A275693 Lexicographically earliest increasing sequence such that the a(n)th term of the sequence has n noncomposite divisors.

1, 2, 4, 6, 7, 30, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 2310, 2311, 2312, 2313, 2314, 2315, 2316, 2317, 2318, 2319, 2320, 2321, 2322, 2323, 2324, 2325, 2326, 2327, 2328, 2329, 2330

Offset: 1

Author

Jaroslav Krizek, Aug 05 2016

Keywords

• nonn

Comments

We let tau_nc(n) = number of noncomposite divisors of n = A083399(n) = A001221(n) + 1 = omega(n) + 1.

Primorial numbers from A002110 are terms.

Examples

			a(1)=1 because tau_nc(1)=1; a(2)=2 because tau_nc(2)=2; a(3) cannot be 3 because tau_nc(3)=2, a(3)=4 (4 is the smallest number x>3); if a(3)=4, a(4) must be the smallest number x>a(3) with 3 noncomposite divisors, a(4)=6; a(6) must be number with 4 noncomposite divisors and must keep increase of the sequence, a(6)=30; a(5)=7 because 7>a(4); a(7) must be the smallest number with 5 noncomposite divisors because a(5)=7, a(7)=210; if a(6)=30, a(30) must be the smallest number x>a(7) with 6 noncomposite divisors and must keep increase of the sequence, a(30)=2310; a(8)-a(29) are numbers from interval 211-232; etc...
		

Links

• Jaroslav Krizek, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A001221, A002110, A083399, A275658.

Formula

tau_nc(a(a(n))) = A083399(a(a(n))) = A001221(a(a(n))) + 1 = omega(a(a(n))) + 1 = n.