### Post by alysdexia on Apr 20, 2019 3:22:26 GMT -8

Had to use a spare account to copy these as James "Buffoonery" Peterson blockd me on Facebook. I bring out the ignorant like no other nonspammer.

www.facebook.com/groups/8273427091/permalink/10156978415357092/

James Peterson shared a link.

April 16 at 11:24 AM

Here's a project I made for testing coinage. I locked it to Base 12, I'm trying to see what numbers would work the best.

docs.google.com/spreadsheets/d/1z7tvT6fxG2NCT3bXzI2gZIyRFMEayyr_LLky5AdJPtc/edit?fbclid=IwAR1sd0rTQHe7CvLPwyxCAJaCBxIGwR4pKUfroFnXP3rfR99YKbPGNEAYZSI#gid=1752403888

James Peterson If we had 5 coins, I'm thinking 60¢, 30¢, 16¢, 10¢, and 6¢

Like · 3d

Paul Rapoport Others have analyzed this problem and come up with different and odd solutions. The Dozenal forum has those conversations…somewhere.

Like · 3d

James Peterson I've read a bunch of them, but I can't find proof of the best coinage. What makes it more difficult is that many of them are from the UK and are using "shillings" and "pence". I'd like a system that is easy to understand and is practical (ie. doesn't use 8 different kinds of coins).

Like · 3d

Paul Rapoport Nothing I've read about dozenal coinage mentions shillings and pence, except as a historical footnote. Certainly the old English system was a shambles (esp. when you get into farthings, crowns, and guineas), if an amiable one.

1 Like · 3d · Edited

Autymn Castleton See my proof below.

Like · 13h

Autymn Castleton Base e works the best.* Notice the American denominations approach this with 5/1, 10/5, 25/10, 100/25, 500/100, 1000/500, 2000/1000, 5000/2000, 10000/5000: average 2·8; without the penny 2·75.

2·7 = 27/10 = (2a+3b)/10; a+b=10 => a=3; b=7.

*e yields the minimum representation if 0 is one of the members; otherwise 2 does:

a = blog_b(n) = bln(n)/ln(b); 0 = a' = (ln(n)ln(b)-bln(n)/b)/ln(b)^2 = ln(n)(ln(b)-1); b = e.

a = (b-1)log_b(n) = (b-1)ln(n)/ln(b); 0 = a' = (ln(n)ln(b)-(b-1)ln(n)/b)/ln(b)^2 = ln(n)(ln(b)-(b-1)/b); b = 1; a'(2) < a'(>2).

Like · 21h · Edited

James Peterson But then, how do you count in base E?

Like, if I have 3 items, each of them $2. How do you write this in base E? How do you write whole numbers?

Like · 9h · Edited

Autymn Castleton James Peterson e, not E. Practically one has to approximate it with a set (better a sequence) of integers like the top of my comment already said. The denominations repeat with 7 3s and 3 2s, like 3, 3, 2, 3, 3, 3, 2, 3, 3, 2. 3 $2 items add as $2, $11, $20.

Sadly the fewest-tokens and fewest-terms problems oppose each other.

Like · 8h · Edited

www.facebook.com/groups/8273427091/permalink/10156978415357092/

James Peterson shared a link.

April 16 at 11:24 AM

Here's a project I made for testing coinage. I locked it to Base 12, I'm trying to see what numbers would work the best.

docs.google.com/spreadsheets/d/1z7tvT6fxG2NCT3bXzI2gZIyRFMEayyr_LLky5AdJPtc/edit?fbclid=IwAR1sd0rTQHe7CvLPwyxCAJaCBxIGwR4pKUfroFnXP3rfR99YKbPGNEAYZSI#gid=1752403888

James Peterson If we had 5 coins, I'm thinking 60¢, 30¢, 16¢, 10¢, and 6¢

Like · 3d

Paul Rapoport Others have analyzed this problem and come up with different and odd solutions. The Dozenal forum has those conversations…somewhere.

Like · 3d

James Peterson I've read a bunch of them, but I can't find proof of the best coinage. What makes it more difficult is that many of them are from the UK and are using "shillings" and "pence". I'd like a system that is easy to understand and is practical (ie. doesn't use 8 different kinds of coins).

Like · 3d

Paul Rapoport Nothing I've read about dozenal coinage mentions shillings and pence, except as a historical footnote. Certainly the old English system was a shambles (esp. when you get into farthings, crowns, and guineas), if an amiable one.

1 Like · 3d · Edited

Autymn Castleton See my proof below.

Like · 13h

Autymn Castleton Base e works the best.* Notice the American denominations approach this with 5/1, 10/5, 25/10, 100/25, 500/100, 1000/500, 2000/1000, 5000/2000, 10000/5000: average 2·8; without the penny 2·75.

2·7 = 27/10 = (2a+3b)/10; a+b=10 => a=3; b=7.

*e yields the minimum representation if 0 is one of the members; otherwise 2 does:

a = blog_b(n) = bln(n)/ln(b); 0 = a' = (ln(n)ln(b)-bln(n)/b)/ln(b)^2 = ln(n)(ln(b)-1); b = e.

a = (b-1)log_b(n) = (b-1)ln(n)/ln(b); 0 = a' = (ln(n)ln(b)-(b-1)ln(n)/b)/ln(b)^2 = ln(n)(ln(b)-(b-1)/b); b = 1; a'(2) < a'(>2).

Like · 21h · Edited

James Peterson But then, how do you count in base E?

Like, if I have 3 items, each of them $2. How do you write this in base E? How do you write whole numbers?

Like · 9h · Edited

Autymn Castleton James Peterson e, not E. Practically one has to approximate it with a set (better a sequence) of integers like the top of my comment already said. The denominations repeat with 7 3s and 3 2s, like 3, 3, 2, 3, 3, 3, 2, 3, 3, 2. 3 $2 items add as $2, $11, $20.

Sadly the fewest-tokens and fewest-terms problems oppose each other.

Like · 8h · Edited