OFFSET
1,1
COMMENTS
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.
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, New York, Springer-Verlag, p. 50, 1996.
L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
Jonathan Vos Post, Table of Polytope Numbers, Sorted, Through 1,000,000. [dead link]
Eric Weisstein's World of Mathematics, Octahedral Number.
Eric Weisstein's World of Mathematics, Semiprime.
FORMULA
EXAMPLE
93 is in this sequence because A005900(93) = (2*93^3 + 93)/3 = 536269 = 31 * 17299, which is semiprime.
MATHEMATICA
Flatten[Position[Table[(2n^3+n)/3, {n, 1000}], _?(PrimeOmega[#]==2&)]] (* Harvey P. Dale, Jun 17 2013 *)
PROG
(PARI) isok(n) = bigomega((2*n^3+n)/3) == 2; \\ Michel Marcus, Dec 14 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Feb 23 2005
EXTENSIONS
More terms from Harvey P. Dale, Jun 17 2013
STATUS
approved