# Challenge

• Apr 21st 2012, 11:14 PM
ophelius
Challenge
Consider a function that behaves like this:
F[0,y]={1,1,1,1,1..}
F[1,y]={1,2,3,4,5..}
F[2,y]={1,3,6,10,15..}
F[3,y]={1,4,10,20,35..}
F[4,y]={1,5,15,35,70..}
...
which is simply F[x,y]=Binomial[x+y,x].

Now the challenge is to come up with an equation that explains this behavior:
G[1,y]={0,1,2,3,4,5....}
G[2,y]={0,1,1,3,3,3,6,6,6,6,10,10,10,10,10,15,15,15,15,1 5,15,21...}
G[3,y]={0,1,1,1,4,4,4,4,4,4,10,10,10,10,10,10,10,10,10,1 0,20,20....}
G[4,y]={0,1,1,1,1,5,5,5,5,5,5,5,5,5,5,15...}
...
It shouldn't be too hard to figure out the pattern, but coming up with a function is the tricky part I can't figure out. Anybody want to take a shot at it? This isn't homework.
• Apr 22nd 2012, 02:32 AM
a tutor
Re: Challenge
Can you invert figurate numbers, in general?
• Apr 22nd 2012, 09:16 AM
ophelius
Re: Challenge
Quote:

Originally Posted by a tutor
Can you invert figurate numbers, in general?

I know what figurate numbers are, but I don't understand the question.
• Apr 22nd 2012, 10:25 AM
a tutor
Re: Challenge
G[3,4] = 4 = F[3,1]
G[3,10] = 10 = F[3,2]
G[3,20] = 20 = F[3,3]

If you solve

n(n+1)(n+2)/6 =4

n(n+1)(n+2)/6 =10

n(n+1)(n+2)/6 =20

you get n=2,n=3 and n=4.
• Apr 22nd 2012, 10:40 AM
ophelius
Re: Challenge
Quote:

Originally Posted by a tutor
G[3,4] = 4 = F[3,1]
G[3,10] = 10 = F[3,2]
G[3,20] = 20 = F[3,3]

If you solve

n(n+1)(n+2)/6 =4

n(n+1)(n+2)/6 =10

n(n+1)(n+2)/6 =20

you get n=2,n=3 and n=4.

Not sure I follow. G[3,4] should equal 4 as well as G[3,5]=4, G[3,6]=4, G[3,7]=4, G[3,8] and G[3,9]=4. G[3,10] should then equal 10, G[3,11]=10, etc.
• Apr 22nd 2012, 12:39 PM
a tutor
Re: Challenge
Yes but if you solve

n(n+1)(n+2)/6=4
n(n+1)(n+2)/6=5
n(n+1)(n+2)/6=6
n(n+1)(n+2)/6=7
n(n+1)(n+2)/6=8
n(n+1)(n+2)/6=9

and round down you get n=2 in every case.
• Apr 22nd 2012, 12:55 PM
ophelius
Re: Challenge
Quote:

Originally Posted by a tutor
Yes but if you solve

n(n+1)(n+2)/6=4
n(n+1)(n+2)/6=5
n(n+1)(n+2)/6=6
n(n+1)(n+2)/6=7
n(n+1)(n+2)/6=8
n(n+1)(n+2)/6=9

and round down you get n=2 in every case.

Ah, I understand. I haven't tried it but I assume the continuation for other values of x in G[x,y] would be:

x=4: n(n+1)(n+2)(n+3)/24
x=5: n(n+1)(n+2)(n+3)(n+4)/120
...
But then there's the problem of isolating an equation for n for any x. The higher x is, the more difficult it is to solve for n. Is this correct?