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        riddles >> medium >> Centred Hexagonal Numbers
        
(Message started by: ThudanBlunder on Jul 18^th, 2007, 5:51pm)

Title: Centred Hexagonal Numbers
Post by ThudanBlunder on Jul 18^th, 2007, 5:51pm

Consider an arithmetic progression 1, a, b and a geometric progression 1, c, d, where a,b,c,d are positive integers and a+b = c+d. Prove that every hex number (http://en.wikipedia.org/wiki/Centered_hexagonal_number) is a possible value of a and that every possible value of a is a hex number.

Title: Re: Centred Hexagonal Numbers
Post by Sameer on Jul 18^th, 2007, 9:21pm

First part

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a = 1 + 3n(n-1)
b = 2a - 1
a + b = 3a - 1 = 3 + 9n(n-1) - 1 = 2 + 9n(n-1) = (3n-2)(3n-1)

c + d = c + c^2 = c(c+1)

Thus if c = 3n-2
and d = c^2 = (3n-2)^2
will always give us integers
Thus all hex numbers are possible values for a

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Title: Re: Centred Hexagonal Numbers
Post by FiBsTeR on Jul 20^th, 2007, 11:27am

And the second:

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Given the arithmetic sequence {1, a, b, ...} and the geometric sequence {1, c, d, ...}, we can rewrite them as {1, a, 2a-1, ...} and {1, c, c², ...}.

Then:
a + b = c + d
a + (2a - 1) = c + (c²)
3a - 1 = c² + c
3a = c² + c + 1

Since we want to show that a must be in the form 1 + 3n(n-1) = 3n² - 3n + 1, we have to find some c = pk + q that will produce a = 3n² - 3n + 1. So let c = pk + q:

3a = (pk + q)² + (pk + q) + 1
3a = p²k² + 2pqk + q² + pk + q + 1
a = [ (p²)k² + (2pq + p)k + (q² + q + 1) ]/3

We want:
p²/3 = 3
(2pq + p)/3 = -3
(q² + q + 1)/3 = 1

A bit of algebra shows that we need p = -3, q = 1.

So thus we take c = -3n + 1:
3a = c² + c + 1
3a = (-3n + 1)² + (-3n + 1) + 1
3a = 9n² - 6n + 1 - 3n + 1 + 1
3a = 9n² - 9n + 3
a = 3n² - 3n + 1
a = 1 + 3n(n-1)

Thus:
If c = 1 (mod 3), then a must be a hexagonal number.

If c = 0 (mod 3), then:
3a = (3n)² + (3n) + 1 = 9n² + 3n + 1, which is not divisible by 3, and a is thus not an integer, which cannot be.

If c = 2 (mod 3), then:
3a = (3n + 2)² + (3n + 2) + 1 = 9n² + 12n + 4 + 3n + 2 + 1 = 9n² + 15n + 7, whichis not divisible by 3, and a is thus not an integer, which cannot be.

Therefore c must be congruent to 1 modulo 3, which as shown above implies that a must be a hex number.
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