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I have a problem that boils down to prove that the succession $\{a_n\}$ tends to infinity, where $$a_n:=1+\sum_{0\leq j<n}D_{2j+1}(n-j)$$ and $D_{m}(n)$ is the number of divisors $d>1$ of $n$ such that $d\equiv 1 \mod{m}.$

First question is how to prove or disprove that $a_n\to \infty.$

As a matter of fact, numerical computations lead me to conjecture that, for example, $a_n\geq \lfloor \log(n)\rfloor.$

Notice that, $a_n\geq d(n),$ the number of divisors (the summand corresponding to $j=0$). If $n-j$ is itself equivalent to 1 $\mod{2j+1},$ $n-j=t(2j+1)+1$ which implies $2n-1=(2t+1)(2j+1).$ Hence $a_n\geq d(2n-1).$ If $n-j$ has $(2j+1)+1$ as a divisor, $n-j=(2j+2)c$ which implies $n+1=(2c+1)(j+1).$ Hence $a_n\geq d_o(n+1),$ the number of odd divisors of $n+1.$ Thus we have $$a_n\geq \max(d(n),d(2n-1),d_o(n+1)).$$ The previous inequalities shows that $a_n$ is ``big'' for almost all $n,$ but it is not possible to conclude that it tends to infinity.

I tried to prove the fact that $a_n\to \infty$ using the estimates given in the paper ``Equidistribution of Divisors and Representations by Binary Quadratic Forms'' by Drmota and Skalba. In this paper the authors prove that for almost all $n,$ if $(n,m)=1,$
$D_{m}(n)\approx \frac{1}{\phi(m)}d(n),$ where $d(n)$ is the number of divisors of $n.$ However, I wasn't able to get anything from this result because, in my sum, there is a dependency between $m$ and $n$ and the cited result is true if $m$ is fixed. Moreover my sum contains terms in which $(n,m)\neq 1.$

To be more precise, the original problem was to prove that the following Diophantine equation $$2xyz+yz+x+z=n$$ in the unknowns $x\geq 0,\, y\geq 1\, z\geq 1$ and with $n\in \mathbb N$ has a number of solutions that tends to infinity when $n$ increases. It is easy to show this problem is equivalent to the fact that $a_n \to \infty.$

Second question: is it more easy to deal with the original problem?

I have a problem that boils down to prove that the succession $\{a_n\}$ tends to infinity, where $$a_n:=1+\sum_{0\leq j<n}D_{2j+1}(n-j)$$ and $D_{m}(n)$ is the number of divisors $d>1$ of $n$ such that $d\equiv 1 \mod{m}.$

First question is how to prove or disprove that $a_n\to \infty.$

As a matter of fact, numerical computations lead me to conjecture that, for example, $a_n\geq \lfloor \log(n)\rfloor-1.$

Notice that, $a_n\geq d(n),$ the number of divisors (the summand corresponding to $j=0$). If $n-j$ is itself equivalent to 1 $\mod{2j+1},$ $n-j=t(2j+1)+1$ which implies $2n-1=(2t+1)(2j+1).$ Hence $a_n\geq d(2n-1).$ If $n-j$ has $(2j+1)+1$ as a divisor, $n-j=(2j+2)c$ which implies $n+1=(2c+1)(j+1).$ Hence $a_n\geq d_o(n+1),$ the number of odd divisors of $n+1.$ Thus we have $$a_n\geq \max(d(n),d(2n-1),d_o(n+1)).$$ The previous inequalities shows that $a_n$ is ``big'' for almost all $n,$ but it is not possible to conclude that it tends to infinity.

I tried to prove the fact that $a_n\to \infty$ using the estimates given in the paper ``Equidistribution of Divisors and Representations by Binary Quadratic Forms'' by Drmota and Skalba. In this paper the authors prove that for almost all $n,$ if $(n,m)=1,$
$D_{m}(n)\approx \frac{1}{\phi(m)}d(n),$ where $d(n)$ is the number of divisors of $n.$ However, I wasn't able to get anything from this result because, in my sum, there is a dependency between $m$ and $n$ and the cited result is true if $m$ is fixed. Moreover my sum contains terms in which $(n,m)\neq 1.$

To be more precise, the original problem was to prove that the following Diophantine equation $$2xyz+yz+x+z=n$$ in the unknowns $x\geq 0,\, y\geq 1\, z\geq 1$ and with $n\in \mathbb N$ has a number of solutions that tends to infinity when $n$ increases. It is easy to show this problem is equivalent to the fact that $a_n \to \infty.$

Second question: is it more easy to deal with the original problem?

I have a problem that boils down to prove that the succession $\{a_n\}$ tends to infinity, where $$a_n:=\sum_{0\leq j<n}D_{2j+1}(n-j)$$ and $D_{m}(n)$ is the number of (positive) divisors $d$ of $n$ such that $d\equiv 1 \mod{m}.$

First question is how to prove or disprove that $a_n\to \infty.$

As a matter of fact, numerical computations lead me to conjecture that, for example, $a_n\geq \lfloor \log(n)\rfloor.$

Notice that, $a_n\geq d(n),$ the number of divisors (the summand corresponding to $j=0$). If $n-j$ is itself equivalent to 1 $\mod{2j+1},$ $n-j=t(2j+1)+1$ which implies $2n-1=(2t+1)(2j+1).$ Hence $a_n\geq d(2n-1).$ If $n-j$ has $(2j+1)+1$ as a divisor, $n-j=(2j+2)c$ which implies $n+1=(2c+1)(j+1).$ Hence $a_n\geq d_o(n+1),$ the number of odd divisors of $n+1.$ Thus we have $$a_n\geq \max(d(n),d(2n-1),d_o(n+1)).$$ The previous inequalities shows that $a_n$ is ``big'' for almost all $n,$ but it is not possible to conclude that it tends to infinity.

I tried to prove the fact that $a_n\to \infty$ using the estimates given in the paper ``Equidistribution of Divisors and Representations by Binary Quadratic Forms'' by Drmota and Skalba. In this paper the authors prove that for almost all $n,$ if $(n,m)=1,$
$D_{m}(n)\approx \frac{1}{\phi(m)}d(n),$ where $d(n)$ is the number of divisors of $n.$ However, I wasn't able to get anything from this result because, in my sum, there is a dependency between $m$ and $n$ and the cited result is true if $m$ is fixed. Moreover my sum contains terms in which $(n,m)\neq 1.$

To be more precise, the original problem was to prove that the following Diophantine equation $$2xyz+yz+x+z=n$$ in the unknowns $x\geq 0,\, y\geq 1\, z\geq 1$ and with $n\in \mathbb N$ has a number of solutions that tends to infinity when $n$ increases. It is easy to show this problem is equivalent to the fact that $a_n \to \infty.$

Second question: is it more easy to deal with the original problem?

I have a problem that boils down to prove that the succession $\{a_n\}$ tends to infinity, where $$a_n:=1+\sum_{0\leq j<n}D_{2j+1}(n-j)$$ and $D_{m}(n)$ is the number of divisors $d>1$ of $n$ such that $d\equiv 1 \mod{m}.$

First question is how to prove or disprove that $a_n\to \infty.$

As a matter of fact, numerical computations lead me to conjecture that, for example, $a_n\geq \lfloor \log(n)\rfloor.$

Notice that, $a_n\geq d(n),$ the number of divisors (the summand corresponding to $j=0$). If $n-j$ is itself equivalent to 1 $\mod{2j+1},$ $n-j=t(2j+1)+1$ which implies $2n-1=(2t+1)(2j+1).$ Hence $a_n\geq d(2n-1).$ If $n-j$ has $(2j+1)+1$ as a divisor, $n-j=(2j+2)c$ which implies $n+1=(2c+1)(j+1).$ Hence $a_n\geq d_o(n+1),$ the number of odd divisors of $n+1.$ Thus we have $$a_n\geq \max(d(n),d(2n-1),d_o(n+1)).$$ The previous inequalities shows that $a_n$ is ``big'' for almost all $n,$ but it is not possible to conclude that it tends to infinity.

I tried to prove the fact that $a_n\to \infty$ using the estimates given in the paper ``Equidistribution of Divisors and Representations by Binary Quadratic Forms'' by Drmota and Skalba. In this paper the authors prove that for almost all $n,$ if $(n,m)=1,$
$D_{m}(n)\approx \frac{1}{\phi(m)}d(n),$ where $d(n)$ is the number of divisors of $n.$ However, I wasn't able to get anything from this result because, in my sum, there is a dependency between $m$ and $n$ and the cited result is true if $m$ is fixed. Moreover my sum contains terms in which $(n,m)\neq 1.$

To be more precise, the original problem was to prove that the following Diophantine equation $$2xyz+yz+x+z=n$$ in the unknowns $x\geq 0,\, y\geq 1\, z\geq 1$ and with $n\in \mathbb N$ has a number of solutions that tends to infinity when $n$ increases. It is easy to show this problem is equivalent to the fact that $a_n \to \infty.$

Second question: is it more easy to deal with the original problem?