The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A058032 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A058032

Largest m such that 2^n / primorial(m) >= 1.
(history; published version)

#15 by OEIS Server at Sat Feb 08 04:51:06 EST 2025

LINKS

Amiram Eldar, <a href="/A058032/b058032_1.txt">Table of n, a(n) for n = 0..10000</a>

#14 by Peter Luschny at Sat Feb 08 04:51:06 EST 2025

STATUS

reviewed

approved

Discussion

Sat Feb 08

04:51

OEIS Server: Installed first b-file as b058032.txt.

#13 by Joerg Arndt at Sat Feb 08 03:51:29 EST 2025

STATUS

proposed

reviewed

#12 by Amiram Eldar at Sat Feb 08 03:16:53 EST 2025

STATUS

editing

proposed

#11 by Amiram Eldar at Sat Feb 08 03:12:54 EST 2025

EXTENSIONS

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

#10 by Amiram Eldar at Sat Feb 08 03:03:06 EST 2025

CROSSREFS

Cf. A045716 (binary order (, A029837) , of primorials, A045716A002110).

Cf. A002110, A045716, A029837.

#9 by Amiram Eldar at Sat Feb 08 03:01:53 EST 2025

EXAMPLE

for For n = 1 and 2, when 2 and 4 is are divided by 2 gives the quotient = 1 or 2, but when they are divided by 6 qthe quotient < 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.

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.

#8 by Amiram Eldar at Sat Feb 08 03:00:05 EST 2025

FORMULA

a(n) = Max{sm| qsign(floor(s)=Sign[Floor[2^n/A002110(nm)))]] = 1}.

#7 by Amiram Eldar at Sat Feb 08 02:58:40 EST 2025

LINKS

Amiram Eldar, <a href="/A058032/b058032_1.txt">Table of n, a(n) for n = 0..10000</a>

#6 by Amiram Eldar at Sat Feb 08 02:58:27 EST 2025

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 *)