

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
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OFFSET

0,3


COMMENTS

Primorial order of powers of 2.


LINKS

Table of n, a(n) for n=0..79.


FORMULA

a(n)=Max{s q(s)=Sign[Floor[2^n/A002110(n)]]=1}


EXAMPLE

for n=1 and 2, when 2 and 4 is divided by 2 gives quotient = 1 or 2, but when divided by 6 q<1, so the largest suitable primorial is the first; thus a(1)=a(2)=1. 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.


CROSSREFS

Cf. binary order (A029837) of primorials, A045716
Cf. A002110, A045716, A029837.
Sequence in context: A087843 A087831 A194327 * A056155 A157684 A194339
Adjacent sequences: A058029 A058030 A058031 * A058033 A058034 A058035


KEYWORD

nonn


AUTHOR

Labos Elemer, Nov 22 2000


STATUS

approved



