A227068 Least number with exactly n divisors less than its square root.

2, 6, 12, 24, 48, 60, 144, 120, 180, 240, 3072, 360, 900, 960, 720, 840, 5184, 1260, 36864, 1680, 2880, 3600, 12582912, 2520, 6480, 61440, 6300, 6720, 805306368, 5040, 14400, 7560, 46080, 983040, 25920, 10080, 32400, 746496, 184320, 15120

Offset: 1

Author

T. D. Noe, Jul 11 2013

Keywords

• nonn

Comments

This is similar to A038549, which counts divisors of n <= sqrt(n). Note that an upper bound on a(n) is 3*2^(n-1), which is attained at n = 2, 3, 4, 5, 11, 23, and 29 -- the number 4 and the primes in A005384 (Sophie Germain primes, p and 2p+1 are prime).

A056924(a(n)) = n and A056924(m) <> n for m < a(n). - Reinhard Zumkeller, Jul 12 2013

Examples

			The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Only 6 of these are < sqrt(60). And 60 is the first such number.
		

Links

• Paul Tek, C program for this sequence

Crossrefs

Cf. A005384, A038549, A056924.

Programs

C

/* See Tek link. */

Haskell

import Data.List (elemIndex); import Data.Maybe (fromJust)
a227068 = (+ 1) . fromJust . (`elemIndex` a056924_list)
-- Reinhard Zumkeller, Jul 12 2013

Mathematica

nn = 22; t = Table[0, {nn}]; found = 0; n = 0; While[found < nn, n++; c = Length[Select[Divisors[n], # < Sqrt[n] &]]; If[c > 0 && c <= nn && t[[c]] == 0, t[[c]] = n; found++]]; t
Map[Function[k, FirstPosition[#, k]], Range@ 22] &@ Table[Count[Divisors@ n, m_ /; m < Sqrt@ n], {n, 10^5}] // Flatten (* Michael De Vlieger, May 13 2016, Version 10 *)

Extensions

a(29) from Paul Tek, Jul 13 2013