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Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ lots of $n$-bonacci numbers. It can even be done in a way that all representations obtained in the process contain only distinct numbers (in line with the product in current MO question).

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded. To this end, consider the modified product (where $y$ registers number of products contributing to $x^k$) \begin{align*} \prod_{i\geq3}(1+y\,x^{T_i}) &=1+yx+yx^2+\cdots+(y^5+y^3+y)x^{81}+(y^6+y^4+y^2)x^{82}+\cdots \end{align*} in which either the powers of $y$ are all of the same parity (e.g. coefficients of $x^{81}$ or $x^{82}$).

For even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.) By contrast, take the case $n=4$ and let $J_k$ be given by $\sum_{k\geq0}J_kx^k=\frac{x^3}{1-x-x^2-x^3-x^4}$. The corresponding product \begin{align*} \prod_{i\geq4}(1+y\,x^{J_i}) &=1+yx+yx^2+\cdots+(y^7+y^4+y)x^{208}+(y^5+y^2)x^{209}+\cdots \end{align*} shows that the powers of $y$ can take up any parity.

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ lots of $n$-bonacci numbers. It can even be done in a way that all representations obtained in the process contain only distinct numbers (in line with the product in current MO question).

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded. To this end, consider the modified product (where $y$ registers number of products contributing to $x^k$) \begin{align*} \prod_{i\geq3}(1+y\,x^{T_i}) &=1+yx+yx^2+\cdots+(y^5+y^3+y)x^{81}+(y^6+y^4+y^2)x^{82}+\cdots \end{align*} in which either the powers of $y$ are all of the same parity (e.g. coefficients of $x^{81}$ or $x^{82}$).

For even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.) By contrast, take the case $n=4$ and let $J_k$ be given by $\sum_{k\geq0}J_kx^k=\frac{x^3}{1-x-x^2-x^3-x^4}$. The corresponding product \begin{align*} \prod_{i\geq4}(1+y\,x^{J_i}) &=1+yx+yx^2+\cdots+(y^7+y^4+y)x^{208}+(y^5+y^2)x^{209}+\cdots \end{align*} shows that the powers of $y$ can take up any parity.

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obbtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ ones. It can even be done in a way that all representations obtained in the process contain only distinct numbers.

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded.

FFor even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.)

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ lots of $n$-bonacci numbers. It can even be done in a way that all representations obtained in the process contain only distinct numbers (in line with the product in current MO question).

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded. To this end, consider the modified product (where $y$ registers number of products contributing to $x^k$) \begin{align*} \prod_{i\geq3}(1+y\,x^{T_i}) &=1+yx+yx^2+\cdots+(y^5+y^3+y)x^{81}+(y^6+y^4+y^2)x^{82}+\cdots \end{align*} in which either the powers of $y$ are all of the same parity (e.g. coefficients of $x^{81}$ or $x^{82}$).

For even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.) By contrast, take the case $n=4$ and let $J_k$ be given by $\sum_{k\geq0}J_kx^k=\frac{x^3}{1-x-x^2-x^3-x^4}$. The corresponding product \begin{align*} \prod_{i\geq4}(1+y\,x^{J_i}) &=1+yx+yx^2+\cdots+(y^7+y^4+y)x^{208}+(y^5+y^2)x^{209}+\cdots \end{align*} shows that the powers of $y$ can take up any parity.

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obbtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ ones. It can even be done in a way that all representations obtained in the process contain only distinct numbers.

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded.

FFor even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.)

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obbtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ ones. It can even be done in a way that all representations obtained in the process contain only distinct numbers.

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded.

FFor even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.)

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$