A114445 Indices of 5-almost prime pentagonal numbers.

11, 35, 40, 42, 51, 54, 59, 63, 67, 80, 87, 92, 100, 115, 120, 125, 126, 131, 132, 136, 159, 165, 167, 168, 175, 184, 189, 200, 204, 210, 215, 217, 222, 225, 227, 231, 232, 242, 247, 250, 251, 255, 259, 260, 261, 270, 280, 282, 283, 285, 287, 291, 295, 304, 306

Offset: 1

Author

Jonathan Vos Post, Feb 14 2006

Keywords

• easy
• nonn

Comments

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].

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Almost Prime.
• Eric Weisstein's World of Mathematics, Pentagonal Number.

Crossrefs

Cf. A000326, A001222, A014614.

Programs

Mathematica

Select[Range[300], PrimeOmega[PolygonalNumber[5, #]] == 5 &] (* Amiram Eldar, Oct 05 2024 *)

Formula

{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}.

Extensions

More terms from Amiram Eldar, Oct 05 2024