[seqfan] Re: n consecutive perfect powers sum to a perfect power
israel at math.ubc.ca
israel at math.ubc.ca
Wed Dec 13 16:19:47 CET 2023
Here's one idea. Most perfect powers are squares. The sum of n squares
starting at x^2 is a quadratic in x:
S_n(x) = Sum_{i=0..n-1} (x+i)^2 = n*x^2 + n*(n-1)*x + n*(2*n-1)*(n-1)/6
and the diophantine equation S_n(x) = y^2 can be solved by standard
methods. S_n(x) = y^3 should be an elliptic curve, and maybe its integral
points can be found as well, but I'm not an expert on that. The remaining
cases to consider have one or more of the consecutive perfect powers or the
sum a higher power (cube or more), and it should be easier to search the
higher powers than to search using all perfect powers.
Cheers,
Robert
On Dec 12 2023, jnthn stdhr wrote:
>Howdy, all.
>
> What is the least perfect power m (in A001597) that is the sum of n
>consecutive perfect powers.
>
> The sequence isn't in the database, and begins 1, 25, 441, 100, 169, 289,
>121, 2395417249, -1, -1, 676, 232324, -1, -1, -1, 64866916, 3721,
>3622354596, 279936, ..., with -1 representing no solution found up to
>~10^10.
>
> For the first few terms, we have:
>
>{1}=1, {9+16}=25, {128+144+169}=441, {16+25+27+32}=100, etc.
>
> Should I add this? If so, up to a(8) only, or include the -1s?
>
> When I first searched for solutions, the maximum value of the set of
>perfect powers was ~10^8, and both a(8) and a(18) came out -1. But when I
>increased the max to 10^10 solutions for those two terms were found. At
>10^8 I was able to get to 100+ terms in a somewhat reasonable time, with
>solutions becoming more and more sparse. At 10^10, things get very bogged
>down, but more solutions are found along the way. Also, the erratic nature
>of the terms seems to persist.
>
> Can a solution always be found if the set of perfect powers is large
>enough?
>
>-Jonathan
>
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>
>
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