+1 is prime). A005900(37) = 33781 = 11 * 37 * 83, three prime factors with same number of digits. A005900(41) = 45961 = 19 * 41 * 59, three prime factors with same number of digits. A005900(57) = 123481 = 19 * 67 * 97, three prime factors with same number of digits. A005900(67) = 200531 = 41 * 67 * 73, three prime factors with same number of digits. A005900(73) = 259369 = 11 * 17 * 19 * 73, four prime factors with same number of digits.

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.

H. K. Kim, On Regular Polytope Numbers, Journal: Proc. Amer.Math. Soc. 131 (2003), 65-75, as PDF file.

J. V. Post, Table of Polytope Numbers, Sorted, Through 1,000,000.

Eric Weisstein's World of Mathematics, Semiprime.

Eric Weisstein's World of Mathematics, Octahedral Number

n such that A005900(n) is an element of A001358. n such that A103981(n) = 2. n such that A001222(A005900(n)) = 2. n such that Bigomega((2*n^3 + n)/3) = 2.

93 is in this sequence because A005900(93) = (2*93^3 + 93)/3 = 536269 = 31 * 17299, which is semiprime.

Cf. A001222, A005900, A103946, A103981.

Sequence in context: A094350 A104857 A055198 this_sequence A030488 A163782 A051677

Adjacent sequences: A103979 A103980 A103981 this_sequence A103983 A103984 A103985

easy,nonn

Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 23 2005