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Max Alekseyev
Max Alekseyev
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First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, with the help of Lemma 4.7 for $p=2$ in this paper we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, with the help of Lemma 4.7 for $p=2$ in this paper we have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

more accurate
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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{n}}\pmod{2^{t+2}},$$$$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

First notice that $$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$

Computing modulo $8$, we have $$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$ Correspondingly, $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$ For $k=2^s m$ with $s\geq 0$ and odd $m$, we have $$a_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$ where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu_2(a_k)$.

Similarly, we have $$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$ and $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$ And again for $k=2^sm$, we have $$b_k = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$ proving the result for $\nu_2(b_k)$.

For the ADDED part, we quite similarly have $$\prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum_{n\geq 1} \frac{q^n}{1-q^{n}}\pmod{2^{t+2}},$$ where $t:=\nu_2(r)$. Then $$a_k^{(r)} = [q^k]\ \prod_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$ implying that $\nu_2(a_k^{(r)}) = \nu_2(2r) = t+1$ when $k$ is a square.

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Max Alekseyev
Max Alekseyev
  • 34.3k
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  • 152
added 12 characters in body
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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152
Source Link
Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152