[seqfan] Re: Sum( floor( k^2/n ), k=1..n)

[israel at math.ubc.ca]

Oops, silly of me: this alone doesn't do it.  We also need formulas
for a(4m), a(4m+1) and a(4m+3), which don't look so easy.

Cheers,
Robert

On Aug 8 2013, israel at math.ubc.ca wrote:

>It looks to me like 
>a(4m+2) = a(2m+1) + (2m+1)^2
>which should allow computation of A(n) in something like log_2(n) steps.
>
>Cheers,
>Robert
>
>
>On Aug 7 2013, Charles Greathouse wrote:
>
>>A014817 is defined as in the title. You can also sum in the opposite
>>direction:
>>a(n) = n^2 - sum_{m=1..n} floor(sqrt(n*m-1))
>>
>>But neither of these are efficient, taking n calculations (exponential in
>>the length of n). Is there a better way to compute this sum (and the
>>related ones in A166387, A166375, and A165993)? I keep having images of
>>Dirichlet's hyperbola method but can't quite work out how to do it
>>efficiently here.
>>
>>Maybe it's not hard and I'm just not looking at it right?
>>
>>Charles Greathouse
>>Analyst/Programmer
>>Case Western Reserve University
>>
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>>
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>>
>>
>