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Agnishom
Today 14:56:09

#### bobbym wrote:

Yes, I see that.

with a(0)=1,a(1)=3

This is for what?

anonimnystefy
Today 14:36:25

That looks like something that would please the OP.

By the way, I just though of a twist to the puzzle to make it just a bit harder.

bobbym
Today 09:17:38

Hi;

Welcome to the forum, 308 is the correct answer.

Batman
Today 08:47:13

What you could do is say the kids got a, b, c, d, d (the twins each got the same) and then seek quadruples satisfiing a+b+c+2d=13 than try each value of d.
d=0: 13 candies 3 kids place dividers in ccccccccccccc (c for CANDY!) and see the number of ways we can put the 3 dividers (3 kids left) so 15 choose 2 = 105
d=1: same, 13 choose 2 = 78
d=2: same logic, 11 choose 2 = 55
d=3: ditto I stop saying same 9 choose 2 = 36
d=4: 7 choose 2 = 21
d=5: 5 choose 2 = 10
d=6: 3 choose 2 = 3
105+78+55+36+21+10+3=308
I got 308

bobbym
2013-01-09 06:00:14

That is not what I meant. GF's are rather obscure and not taught. He/she will probably expect a solution in terms of ncr's.

anonimnystefy
2013-01-09 05:58:00

bobbym
2013-01-09 05:53:39

Hi;

Okay, somehow though I do not think these answers are what the OP will require.

anonimnystefy
2013-01-09 05:42:42

Partial fractions in Maxima and expanding by hand.

bobbym
2013-01-09 04:55:59

Hi;

Did them by maxima?

anonimnystefy
2013-01-09 04:45:54

Partial fractions, then expanding.

bobbym
2013-01-09 04:43:48

Hi;

I know how I got mine, may I ask how you got yours?

anonimnystefy
2013-01-09 04:42:17

I think the two are same... I just put mine in the form above because of the common expression 1+(-1)^n. It is noce to know that that part is 0 when n is odd.

bobbym
2013-01-09 04:21:51

Hi;

Yes, I believe it is. I got,

anonimnystefy
2013-01-09 04:14:29

So my formula in post #9 is correct?

bobbym
2013-01-09 04:11:56

Yes, that is correct.

I meant coming up with an analytical form for the coeffs of the expansion of the polynomial in post #3. But I guess that is not really that important.