Detailed mathematical approach

Problem Summary

Project Euler 812 asks for the counting function \(S(n)\) associated with a family of dynamical-polynomial configurations, with the final result taken modulo \(998244353\). The key point is that the admissible objects do not need to be generated one by one. They can be encoded inside a product of generating functions, and \(S(n)\) is recovered as a single coefficient.

The implementation reorganizes all positive integers by their odd part. Every index can be written uniquely as \(2^u b\) with \(b\) odd, so the problem splits into independent dyadic chains \(b,2b,4b,\dots\). Counting the allowed contributions of all chains and extracting the coefficient of \(x^n\) gives the desired value.

Mathematical Approach

The arithmetic input for the generating function is the degree contribution

$c(1)=1,\qquad c(2)=1,\qquad c(m)=\frac{\varphi(m)}{2}\quad (m\ge 3),$

where \(\varphi\) is Euler's totient function. These values are exactly the numbers precomputed before the coefficient DP begins.

Step 1: Split Every Index into an Odd Base and a Power of Two

Every positive integer has a unique form

$m=2^u b,\qquad b\ \text{odd}.$

For a fixed odd base \(b\), the relevant degrees form the chain

$c(b),\ c(2b),\ c(4b),\ c(8b),\dots.$

If \(b\ge 3\) is odd and we define

$a_b=\frac{\varphi(b)}{2},$

then multiplicativity of \(\varphi\) implies

$c(b)=a_b,\qquad c(2b)=a_b,\qquad c(2^u b)=2^{u-1}a_b\quad (u\ge 2).$

So each ordinary odd-base chain begins as

$a_b,\ a_b,\ 2a_b,\ 4a_b,\dots.$

Step 2: Convert One Ordinary Chain into a Partition Factor

For \(b\ge 3\), define cumulative chain weights

$W_{b,j}=\sum_{u=0}^{j} c(2^u b).$

Because the degrees follow the pattern above, these cumulative sums collapse to

$W_{b,0}=a_b,\qquad W_{b,1}=2a_b,\qquad W_{b,2}=4a_b,\qquad W_{b,j}=2^j a_b.$

That means the entire chain contributes the ordinary generating factor

$F_b(x)=\prod_{j\ge 0}\frac{1}{1-x^{W_{b,j}}} =\prod_{j\ge 0}\frac{1}{1-x^{2^j a_b}}.$

Each factor \((1-x^w)^{-1}\) says that a weight \(w\) may be used any number of times, which is why the code updates coefficients with a forward unbounded-knapsack pass.

Step 3: Handle the Exceptional Chain with Odd Base \(1\)

The chain starting from \(1\) is not treated by the same unrestricted rule. After the first two low-degree terms are separated, the tail comes from \(4,8,16,\dots\), whose degree contributions are

$c(4)=1,\qquad c(8)=2,\qquad c(16)=4,\qquad \dots.$

Its cumulative tail weights are therefore

$T_j=1+\sum_{u=2}^{j+2} c(2^u)=2^{j+1}\qquad (j\ge 0).$

The implementation builds two tail products, a symmetric one and an alternating one:

$P_+(x)=\prod_{j\ge 1}\frac{1}{1-x^{2^j}},\qquad P_-(x)=\prod_{j\ge 1}\frac{1}{1+x^{2^j}}.$

They are combined into the core polynomial

$C(x)=\frac{(1+x)P_+(x)+(1-x)P_-(x)}{2}.$

Finally the exceptional chain contributes

$F_1(x)=\frac{C(x)}{(1-x)(1-x^2)}.$

The two denominators correspond exactly to the two cumulative coefficient passes applied after the \(P_+\) and \(P_-\) combination.

Step 4: Multiply the Independent Chains

All odd bases are independent, so the full generating function is

$F(x)=F_1(x)\prod_{\substack{b\ge 3\\ b\ \text{odd}}}\ \prod_{j\ge 0}\frac{1}{1-x^{2^j a_b}}.$

The answer is the coefficient

$S(n)=[x^n]F(x)\pmod{998244353}.$

Only weights at most \(n\) matter, because a larger weight can never affect the coefficient of \(x^n\).

Step 5: Worked Example for \(n=2\)

This is the smallest nontrivial checkpoint, and the implementations return \(S(2)=6\).

For the exceptional chain, only the weight \(2\) can matter up to degree \(2\). Hence

$P_+(x)=1+x^2+O(x^3),\qquad P_-(x)=1-x^2+O(x^3),$

so

$C(x)=1+O(x^3),\qquad F_1(x)=\frac{1}{(1-x)(1-x^2)}+O(x^3)=1+x+2x^2+O(x^3).$

Among ordinary odd bases, \(b=3\) has \(a_3=1\), so its factor is also

$F_3(x)=\frac{1}{(1-x)(1-x^2)}+O(x^3)=1+x+2x^2+O(x^3).$

Next, \(b=5\) has \(a_5=2\), giving

$F_5(x)=\frac{1}{1-x^2}+O(x^3)=1+x^2+O(x^3).$

Every larger odd base starts at degree at least \(3\), so it cannot change the coefficient of \(x^2\). Therefore

$F(x)=(1+x+2x^2)(1+x+2x^2)(1+x^2)+O(x^3) =1+2x+6x^2+O(x^3),$

and the coefficient of \(x^2\) is indeed

$S(2)=6.$

How the Code Works

The C++, Python, and Java implementations follow the same pipeline. First they choose a sieve range proportional to \(n\) and precompute \(\varphi(m)\) for every needed \(m\). From that table they build the degree sequence \(c(m)\).

Next they construct the exceptional odd-base-\(1\) factor. They generate the usable weights \(2,4,8,\dots\), build one coefficient table for \(P_+(x)\) and another for \(P_-(x)\), combine them through the formula for \(C(x)\), and then apply two cumulative passes to account for division by \((1-x)(1-x^2)\).

After that they iterate over odd bases \(b\ge 3\). For each base they accumulate the chain weights \(a_b,2a_b,4a_b,\dots\) until the next one would exceed \(n\). Every such weight performs a forward coefficient update, which is exactly multiplication by \((1-x^w)^{-1}\). When all chains are processed, the coefficient of degree \(n\) is returned modulo \(998244353\).

Complexity Analysis

Let \(M=\Theta(n)\) be the sieve limit chosen by the implementation. Computing all totients up to \(M\) costs \(O(M\log\log M)\) time and \(O(M)\) memory. The exceptional chain contributes only \(O(\log n)\) usable weights. For the ordinary odd bases, let \(W\) be the total number of cumulative weights \(2^j a_b\) that do not exceed \(n\). Each such weight triggers one forward pass over a coefficient table of length \(n+1\), so the DP phase costs \(O(nW)\) time and \(O(n)\) memory. This is a pseudo-polynomial algorithm with linear memory and a one-dimensional state array.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=812
2. Euler's totient function: Wikipedia - Euler's totient function
3. Generating function: Wikipedia - Generating function
4. Binary partition: Wikipedia - Binary partition
5. Knapsack problem: Wikipedia - Knapsack problem