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Topic: Arithmetic progressions and partitions (Read 1838 times) |
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Pietro K.C.
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Arithmetic progressions and partitions
« on: Mar 17th, 2003, 11:06am » |
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Can the positive integers {1, 2, 3, ...} be partitioned into a finite number of sets S1, ... , Sk, each of which is an arithmetic progression - that is,
S1 = {a1, a1 + d1, ...}
...
Sk = {ak, ak + dk, ...}
- and such that all the di's are nonzero and distinct? Note that if we allowed equal di's, the answer would be a trivial yes; just take S1 = "odd integers" and S2 = "even integers".
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| « Last Edit: Mar 25th, 2003, 2:07pm by Pietro K.C. » |
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towr
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Re: Arithmetic progressions and partitions
« Reply #1 on: Mar 17th, 2003, 1:25pm » |
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If I understand right we've got
if k needn't be larger than 1, there's the trivial solution with d1=1
if the number of partitions didn't have to be finite you could use
a1=1 and d1=2
a2=2 and d2=4
a3=4 and d3=8
a4=8 and d4=16
a5=16 and d5=32
etc
But at the moment I don't see how it could work for a finite number of partitions (where k > 1). It seems to me some of the sets will allways overlap..
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towr
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Re: Arithmetic progressions and partitions
« Reply #2 on: Mar 25th, 2003, 1:02pm » |
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how about another trivial case,
S1 = {1} (a1 = 1, d1 = 0)
S2 = {2,3,4 ..} (a2 = 2, d2 = 1)
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Pietro K.C.
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Re: Arithmetic progressions and partitions
« Reply #3 on: Mar 25th, 2003, 2:06pm » |
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OK, sorry, I forgot to mention the di are nonzero. I'm going to change my post. But I really mean this problem; aside from the trivial cases, it is very interesting, and has a beautiful solution.
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Icarus
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Re: Arithmetic progressions and partitions
« Reply #4 on: Mar 25th, 2003, 6:40pm » |
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So far every time I look at this I get to the same point, and then run out of time before I can make any definite improvement. I'm posting this much simply so I don't have look it up again:
Because of the Chinese Remainder Theorem, no pair di, dk can be relatively prime (otherwise, the congruences x=ai mod di and x=ak mod dk have mutual solutions, some of which will occur in both Si and Sk).
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Eigenray
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Re: Arithmetic progressions and partitions
« Reply #5 on: Apr 24th, 2003, 10:10pm » |
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No, they cannot.
Say S_i = { ai, ai + di, ai + 2di, ... }.
If the di's are all distinct and positive, we may assume
1 < d1 < d2 < . . . < dk.
If N is the disjoint union of S_1,...S_k, we have:
sum_{n \in N} x^n = sum_{i=1}^{k} sum_{n \in S_i} x^n
(*) x/(1-x) = sum_{i=1}^{k} x^ai/(1-x^di)
But as x->e^(2pi*i/dk), the function
x^ai/(1-x^di)
remains bounded unless dk | di, which happens only when i=k.
Thus the left side of (*) remains bounded, while the right side has a pole; therefore they cannot be equal.
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Hippo
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Re: Arithmetic progressions and partitions
« Reply #6 on: Nov 25th, 2007, 2:17am » |
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on Apr 24th, 2003, 10:10pm, Eigenray wrote:No, they cannot.
Say S_i = { ai, ai + di, ai + 2di, ... }.
If the di's are all distinct and positive, we may assume
1 < d1 < d2 < . . . < dk.
If N is the disjoint union of S_1,...S_k, we have:
sum_{n \in N} x^n = sum_{i=1}^{k} sum_{n \in S_i} x^n
(*) x/(1-x) = sum_{i=1}^{k} x^ai/(1-x^di)
But as x->e^(2pi*i/dk), the function
x^ai/(1-x^di)
remains bounded unless dk | di, which happens only when i=k.
Thus the left side of (*) remains bounded, while the right side has a pole; therefore they cannot be equal. |
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Soory for opening this ancient topic, but I cannot resist: It may be confusing when you use natural number i in the context e^{i phi}. 
BTW: nice reasoning
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