# [seqfan] Re: : The alternating sum in A196020

Max Alekseyev maxale at gmail.com
Sun Nov 17 21:15:40 CET 2013

Here is the proof for this conjecture for A196020.

First, we have the following explicit formula:
If n==k/2 (mod k) and n>=k(k+1)/2, T(n,k) = 2*n/k - k;
otherwise T(n,k) = 0.

Alternatively, we can say that if n==k/2 (mod k) , T(n,k) = f(2*n/k - k),
where f(x) = (x+abs(x)) /2.
(in other words, f(x) = x if x >= 0, and f(x) = 0 if x < 0)

Let n = 2^t*s, where s is odd.
Then n==k/2 (mod k) is equivalent to k = m or k = 2^(t+1)*m, where m divides s.

Now, the alternating sum for row n is

\sum_{m|s} f(2^(t+1)*s/m - m) - f(s/m - 2^(t+1)*m)
= \sum_{p*q=s} f(2^(t+1)*q - p) - f(q - 2^(t+1)*p)
= \sum_{p*q=s} f(2^(t+1)*q - p) - f(p - 2^(t+1)*q)
= \sum_{p*q=s} 2^(t+1)*q - p
(here we used the property that f(x) - f(-x) = x for any x)
= 2^(t+1)*sigma(s) - sigma(s) = (2^(t+1)-1)*sigma(s)
= sigma(n).

Regards,
Max

On Sat, Nov 16, 2013 at 12:10 PM, Omar E. Pol <info at polprimos.com> wrote:
> Dear Professors,
>
> The entry A196020 contains a conjecture:
> It appears that the alternating sum of row n equals the sum of divisors of n.
> Does anybody knows how to prove o disprove this conjecture?
>
> https://oeis.org/A196020
> https://oeis.org/A000203
>
> Best regards.
>
> Omar E. Pol
>
>
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