A051466 Largest product of primorials less than A025487(n) that divides A025487(n).

1, 2, 2, 4, 6, 8, 12, 6, 16, 12, 24, 30, 32, 36, 48, 60, 64, 72, 60, 96, 30, 72, 120, 128, 144, 180, 192, 210, 216, 240, 256, 288, 360, 384, 420, 432, 180, 480, 512, 360, 576, 420, 432, 720, 768, 840, 864, 900, 960, 1024, 1080, 1152, 210, 1260, 1296, 1440

Offset: 2

Author

Alford Arnold

Keywords

• easy
• nice
• nonn

Comments

Note that A036041(A025487(n)) = A036041(a(n)) + 1 since A025487(n)/a(n) is prime.

Examples

			A025487 = 1, 2, 4, 6, 8, 12, 16, 24, 30, 32, 36, ...; a(n)= 1, 2, 2, 4, 6, 8, 12, 6, 16, 12, ... . (12 divides 36, but 16 through 32 do not.)
A025487(38) = 900 = 5#*5#. The largest product of primorials that divides this number will be 5#*3# = 180 = a(38). - _Charlie Neder_, Oct 20 2018
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A036041, A034776.

Programs

Haskell

a051466 n = a051466_list !! (n-2)
a051466_list = f [head a025487_list] $ tail a025487_list where
   f us (v:vs) = fromJust (find (\x -> mod v x == 0) us) : f (v : us) vs
-- Reinhard Zumkeller, Jul 17 2013

Mathematica

(* First, load second program at A025487, then: *)
With[{s = Union@ Flatten@ f[5]}, Table[SelectFirst[Reverse@ Take[s, n - 1], Mod[s[[n]], #] == 0 &], {n, 2, Length@ s}]] (* Michael De Vlieger, Dec 27 2019 *)

Formula

a(n) = A025487(n) / p, where p is the largest prime such that p^A051282(n) | A025487(n). - Charlie Neder, Oct 12 2018

Extensions

Offset updated by Matthew Vandermast, Jul 03 2012

Name edited by Charlie Neder, Oct 20 2018

Name clarified by Antti Karttunen, Dec 24 2019