%I #23 Feb 16 2025 08:32:56
%S 2,5,6,9,13,17,19,21,23,31,33,53,71,87,89,93,113,123,127,157,163,167,
%T 177,181,197,201,219,229,237,321,327,347,373,393,401,409,417,419,449,
%U 487,489,503,509,519,523,537,541,563,571,577,597,599,633,647,699,751
%N Indices of octahedral numbers (A005900) which are semiprimes.
%C 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.
%D J. H. Conway and R. K. Guy, The Book of Numbers, New York, Springer-Verlag, p. 50, 1996.
%D L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
%H Amiram Eldar, <a href="/A103982/b103982.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Harvey P. Dale)
%H 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 (2003), 65-75.
%H Jonathan Vos Post, <a href="http://magicdragon.com/poly.html">Table of Polytope Numbers, Sorted, Through 1,000,000</a>. [dead link]
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/OctahedralNumber.html">Octahedral Number</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Semiprime.html">Semiprime</a>.
%F Numbers k such that A005900(k) is a term of A001358.
%F Numbers k such that A103981(k) = 2.
%F Numbers k such that A001222(A005900(k)) = 2.
%F Numbers k such that Bigomega((2*k^3 + k)/3) = 2.
%e 93 is in this sequence because A005900(93) = (2*93^3 + 93)/3 = 536269 = 31 * 17299, which is semiprime.
%t Flatten[Position[Table[(2n^3+n)/3,{n,1000}],_?(PrimeOmega[#]==2&)]] (* _Harvey P. Dale_, Jun 17 2013 *)
%o (PARI) isok(n) = bigomega((2*n^3+n)/3) == 2; \\ _Michel Marcus_, Dec 14 2015
%Y Cf. A001222, A005900, A103946, A103981.
%K easy,nonn,changed
%O 1,1
%A _Jonathan Vos Post_, Feb 23 2005
%E More terms from _Harvey P. Dale_, Jun 17 2013