# [seqfan] Completeness of A278373

David Wilson davidwwilson at comcast.net
Thu Dec 1 05:19:53 CET 2016

```I submitted sequence A278373 of numbers of the form f(n) =
sigma(n)+phi(n)-2n.

I have also submitted companion sequence A278374 with f(A278374(n)) =
A278373(n), which can be used to verify that all elements in A278373 are
indeed of the specified form.

For each I have submitted a b-file of 10000 elements.

I have also submitted a 10000-element b-file for A056996, the complement of
A278373.

I believe approval of all of these sequences is being held up by a question
of the completeness of A278373, asked by Charles Greathouse in a discussion
note on that sequence. Specifically, Greathouse asks if there might be
additional odd values that belong in A278373.

The answer is, I do not believe so.

You can easily show that if f(n) = sigma(n)+phi(n)-2n is odd, then n must be
of the form k^2 or 2k^2.

Numerical evidence strongly suggests the following:

If k^2 > p^4 for prime p, then either k is prime and f(k^2) = 1, or else
f(k^2) > f(p^4) = p^2+p+1.
If 2k^2 > p^4 for prime p, then f(2k^2) > f(p^4) = p^2+p+1.

I'm convinced these statements are true, they look susceptible to proof by
inductive arguments over the prime signatures of k^2 or 2k^2.

If this is indeed correct, then we need only compute all f(k^2) and f(2k^2)
for all arguments <= p^4 to obtain all odd values of f <= f(p^4) = p^2+p+1.
Computing f(n) for n > p^4 will not produce odd values < p^2+p+1. For
example, if we compute all values of f(k^2) with k^2 <= 101^4 and f(2k^2)
with 2k^2 <= 101^4, we obtain odd all values of f <= 101^2+101+1 = 10303.
This is how I computed the b-files for A278373 and A056996.

```