[seqfan] Re: A062682

israel at math.ubc.ca israel at math.ubc.ca
Mon Nov 19 09:06:44 CET 2012


Here's a very rough heuristic analysis.  You may take it with a grain of
salt.  But it doesn't support the idea that cubics in general should have
values with arbitrarily large numbers of solutions. 

Suppose for each (x,y) we randomly and independently choose an integer in 
[0,max(x,y)^d] as P(x,y). Then the probability that P(x,y) = z is 
1/max(x,y)^d if max(x,y)^d >= z. There are approximately 2n pairs with 
max(x,y) = n, so the expected number of "hits" on z is 
sum_{n=z^(1/d)}^infty 2n/n^d. For d=2 this is infinity: for each z, 
P(x,y)=z would be expected to have infinitely many solutions. Otherwise it 
is of order z^(-(d-2)/d). Now if we imagine a Poisson process with mean 
lambda ~= z^(-(d-2)/d) for the number of hits, the probability of k or more 
hits is on the order of lambda^k. In the case d=3, lambda ~= z^(-1/3). 
Since sum_z z^(-k/3) is infinity for k <= 3 but is finite for k > 3, this 
would imply that there are infinitely many z with 3 hits, but only finitely 
many with 4 or more hits. In the case d=4, lambda ~= z^(-1/2), you would 
have infinitely many with 2 hits, but only finitely many with 3 or more. 
And for d>=5, there are only finitely many with 2 or more.


Robert Israel
University of British Columbia


On Nov 18 2012, David Wilson wrote:

>For cubic or lower degree polynomials, p(x, y) = c in general seems to 
>have an unbounded number of solutions, e.g. there are c for which a^3 + 
>b^3 = c has an arbitrarily large number of solutions, vis A011541.
>
>For quartic or higher polynomials, this does not appear to be the case. 
>For example, a^4 + b^4 = c seems to have at most two distinct solutions 
>in a, b for any c, vis A018768.  At the time I computed A018768, I 
>computed it way  beyond the published values, but found no values with 
>more than two representations. To my knowledge, no one has ever proved 
>that there cannot be three representations, but no one has ever found an 
>example.
>
>sum(a..b, i^3) = p(b) - p(a-1) where p(n) = (n*(n+1)/2)^2, a quartic.  
>Thus I might expect sum(a..b, i^3) = c to have a bounded number of 
>distinct solutions.
>
>On 11/18/2012 11:02 AM, Uwe Lauth wrote:
>> I wrote a C program to find more missing numbers
>> n = sum(i=a..b i^3) = sum(i=a'..b' i^3) where b'>b
>> Up to b=2450, nothing new was found.
>>
>> All numbers I found had only two different ways of this sum.
>> What is the first number that has three sums?
>>
>> Uwe
>>
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