Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?

Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.

The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.

If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.

The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps: $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$ If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.

This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form: $$n = x_0^2 + 2x_1^2 + 8 x_3^2 \; , \; \; \mbox{with} \; \; x_0, x_1, x_3 \geq 0 \;?$$ Some play with the parities of binomial coefficients reduces this to the question I led with.

Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?

Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.

The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.

If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.

The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps: $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$ If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.

This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form: $$n = x_0^2 + 2x_1^2 + 8 x_3^2 \; , \; \; \mbox{with} \; \; x_0, x_1, x_3 \geq 0 \;?$$ Some play with the parities of binomial coefficients reduces this to the question I led with.

Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?

Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.

The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.

If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.

The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps: $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$ If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.

This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form: $$n = x_0^2 + 2x_1^2 + 8 x_3^2?$$ Some play with the parities of binomial coefficients reduces this to the question I led with.

Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?

Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.

The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.

If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.

The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps: $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$ If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.

This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form: $$n = x_0^2 + 2x_1^2 + 8 x_3^2 \; , \; \; \mbox{with} \; \; x_0, x_1, x_3 \geq 0 \;?$$ Some play with the parities of binomial coefficients reduces this to the question I led with.