Re: Sum with floor is sum with # divisors function
- From: "Proginoskes" <proginoskes@xxxxxxxxxxxxx>
- Date: 23 May 2005 21:47:27 -0700
Leroy Quet wrote:
> I highly doubt this is not well known.
> I myself probably already posted a generalization
> of it in the far past,
> altough I cannot find anything by doing a search.
>
> In any case, it has been quite a while since I posted
> anything not a game to sci.math,
> so I will post this (easily found if true) identity
> anyway.
>
> If d(m) is the number of positive divisors of m,
> and sigma(m) is the sum of the positive divisors of m,
>
> then
> [non-linear mode version clipped]
> sum{j=1 to n} (2 j d(j) - sigma(j))
> =
> sum{j=1 to n} j (floor(n/j))^2
So what do you want us to do? Prove it? It's not actually hard to
prove:
sum (j=1..n, 2 j d(j))
= sum (j=1..n, 2 j sum (k | j, 1))
= sum (j=1..n, k | j, 2 j)
= sum (k=1..n, m=1..floor(n/k), 2 k m) [write j = k m]
= sum (k=1..n, 2 k sum (m=1..floor(n/k), m))
= sum (k=1..n, k floor(n/k) [floor (n/k) + 1]).
Similarly,
sum (j=1..n, sigma(j))
= sum (j=1..n, sum (k | j, k))
= sum (j=1.., k | j, k)
= sum (k=1..n, m=1..floor(n/k), k)
= sum (k=1..n, k floor(n/k))
Subtracting the first equation from the second yields
sum (j=1..n, 2 j d(j) - sigma(j))
= sum (k=1..n, k [floor(n/k)]^2)
Now, what about sum (k=1..n, k [floor(n/k)]^3)?
--- Christopher Heckman
.
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