MHB Compute $a_{1996}$ for $\prod_{k=1}^{1996} (1+kx^{3^k})$

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The discussion focuses on computing the coefficient \( a_{1996} \) from the product \( \prod_{k=1}^{1996} (1+kx^{3^k}) \). Participants confirm the correctness of an amended solution, acknowledging a previous subtraction error identified by a user named Opalg. The conversation highlights the collaborative nature of problem-solving within the forum, with gratitude expressed for assistance received. The solution is noted as potentially unverified, but participants are confident in its validity. Overall, the thread emphasizes the importance of community support in mathematical problem-solving.
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Let $\displaystyle \prod_{k=1}^{1996} (1+kx^{3^k})=1+a_1x^{n_1}+a_2x^{n_2}+\cdots+a_mx^{n_m}$, where $a_1,\,a_2,\,\cdots$ are nonzero and $n_1<n_2<\cdots<n_m$.

Compute $a_{1996}$.
 
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Note: solution not verified, possible off-by-one error, but it definitely works.

We express $a_n$ as a recurrence by induction on $k$. First consider the base case $k = 1$, which gives us $a_0 = a_1 = 1$. Suppose that for some $k - 1 \in \mathbb{N}$, we have nonzero polynomial coefficients $a_0, a_1, \cdots, a_{n - 1}$ for some $n$, then it follows that:
$$(1 + k x^{3^k}) \sum_{i = 0}^{n - 1} a_i x^{n_i} = \sum_{i = 0}^{n - 1} a_i x^{n_i} + k \sum_{i = 0}^{n - 1} a_i x^{3^k + n_i}$$
We observe that $3^k$ is larger than all exponents $n_i$ which are at most $3^k - 1$, being a sum of powers of $3$ less than $k$, and so the coefficients $a_0$ to $a_{n - 1}$ remain unchanged by the multiplication by $1 + k x^{3^k}$. Furthermore, as a result the polynomial for $k$ must then have $2n$ nonzero terms (twice as many). These two facts together allows us to derive $k$ from $n$ as $k = \lceil \log_2(n + 1) \rceil$ and hence express $a_n$ as a simple recurrence:
$$a_n = k a_{n - 2^{k - 1}} ~ ~ ~ \text{where} ~ k = \lceil \log_2(n + 1) \rceil, a_0 = a_1 = 1$$
So that:
$$a_{1996} = 11 a_{1996 - 1024} = 11 a_{972}$$
$$a_{972} = 10 a_{972 - 512} = 10 a_{460}$$
$$a_{460} = 9 a_{460 - 256} = 9 a_{204}$$
$$a_{204} = 8 a_{204 - 128} = 8 a_{76}$$
$$a_{76} = 7 a_{76 - 64} = 7 a_8$$
$$a_{8} = 4 a_{8 - 8} = 4 a_0$$
And so we conclude that:
$$a_{1996} = 11 \times 10 \times 9 \times 8 \times 7 \times 4 \times a_0 = 221760$$
 
Amended solution:

We express $a_n$ as a recurrence by induction on $k$. First consider the base case $k = 1$, which gives us $a_0 = a_1 = 1$. Suppose that for some $k - 1 \in \mathbb{N}$, we have nonzero polynomial coefficients $a_0, a_1, \cdots, a_{n - 1}$ for some $n$, then it follows that:
$$(1 + k x^{3^k}) \sum_{i = 0}^{n - 1} a_i x^{n_i} = \sum_{i = 0}^{n - 1} a_i x^{n_i} + k \sum_{i = 0}^{n - 1} a_i x^{3^k + n_i}$$
We observe that $3^k$ is larger than all exponents $n_i$ which are at most $3^k - 1$, being a sum of powers of $3$ less than $k$, and so the coefficients $a_0$ to $a_{n - 1}$ remain unchanged by the multiplication by $1 + k x^{3^k}$. Furthermore, as a result the polynomial for $k$ must then have $2n$ nonzero terms (twice as many). These two facts together allows us to derive $k$ from $n$ as $k = \lceil \log_2(n + 1) \rceil$ and hence express $a_n$ as a simple recurrence:
$$a_n = k a_{n - 2^{k - 1}} ~ ~ ~ \text{where} ~ k = \lceil \log_2(n + 1) \rceil, a_0 = a_1 = 1$$
So that:
$$a_{1996} = 11 a_{1996 - 1024} = 11 a_{972}$$
$$a_{972} = 10 a_{972 - 512} = 10 a_{460}$$
$$a_{460} = 9 a_{460 - 256} = 9 a_{204}$$
$$a_{204} = 8 a_{204 - 128} = 8 a_{76}$$
$$a_{76} = 7 a_{76 - 64} = 7 a_{12}$$
$$a_{12} = 4 a_{12 - 8} = 4 a_4$$
$$a_{4} = 3 a_{4 - 4} = 3 a_0$$
And so we conclude that:
$$a_{1996} = 11 \times 10 \times 9 \times 8 \times 7 \times 4 \times 3 \times a_0 = 665280$$
 
Bacterius said:
Amended solution:

We express $a_n$ as a recurrence by induction on $k$. First consider the base case $k = 1$, which gives us $a_0 = a_1 = 1$. Suppose that for some $k - 1 \in \mathbb{N}$, we have nonzero polynomial coefficients $a_0, a_1, \cdots, a_{n - 1}$ for some $n$, then it follows that:
$$(1 + k x^{3^k}) \sum_{i = 0}^{n - 1} a_i x^{n_i} = \sum_{i = 0}^{n - 1} a_i x^{n_i} + k \sum_{i = 0}^{n - 1} a_i x^{3^k + n_i}$$
We observe that $3^k$ is larger than all exponents $n_i$ which are at most $3^k - 1$, being a sum of powers of $3$ less than $k$, and so the coefficients $a_0$ to $a_{n - 1}$ remain unchanged by the multiplication by $1 + k x^{3^k}$. Furthermore, as a result the polynomial for $k$ must then have $2n$ nonzero terms (twice as many). These two facts together allows us to derive $k$ from $n$ as $k = \lceil \log_2(n + 1) \rceil$ and hence express $a_n$ as a simple recurrence:
$$a_n = k a_{n - 2^{k - 1}} ~ ~ ~ \text{where} ~ k = \lceil \log_2(n + 1) \rceil, a_0 = a_1 = 1$$
So that:
$$a_{1996} = 11 a_{1996 - 1024} = 11 a_{972}$$
$$a_{972} = 10 a_{972 - 512} = 10 a_{460}$$
$$a_{460} = 9 a_{460 - 256} = 9 a_{204}$$
$$a_{204} = 8 a_{204 - 128} = 8 a_{76}$$
$$a_{76} = 7 a_{76 - 64} = 7 a_{12}$$
$$a_{12} = 4 a_{12 - 8} = 4 a_4$$
$$a_{4} = 3 a_{4 - 4} = 3 a_0$$
And so we conclude that:
$$a_{1996} = 11 \times 10 \times 9 \times 8 \times 7 \times 4 \times 3 \times a_0 = 665280$$

Thanks for participating, Bacterius! Yes, your amended solution is correct and the subtraction error was spotted by Opalg and so I want to thank Opalg because he has helped me numerous times in the past even if he was busy at the time.
 
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