Following is an experimental math claim.

We denote $(a,b)=\gcd(a,b)$.

Let $$G(a)=\sum_{i=1}^{a-1}(-1)^i(a,i).$$


$$ G(a) = \begin{cases} 0, & \text{if $a\equiv 1\pmod4$} \\ \text{odd}, & \text{if $a\equiv 2\pmod4$} \\ 0, & \text{if $a\equiv 3\pmod4$} \\ \text{even}, & \text{if $a\equiv 0\pmod4$.} \end{cases}$$

Can it be shown that for every $a\in\mathbb{Z}_{\ge2}$, $G(a)\ge-1$?


$$\begin{array}{|c |c |} \hline a & G(a) \\ \hline 2 & -1 \\ \hline 3 & 0 \\ \hline 4 &0 \\ \hline 5 &0 \\ \hline 6 &-1 \\ \hline 7 &0 \\ \hline 8 &4 \\ \hline 9 &0 \\ \hline \vdots &\vdots \\ \hline \end{array}$$

source code PARI/GP

for(a=1,10000,if(sum(i=1,a-1, (-1)^i*gcd(a, i))<-1,print([a])))
  • $\begingroup$ This is somewhat like oeis.org/A106475 "An alternating sum of greatest common divisors". $\endgroup$ May 16 at 0:56
  • 2
    $\begingroup$ Also, $-(G(a) + (-1)^aa)$ gives oeis.org/A199084 $\endgroup$ May 16 at 14:23
  • 1
    $\begingroup$ Added this sequence as oeis.org/A344373 along with two related sequences. $\endgroup$ May 16 at 15:06
  • 1
    $\begingroup$ This question raises several further questions about iterates (which are probably very hard). For example, is 12 the unique $a$ with $G(G(G(a)))=0$, $G(G(a))\ne0$? Or, there seem to be many $a$ with $G(a)$, $G(G(a))$, $G(G(G(a)))$, ... never stabilizing. Can one actually prove that there is some such? In fact, it seems that $-1$, $0$ and $16$ are the only sinks. Is this so? Are there any periodic orbits of iterates? There also seem to be many $x$ with $G^{-1}(x)$ empty. Can one characterize those? $\endgroup$ May 16 at 16:02

Yes, it is true. Further we suppose that $a=2b$ is even, as for odd $a$ the sum $G(a)$ equals 0 due to pairing $\{i,a-i\}$.

We start with $$(a,i)=\sum_{d|(a,i)}\varphi(d)=\sum_{d|a} \varphi(d)\cdot \mathbf{1}_{d|i}.$$ Therefore $$ G(a)=\sum_{i=1}^{a-1}(-1)^i(a,i)=\sum_{d|a}\varphi(d)\sum_{i=1}^{a-1}(-1)^i \mathbf{1}_{d|i}. $$ The inner sum $\sum_{i=1}^{a-1}(-1)^i \mathbf{1}_{d|i}$ equals $a/d-1$ if $d$ is even and equals $-1$ if $d$ is odd (since it is a sum of $a/d-1=2b/d-1$ alternating $\pm 1$'s). Thus $$G(a)=\sum_{d|a,d\,\text{is even}} \varphi(d)(a/d-1)-\sum_{d|a,d\,\text{is odd}}\varphi(d).$$ Denote $b=2^sc$ for non-negative integer $s$ and odd $c$. Then odd divisors of $a$ are exactly divisors of $c$, and we have $$ \sum_{d|a,d\,\text{is odd}}\varphi(d)=\sum_{d|c}\varphi(d)=c. $$ If 4 divides $a$, then $a/(2d)-1\geqslant 1$ for each divisor $d$ of $c$, and we get $$ \sum_{d|a,d\,\text{is even}} \varphi(d)(a/d-1)\geqslant \sum_{d|c} \varphi(2d)(a/(2d)-1)\geqslant \sum_{d|c} \varphi(2d)=\sum_{d|c} \varphi(d)=c, $$ thus $G(a)\geqslant 0$. Finally, it remains to consider the case when $b=c$ is odd. Then we have $$ \sum_{d|a,d\,\text{is even}} \varphi(d)(a/d-1)=\sum_{d|c} \varphi(2d)(a/(2d)-1)= \sum_{d|c}\varphi(d)(c/d-1)=-c+c\sum_{d|c} \varphi(d)/d, $$ and $G(a)\geqslant -1$ reads as $$c\sum_{d|c} \varphi(d)/d\geqslant 2c-1.$$ This may be proved by $$ \sum_{d|c} \varphi(d)/d=1+\sum_{d|c,d>1} \varphi(d)/d\geqslant 1+ \sum_{d|c,d>1} \varphi(d)/c=1+(c-1)/c=(2c-1)/c. $$

  • $\begingroup$ Thanks for answer. please explain, what this ${\bf 1}_{d|i}$ represent ? $\endgroup$
    – Pruthviraj
    May 16 at 4:11
  • 2
    $\begingroup$ That is the indicator function: 1 if $d$ divides $i$ and 0 otherwise. $\endgroup$ May 16 at 4:35

$$G(a)+(-1)^aa=\sum_{i=1}^{a} (-1)^i (a,i)=\sum_{d|a} d \sum_{\substack{ l \\ (l, a/d)=1}} (-1)^{dl}=T(a).$$

When, $a$ is odd, $d, a/d$ is also odd. Hence, $$\sum_{\substack{l, \\ (l,a/d)=1 \\ d<a}} (-1)^l =0.$$ So, $T(a)=(-1)^aa \rightarrow G(a)=0$.

Now, for $a=2^{r_0}m, m=\prod_{i=1}^s p_i^{r_i}$ odd, $$ T(a)=\sum_{\text{$d$ even}} d\phi(a/d) - \sum_{\text{$d$ odd}} d\phi(a/d) $$ (because, for $d$ even $(-1)^d=1$, and for $d$ odd, $a/d$ even, so $l, dl$ is odd). \begin{align*} & {}=a\sum_{l\mid\frac{a}{2}} \frac{\phi(l)}{l}-\sum_{l \mid m}\frac{\phi(2^rl)}{2^rl} \\ & {}=a(\sum_{l\mid\frac{a}{2}} \frac{\phi(l)}{l}-\frac{1}{2}\sum_{l|m}\frac{\phi(l)}{l}). \end{align*} Now, $\sum_{d\mid n} \frac{\phi(d)}{d}=\prod_{i} (1+(1-\frac{1}{p_i})r_i)$, where $n=\prod_i p_i^{r_i}$.

Using this to the sum we get, the explicit formula for $G(a)$: $$ G(a)=\frac{ar_0}{2p_1p_2\dotsm p_s}\prod_{i=1}^{s} (p_i(r_i+1)-r_i) -a. $$

It's easy to see from the formula that $G(a)$ is odd when $r_0=1$ and even when $r_0>1$, because all $p_i$ are odd, hence each $p_i(r_i+1)-r_i$ is odd, and $a$ is even.

More compactly, $G(a)=a[\frac{r_0}{2}\prod_{i=1}^{s} (r_i+1-\frac{r_i}{p_i})-1]$.

Surely, $G(a) \geq -1$. It's $-1$, for $a=2p$ types, e.g. $a=6$.


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