Detailed mathematical approach

Problem Summary

A partition of \(n\) is called special when every part is distinct and every even part is divisible by \(4\). Equivalently, the allowed part sizes are all odd numbers together with multiples of \(4\), while numbers congruent to \(2 \pmod 4\) are forbidden. If \(P(n)\) denotes the number of such partitions, the task is to compute

$S(10^7)=\sum_{n=1}^{10^7} P(n)\pmod{10^9+7}.$

A direct dynamic program over all allowed parts works for small checks but is too slow for \(10^7\). The implemented solution rewrites the generating function into a product whose coefficients can be obtained from a square-based recurrence and a prefix count of generalized hexagonal numbers.

Mathematical Approach

Let \(M=10^9+7\). The key is to express the special-partition generating function in a form where one factor advances in steps of \(4\), and the other factor is just a sparse indicator series.

Step 1: Write the generating function

Because parts are distinct, each allowed size \(m\) contributes a factor \(1+q^m\). Since the allowed parts are \(4r-3\), \(4r-1\), and \(4r\), we get

$F(q)=\sum_{n\ge 0} P(n) q^n=\prod_{r\ge 1}(1+q^{4r-3})(1+q^{4r-1})(1+q^{4r}).$

This already encodes the whole problem, but extracting all coefficients up to \(10^7\) directly is still expensive.

Step 2: Split the product with Jacobi's triple product

Jacobi's triple product gives the identity

$\sum_{t\in\mathbb Z} q^{t(2t-1)}=\prod_{r\ge 1}(1-q^{4r})(1+q^{4r-1})(1+q^{4r-3}).$

The exponents \(t(2t-1)\) are the generalized hexagonal numbers:

$\mathcal H=\{t(2t-1): t\in\mathbb Z\}=\{0,1,3,6,10,15,21,\dots\}.$

Dividing the original product by the factor \(\prod_{r\ge 1}(1-q^{4r})\) isolates the remaining contribution of the multiples of \(4\):

$F(q)=\left(\prod_{r\ge 1}\frac{1+q^{4r}}{1-q^{4r}}\right)\left(\sum_{t\in\mathbb Z} q^{t(2t-1)}\right).$

Now define

$A(x)=\prod_{r\ge 1}\frac{1+x^r}{1-x^r}=\sum_{k\ge 0} a_k x^k.$

Then the special-partition series becomes

$F(q)=\left(\sum_{k\ge 0} a_k q^{4k}\right)\left(\sum_{t\in\mathbb Z} q^{t(2t-1)}\right).$

Step 3: Derive the square-based recurrence for \(a_k\)

The reciprocal Jacobi theta identity says

$1+2\sum_{m\ge 1}(-1)^m x^{m^2}=\prod_{r\ge 1}(1-x^{2r})(1-x^{2r-1})^2.$

On the other hand, for each \(r\),

$\frac{1+x^r}{1-x^r}=\frac{1-x^{2r}}{(1-x^r)^2}.$

Multiplying over all \(r\) shows that

$A(x)=\frac{1}{1+2\sum_{m\ge 1}(-1)^m x^{m^2}}.$

Therefore

$A(x)\left(1+2\sum_{m\ge 1}(-1)^m x^{m^2}\right)=1.$

Comparing coefficients of \(x^n\) yields

$a_0=1,$

$a_n+2\sum_{m^2\le n}(-1)^m a_{n-m^2}=0 \qquad (n\ge 1),$

so

$a_n=2\sum_{m^2\le n}(-1)^{m+1} a_{n-m^2}\pmod M.$

This is exactly the recurrence used by the implementation.

Step 4: Convert the product into the cumulative sum

Let

$H(T)=\#\{h\in\mathcal H: h\le T\}.$

Since

$F(q)=\left(\sum_{k\ge 0} a_k q^{4k}\right)\left(\sum_{h\in\mathcal H} q^h\right),$

the coefficient of \(q^n\) is

$P(n)=\sum_{\substack{k\ge 0,\ h\in\mathcal H\\ 4k+h=n}} a_k.$

Summing this for \(0\le n\le N\) gives

$\sum_{n=0}^{N} P(n)=\sum_{k=0}^{\lfloor N/4\rfloor} a_k\,H(N-4k).$

The term \(P(0)=1\) is the empty partition, which must be excluded because the problem starts at \(1\). Hence

$\boxed{S(N)=\sum_{k=0}^{\lfloor N/4\rfloor} a_k\,H(N-4k)-1 \pmod M.}$

Worked Example: \(N=10\)

The generalized hexagonal numbers up to \(10\) are

$\mathcal H\cap[0,10]=\{0,1,3,6,10\},$

so \(H(10)=5\), \(H(6)=4\), and \(H(2)=2\).

From the recurrence,

$a_0=1,\qquad a_1=2,\qquad a_2=4.$

Therefore

$S(10)=a_0H(10)+a_1H(6)+a_2H(2)-1=1\cdot 5+2\cdot 4+4\cdot 2-1=20.$

Indeed, the first ten values are

$\bigl(P(1),P(2),\dots,P(10)\bigr)=(1,0,1,2,2,1,2,4,4,3),$

and their sum is \(20\).

How the Code Works

The C++, Python, and Java implementations all follow the same mathematical pipeline. First they compute the coefficient table \(a_k\) up to \(\lfloor N/4\rfloor\). To do that efficiently, they precompute all squares \(m^2\le \lfloor N/4\rfloor\) and apply the alternating recurrence above.

Next they mark every generalized hexagonal number \(t(2t-1)\le N\), which is the same as marking both \(n(2n-1)\) and \(n(2n+1)\) for positive \(n\), together with \(0\). A prefix-sum array then turns each query \(H(T)\) into \(O(1)\) time.

Finally they evaluate

$\sum_{k=0}^{\lfloor N/4\rfloor} a_k\,H(N-4k)-1 \pmod M$

in one pass. No large two-dimensional partition table is needed; all heavy work is compressed into a one-dimensional recurrence plus a sparse counting array.

Complexity Analysis

Let \(K=\lfloor N/4\rfloor\). Building the coefficient table requires, for each \(n\le K\), iterating over all squares up to \(n\), so the time cost is

$O\!\left(\sum_{n=1}^{K}\sqrt{n}\right)=O(K\sqrt{K}).$

Marking generalized hexagonal numbers takes \(O(\sqrt{N})\) updates, building the prefix array takes \(O(N)\), and the final accumulation takes \(O(K)\). Thus the total running time is dominated by \(O(K\sqrt{K})=O(N^{3/2})\), while memory usage is \(O(N)\) because the prefix array over \(0,\dots,N\) is the largest structure.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=614
2. Integer partitions: Wikipedia - Partition (number theory)
3. Jacobi triple product: Wikipedia - Jacobi triple product
4. Theta functions: Wikipedia - Theta function
5. Polygonal numbers: Wikipedia - Polygonal number