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The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\gcd(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.))

2. If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$

3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.

The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\gcd(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.))

2. If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$

3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.

The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\mbox{gcd}(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.))

2. If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$

3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.

The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\gcd(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.))

2. If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$

3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.

The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\mbox{gcd}(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$
2. If $n$ is an odd number not of the form "prime(k)+phi(prime(k))" (see A166257), then $s_n < 0$
3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.

The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious).

Consider the sum: $$s_n := \sum_{i=1}^n (-1)^{i-1}\mbox{gcd}(n+1,i).$$

Simple numerical experiments suggest the following:

1. If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.))

2. If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$

3. otherwise, of course, $s_n=0$.

How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations.