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)) = 5}. {a(n)} = {k such that k*(3*k-1)/2 has exactly 5 prime factors}. {a(n)} = {k such that A000326(k) is an element of A014614}.

a(1) = 11 because P(11) = PentagonalNumber(11) = 11*(3*11-1)/2 = 176 = 2^4 * 11 is a 4-almost prime (the prime factors need not be distinct).

a(2) = 35 because P(35) = 35*(3*35-1)/2 = 1820 = 2^2 * 5 * 7 * 13 is a 5-almost

prime.

a(13) = 100 because P(100) = 100*(3*100-1)/2 = 14950 = 2 * 5^2 * 13 * 23 is a 5-almost

prime.

Cf. A000326, A001222, A014614.

Sequence in context: A041232 A050287 A096762 this_sequence A054475 A029540 A118554

Adjacent sequences: A114442 A114443 A114444 this_sequence A114446 A114447 A114448

easy,nonn

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