%I #17 Dec 15 2019 11:34:38

%S 22,35,51,145,247,287,1247,1717,2147,2501,3151,4187,5017,7957,11051,

%T 13207,15251,16801,17767,20827,26867,33227,49051,63551,68587,71177,

%U 76501,81317,96647,112477,118301,128627,147737,159251,182527,232657,237407,241001,250717

%N Semiprimes of the form n*(3*n-1)/2.

%C Semiprimes among pentagonal numbers A000326 = { (3*n^2-n)/2; n >= 0 }.

%C We can have an odd prime n = 2k + 1 and (3n - 1)/2 = 3k + 1 also prime, i.e., k in A130800, or n = 2p with p prime and 3n - 1 = 6p - 1 also prime, i.e., p in A158015. Considering the ratio of the two prime factors, the two possibilities are mutually exclusive, so this is the disjoint union of {A033570(n)=(2n+1)(3n+1); n in A130800} = A255584 and {p*(6p-1); p in A158015}. - _M. F. Hasler_, Dec 13 2019

%H K. D. Bajpai, <a href="/A245365/b245365.txt">Table of n, a(n) for n = 1..10400</a>

%e n=6: (3*n^2-n)/2 = 51 = 3 * 17 which is semiprime. Hence, 51 appears in the sequence.

%e n=10: (3*n^2-n)/2 = 145 = 5 * 29 which is semiprime. Hence, 145 appears in the sequence.

%t Select[Table[(3*n^2 - n)/2, {n, 500}], PrimeOmega[#] == 2 &]

%o (PARI) select(n->bigomega(n)==2, vector(1000, n, (3*n^2-n)/2)) \\ _Colin Barker_, Jul 20 2014

%Y Cf. A001358, A000326.

%K nonn

%O 1,1

%A _K. D. Bajpai_, Jul 19 2014