Distributing into Unlabeled Containers: Alternative Solution¶

Pentagonal recurrence and removing large parts¶

Codex (GPT-6) applies the classical Euler pentagonal identity and truncated generating functions in alternative.cpp. The main solution explains conjugation and knapsack. Keep only coefficients of degree at most \(n\) throughout.

Unrestricted integer partitions have generating function

\[ P(x)=\prod_{j\ge1}\frac1{1-x^j} \]

Euler's pentagonal identity gives its inverse; see NIST DLMF 27.14:

\[ \prod_{j\ge1}(1-x^j)=1+\sum_{k\ge1}(-1)^k\left(x^{k(3k-1)/2}+x^{k(3k+1)/2}\right) \]

Their product is one. Comparing each positive-degree coefficient \(s\) gives

\[ p_s=\sum_{k\ge1}(-1)^{k-1}\left(p_{s-k(3k-1)/2}+p_{s-k(3k+1)/2}\right) \]

Here \(p_0=1\) and negative-index entries are zero. There are only \(O(\sqrt s)\) relevant pentagonal numbers for each \(s\) , so all unrestricted counts take \(O(n\sqrt n)\) time.

Removing parts beyond the container bound¶

By conjugation, the target generating function is

\[ P_m(x)=\prod_{j=1}^{m}\frac1{1-x^j}=P(x)\prod_{j=m+1}^{n}(1-x^j)\pmod{x^{n+1}} \]

Factors of degree above \(n\) cannot affect the answers. To multiply by \(1-x^j\) , process \(s=n,n-1,\ldots,j\) and set \(p_s\leftarrow p_s-p_{s-j}\) . Descending order is essential to use the coefficients from before this multiplication. Each pass cancels exactly the geometric-series factor allowing part size \(j\) .

The pentagonal recurrence follows from an exact generating-function identity. Cancelling those factors leaves precisely the partitions with parts at most \(m\) . Conjugation gives all answers to the container problem. Normalize every addition and subtraction modulo the specified modulus.

Time is \(O(n\sqrt n+\max(n-m,0)^2)\) and space is \(O(n)\) . For \(m\ge n\) , no factors need cancellation and time is \(O(n\sqrt n)\) . For small \(m\) , cancellation is nearly quadratic and knapsack is more suitable. This is a complementary method with a different useful parameter range, tested over the complete bounds.