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Topic: Centred Hexagonal Numbers (Read 275 times) |
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ThudnBlunder
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Centred Hexagonal Numbers
« on: Jul 18th, 2007, 5:51pm » |
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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 is a possible value of a and that every possible value of a is a hex number.
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| « Last Edit: Jul 18th, 2007, 6:11pm by ThudnBlunder » |
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Sameer
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Re: Centred Hexagonal Numbers
« Reply #1 on: Jul 18th, 2007, 9:21pm » |
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First part
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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FiBsTeR
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Re: Centred Hexagonal Numbers
« Reply #2 on: Jul 20th, 2007, 11:27am » |
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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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| « Last Edit: Jul 20th, 2007, 12:03pm by FiBsTeR » |
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