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Topic: Pascal's Triangle (II) (Read 802 times) |
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ThudnBlunder
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Pascal's Triangle (II)
« on: Apr 17th, 2003, 6:49am » |
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In Pascal's Triangle (I) it was required to find three consecutive BCs in any row of the triangle that are in the ratio 1 : 2 : 3 (LZJ solved it in a jiffy!)
Here we need to consider three consecutive BCs in any row of the triangle that are
in the ratio a : b : c
where a,b,c are positive integers such that a < b =< c and GCD(a,b,c) = 1
That is, BC[n,k] : BC[n,k+1] : BC[n,k+2] = a : b : c
(i) Express n and k in terms of a, b, and c.
(ii) What general restrictions apply to a, b, and c?
(BC = Binomial Coefficient)
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| « Last Edit: Aug 28th, 2003, 3:35am by ThudnBlunder » |
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LZJ
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Re: Pascal's Triangle (II)
« Reply #1 on: Apr 17th, 2003, 6:03pm » |
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All right:
(i)
BC[n,k+1] / BC[n,k]
= (n - k) / (k + 1) = b/a
Therefore, an = (a + b)k + b--------(1)
BC[n,k+2] / BC[n,k+1]
= (n - k - 1) / (k + 2) = c/a
Therefore, bn = (b + c)k + b + 2c---(2)
Solving the equations, we get:
k = (ab + 2ac - b2) / (b2 -ac)
n = (ab + 2ac + bc) / (b2 - ac)
(ii) General restrictions? Dunno, unless its something like
(b2 - ac) must be greater than 0 or something.
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ThudnBlunder
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Re: Pascal's Triangle (II)
« Reply #2 on: Apr 18th, 2003, 4:03am » |
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Quote:Solving the equations, we get:
k = (ab + 2ac - b2) / (b2 -ac)
n = (ab + 2ac + bc) / (b2 - ac)
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Correct, LZJ, although I prefer the form
a(b + c) + c(a + b)
n = -------------------
b2 - ac
a(b + c)
k = --------- - 1
b2 - ac
Obviously, b2 > ac
And both n and k are integers iff the following conditions are satisfied:
(1) b2 - ac divides a(b + c)
(2) b2 - ac divides c(a + b)
Using the fact that k >= 0, we can derive another resriction on c in terms of a and b.
Can you find it?
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| « Last Edit: Aug 28th, 2003, 3:34am by ThudnBlunder » |
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LZJ
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Re: Pascal's Triangle (II)
« Reply #3 on: Apr 18th, 2003, 7:24am » |
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c >= b(b-a)/2a
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ThudnBlunder
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Re: Pascal's Triangle (II)
« Reply #4 on: Apr 18th, 2003, 10:14am » |
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Quote:
Right, and we already have b2 -ac >= 1
b2 - 1 b(b -a)
Hence ------- >= c >= --------
a 2a
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ThudnBlunder
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Re: Pascal's Triangle (II)
« Reply #5 on: Apr 18th, 2003, 12:37pm » |
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I played around with this problem some years ago while I was a graduate student in London.
I considered the following special cases and wrote a Pascal (of course) program to produce the results:
1] a,b,c are in arithmetic progression (2b = a + c)
2] a,b,c are in geometric progression (b2 = ac)
3] a,b,c are in harmonic progression (2/b = 1/a + 1/c)
4] b = a (see c = b)
5] b = a2
6] b = Na, N > 1
7] c = a
8] c = a2 (unfinished)
9] c = a + b
10] c = ab
11] c = b
12] c = b2
13] c = Na, N > 1 (unfinished)
14] c = Nb, N > 1
Cases 8 and 13 remain unfinished in the sense that I have hitherto been unable to express n,k,a,b,c explicitly
in terms of one or more dummy variables, as in the following simple example:
when c = a, {a,b,c} = {m,m+1,m} giving {n,k} = {2m,m-1}
and all solutions satisfy n = 2(k + 1), m = 1,2,3,....
Besides which, for case 8 (c = a2) I could find only 3 solutions for a =< 3,000:
{a,b,c} = {1,2,1} giving {n,k} = {2,0}
{a,b,c} = {2,3,4} giving {n,k} = {34,13}
{a,b,c} = {13,47,169} giving {n,k} = {1079,233}
Are these the only solutions?
In my opinion, the case c = a + b is the most interesting.
There are solutions only when
a = F(2m), b = F(2m + 1), c = F(2m + 2), m = 1,2,3,……..
where F(m) = the mth Fibonacci number, with F(1) = F(2) = 1.
Then we get
n = [F(2m + 2)F(2m + 3)] - 1
k = [F(2m)F(2m + 3)] - 1
All solutions satisfy F(2m)n = F(2m + 2)k + F(2m + 1), m = 1,2,3,….
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| « Last Edit: Apr 20th, 2003, 6:51pm by ThudnBlunder » |
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