show/hide this revision's text 3 strengthened assumptions about the function $f(n)$ and $g(n)$

An additive function $f(n)$ is said to be totally/completely additive if

$$f(ab) = f(a) + f(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if $f$ is a totally/completely additive function, then $f(1) = 0$.

On the other hand, a multiplicative function $g(n)$ is said to be totally/completely multiplicative if

$$g(ab) = g(a)g(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if g is a totally/completely multiplicative function, then $g(1) = 1$.

So now suppose there is a function $h(n)$ such that

$$h(n) = \frac{1}{2}[f(n) + g(n)]$$

where $f$ is totally additive and $g$ is totally multiplicative. (I removed the erroneous assumption on the function $h(n)$, thank you Gerry for pointing that out!)

Then

$$h(1) = \frac{1}{2}(f(1) + g(1)) = \frac{1}{2}$$

Now, let

$$i(n) = h(n + 1)$$

so that

$$i(0) = h(1) = \frac{1}{2}$$

My question is:

How do we then characterize such number-theoretic functions $i(n)$?

    Post Closed as "too localized" by Denis Serre, Chandan Singh Dalawat, Kevin Buzzard, Pete L. Clark, S. Carnahan

show/hide this revision's text 2 removed an erroneous assumption on the function h(n)

An additive function $f(n)$ is said to be totally/completely additive if

$$f(ab) = f(a) + f(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if $f$ is a totally/completely additive function, then $f(1) = 0$.

On the other hand, a multiplicative function $g(n)$ is said to be totally/completely multiplicative if

$$g(ab) = g(a)g(b)$$

holds for all positive integers $a$ and $b$.

Additionally, if g is a totally/completely multiplicative function, then $g(1) = 1$.

So now suppose there is a function $h(n)$ such that

$$h(n) = \frac{1}{2}[f(n) + g(n)]$$

where $f$ is additive and $g$ is multiplicative. In addition, assume that $h$ is both (totally/completely) additive AND multiplicative.I removed the erroneous assumption on the function $h(n)$, thank you Gerry for pointing that out!)

Then

$$h(1) = \frac{1}{2}(f(1) + g(1)) = \frac{1}{2}$$

Now, let

$$i(n) = h(n + 1)$$

so that

$$i(0) = h(1) = \frac{1}{2}$$

My question is:

How do we then characterize such number-theoretic functions $i(n)$?

show/hide this revision's text 1