# Triangular numbers

Just in case you want to impress your friends. The costs of arts and abilities in Ars Magica are cooler than regular square numbers, they are triangular numbers.

en.wikipedia.org/wiki/Triangular_numbers

Aren't we the coolest nerds in the world?

Why yes. Yes, we are the coolest nerds.

I remember teaching this to my students last year, then going to my Ars Magica book and making the connection. It was insight-flash moment, promptly followed by feeling silly for not realizing earlier.

I thought everybody knew that...

Well, I didn't

So did I....

Xavi

I never knew there was something called a "triangular number" before I bumped into that. They don't teach that in the schools I went to.

I always called the ArM scores "pyramid numbers" though I'm not sure where that came from. I'm fairly certain I'm not smart enough to have come up with it on my own. Then again, they aren't "pyramid" exactly, more triangular, which explains the name on the wiki.

Rich

Well, I can verify that ArM4 has the following phrases...

'Abilities are purchased on a pyramid system'
'The score is purchased using a pyramid scale...'
'This pyramid purchase scale is used at several other points in covenant creation.'
'...at double the pyramid cost of the bonus or penalty...'
'Buying a reputation costs a number of pyramid points...'
'Increasing or decreasing the aura costs a number of pyramid points...'
etc.

I don't know if the use of the phrase has carried over to ArM5, since I have the physical books, which aren't nearly as searchable as the 4th ed .pdf I have, and I wasn't in ArM before 4th, so I don't know when the line started using the term either.

Ah. Well that explains it then

Read a book why don't I?

Well, Art costs are triangular numbers. Ability costs are multiples of triangular numbers.

Since D&D Leveling is also done at multiples of triangular numbers (multiplied by 1000 instead of 5) it is not all that unique. Though I am pretty sure Ars Magica did before D&D.

From GenConUK a few years ago, when Jonathan Tweet and I shared a panel.

Floor: Is it true that ArM was basically your house rules for D&D?

JT: No. We were really playing RuneQuest at the time.

Me: Is D&D 3E basically your house rules for ArM?

JT: Yes.

OK, so he was at least partially joking... But the mechanical similarities between D&D 3E and ArM are not coincidental, and the direction of causation is ArM to D&D, not the other way around.

I think it not a coincidence that triangular numbers are one of the simpler and more obvious arithmetic progressions and so are likely to turn up in games which need numerical progressions.
They were used as an example arithmetic progression in my Secondary school maths many years ago. They seem to be a sensible and very liekly result of the mecahnics for paying a cost in an RPG To improve something from one integer value to another. Particularly where improvement does not have a strictly linear effect on the effectiveness of the ability. The usual alternative being a linear progression of Nx Points getting an increase of x regardless of the value being improved
I seem to recall the use of such numbers with a multiplier in the previous edition of the story teller system(experience), Shadowrun (2nd and 3rd edition karma),

OR...

Cost equals the level you are buying * 1 for Arts, * 5 for Abilities. Must buy each level separately.

I know which equation I'm going to force my players to use!

Definitely the former. But then, I'm an evil Storyguide. 8)

DK, you make me do that and i will have Saul explain Pyrrhic victories to you.

n*(n+1)/2 isn't that difficult to deal with.

You mean that thing from Miami Vice?

How? Since they all give the same answer, how would you keep them from using whichever equation they prefer?

Exactly...
BTW, back when I was a casino dealer (a croupier) in Reno, NV. I was once asked to verify a rumor. Some of the dealers said that they had heard that the numbers on a roulette wheel (1...36) added up to 666, and wished they knew how to verify it. I guess they did not want to know badly enough to use a calculator. I was asked to figure this out because I had a reputation for being slightly brilliant, otherwise known as "Good at Jeopardy".
I had never heard of a triangular number and had never passed an algebra class, and I figured out n[(n+1)/2].