[seqfan] Re: A134675 = A007434 + A001065

Max Alekseyev maxale at gmail.com
Wed Jan 7 17:23:39 CET 2015


This identity is true and easily follows from the following formula
for T=A134673:
T(n,k)=k*mu(n/k)+1 if k divides n and k<n, T(n,k)=n for k=n, and
T(n,k)=0 otherwise.

We have:
A134674(n,k) = \sum_{j=n+1-k}^n T(n,j)
and
A134675(n) = \sum_{k=1}^n A134674(n,k) = \sum_{k=1}^n \sum_{j=n+1-k}^n T(n,j)
= \sum_{j=1}^n j*T(n,j) = \sum_{j|n, j<n} j*(j*mu(n/j)+1) + n^2
= \sum_{j|n} j^2*mu(n/j) + (sigma(n) - n) = A007434(n) + A001065(n).

Regards,
Max



On Wed, Jan 7, 2015 at 4:20 PM,  <john.mason at lispa.it> wrote:
> Dear friends
> I noticed that (at least for the first 1000 terms), A134675 = A007434 +
> A001065, though A134675 makes no mention of the other two.
>
> Definitions:
> A007434 : Jordan function J_2(n) (a generalization of phi(n)).
> A001065 : Sum of proper divisors (or aliquot parts) of n: sum of divisors
> of n that are less than n
> A134675 : Row sums of triangle A134674
> (where A134674 = A134673 * A000012,
>         and A134673 = A051731 + A127448 - I, I = Identity matrix,
>                 and so on).
>
> I hesitate in editing the "formula" section of A134675, as I have no idea
> WHY the addition works.
>
> Is OEIS interested in me finding other non-cross-referenced sums?
>
> john
>
>
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>
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