1,1

P(2) = 5 is the only prime pentagonal number, all other factor as P(k) = (k/2)*(3*k-1) or k*((3*k-1)/2) and thus have at least 2 prime factors. P(k) is semiprime iff [k prime and (3*k-1)/2 prime] or [k/2 prime and 3*k-1 prime].

Eric Weisstein's World of Mathematics, Pentagonal Number.

Eric Weisstein's World of Mathematics, Almost Prime.

{a(n)} = {k such that A001222(A000326(k)) = 7}. {a(n)} = {k such that k*(3*k-1)/2 has exactly 7 prime factors}. {a(n)} = {k such that A000326(k) is an element of A046308}.

a(1) = 27 because P(27) = PentagonalNumber(27) = 27*(3*27-1)/2 = 1080 = 2^3 * 3^3 * 5 is a 7-almost prime.

a(2) = 43 because P(43) = 43*(3*43-1)/2 = 2752 = 2^6 * 43 is a 7-almost

prime.

a(7) = 180 because P(180) = 180*(3*180-1)/2 = 48510 = 2 * 3^2 * 5 x 7^2 * 11 is a 7-almost prime.

Cf. A000326, A001222, A046308.

Sequence in context: A039614 A117103 A124940 this_sequence A141229 A121614 A046340

Adjacent sequences: A114443 A114444 A114445 this_sequence A114447 A114448 A114449

easy,nonn

Jonathan Vos Post (jvospost3(AT)gmail.com), Feb 14 2006