A058032 Largest m such that 2^n / primorial(m) >= 1.

0, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18

Offset: 0

Author

Labos Elemer, Nov 22 2000

Keywords

• nonn

Comments

Primorial order of powers of 2.

Examples

			For n = 1 and 2, when 2 and 4 are divided by 2 the quotient = 1 or 2, but when they are divided by 6 the quotient < 1, so the largest suitable primorial is the first; thus a(1) = a(2) = 1.
For n = 11, 2^11 = 2048. The largest primorial P, such that 2048/P > 1 is P = 210, the 4th = A002110(4). So a(11) = 4.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A045716 (binary order, A029837, of primorials, A002110).

Programs

Mathematica

seq[lim_] := Module[{m = -1, p = 1, primo = 1, pow = 1, s = {}}, Do[While[pow < primo, AppendTo[s, m]; pow *= 2]; p = NextPrime[p]; primo *= p; m++, {lim}]; s]; seq[20] (* Amiram Eldar, Feb 08 2025 *)

Formula

a(n) = Max{m| sign(floor(2^n/A002110(m))) = 1}.

Extensions

a(0) inserted by Amiram Eldar, Feb 08 2025