Detailed mathematical approach

Problem Summary

The implementation searches for the smallest integer \(n>42\) such that a derived quantity \(t(n)\) is divisible by \(10^6\). Instead of building enormous exact integers, it works with a seventh-order linear recurrence and tracks only residues modulo \(10^6\).

The recurrence starts from

$a_0=1,\quad a_1=1,\quad a_2=1,\quad a_3=2,\quad a_4=2,\quad a_5=3,\quad a_6=4,$

and the tested quantity is

$t(n)=2^{\lfloor n/2\rfloor}-a_n.$

So the task is to find the first \(n>42\) with

$t(n)\equiv 0 \pmod{10^6}.$

Mathematical Approach

The key observation is that the problem is entirely about divisibility by \(10^6\), so both the recurrence and the power term can be evolved in modular arithmetic from the very beginning.

Step 1: Define the Sequence and the Target Quantity

For \(n\ge 7\), the sequence satisfies

$a_n=a_{n-1}+2a_{n-2}-2a_{n-3}-a_{n-4}+a_{n-5}+a_{n-6}-a_{n-7}.$

The target expression is the difference between a parity-driven power of two and the recurrence value:

$t(n)=2^{\lfloor n/2\rfloor}-a_n.$

Nothing else is needed for the search: once \(a_n\) and \(2^{\lfloor n/2\rfloor}\) are known, the divisibility test is immediate.

Step 2: Reduce the Entire Search Modulo \(10^6\)

Let

$M=10^6.$

Because divisibility by \(M\) depends only on residues modulo \(M\), we may replace every exact value by its residue:

$a_n \equiv \overline{a}_n \pmod{M},\qquad 2^{\lfloor n/2\rfloor}\equiv \overline{p}_n \pmod{M}.$

Then

$t(n)\equiv \overline{p}_n-\overline{a}_n \pmod{M},$

so the search condition becomes

$\overline{p}_n-\overline{a}_n\equiv 0\pmod{M}.$

This means the exact integers are unnecessary. The implementation can run forever using bounded integers in the interval \([0,M)\).

Step 3: Update the Power Term from Parity Alone

Define

$p_n=2^{\lfloor n/2\rfloor}.$

Since \(\lfloor n/2\rfloor\) increases only when \(n\) changes from odd to even, the sequence \(p_n\) obeys a very simple rule:

$p_n=\begin{cases} p_{n-1}, & n\text{ odd},\\ 2p_{n-1}, & n\text{ even}. \end{cases}$

Therefore the modular power residue needs only one multiplication by \(2\) on even steps, and no change on odd steps.

Step 4: Compress the Recurrence to Seven Residues

The recurrence for \(a_n\) depends only on the previous seven values. Consequently, the full history is irrelevant once the latest seven residues are known.

If we keep the block

$\left(\overline{a}_{n-1},\overline{a}_{n-2},\dots,\overline{a}_{n-7}\right),$

then the next residue follows from the same linear combination, interpreted modulo \(M\):

$\overline{a}_n\equiv \overline{a}_{n-1}+2\overline{a}_{n-2}-2\overline{a}_{n-3}-\overline{a}_{n-4}+\overline{a}_{n-5}+\overline{a}_{n-6}-\overline{a}_{n-7}\pmod{M}.$

That is why a circular buffer of length \(7\) is enough. The search is linear in the answer but constant in memory.

Step 5: Normalize Negative Intermediate Values

The recurrence contains negative coefficients, so intermediate sums may be negative before reduction. Mathematically, the correct residue is always the unique value in \([0,M)\) congruent to that sum.

In formula form, one may write

$\operatorname{norm}(x)=((x\bmod M)+M)\bmod M.$

This keeps every stored recurrence value and every tested difference inside the canonical residue range.

Worked Example

The first few recurrence steps are small enough to compute exactly. Using the initial values, we obtain

$\begin{aligned} a_7&=4+2\cdot 3-2\cdot 2-2+1+1-1=5,\\ a_8&=5+2\cdot 4-2\cdot 3-2+2+1-1=7,\\ a_9&=7+2\cdot 5-2\cdot 4-3+2+2-1=9,\\ a_{10}&=9+2\cdot 7-2\cdot 5-4+3+2-2=12. \end{aligned}$

The power term is

$p_7=2^3=8,\qquad p_8=2^4=16,\qquad p_9=2^4=16,\qquad p_{10}=2^5=32.$

Hence

$t(7)=8-5=3,\qquad t(8)=16-7=9,\qquad t(9)=16-9=7,\qquad t(10)=32-12=20.$

Two larger checkpoints that agree with the exact recurrence are

$t(20)=824,\qquad t(42)=1{,}999{,}923.$

Those values confirm that the recurrence and the power-tracking rule are synchronized correctly before the final modular scan continues beyond \(42\).

How the Code Works

The C++, Python, and Java implementations all follow the same structure. They initialize the seven starting values of the recurrence and a running residue for \(2^{\lfloor n/2\rfloor}\). The main loop advances \(n\) one step at a time. On even \(n\), the power residue is doubled modulo \(10^6\). Once \(n\ge 7\), the next recurrence value is computed from the previous seven stored residues and written back into the recycled slot of a circular buffer.

After that update, the implementation compares the power residue and the recurrence residue modulo \(10^6\). If their difference is congruent to \(0\), then \(t(n)\) is divisible by \(10^6\). The loop returns the first such \(n\) with \(n>42\). Because only seven recurrence entries and one power residue are retained, the code stays compact and memory-efficient.

Complexity Analysis

Let \(N_*\) be the first index greater than \(42\) satisfying \(t(N_*)\equiv 0\pmod{10^6}\). Each loop iteration performs a constant amount of arithmetic: one parity check, at most one multiplication by \(2\), one recurrence update using seven cached residues, and one modular comparison. Therefore the running time is \(O(N_*)\), while the memory usage is \(O(1)\).

Footnotes and References

1. Problem page: Project Euler 710
2. Linear recurrences: Wikipedia - Linear recurrence with constant coefficients
3. Modular arithmetic: Wikipedia - Modular arithmetic
4. Circular buffer: Wikipedia - Circular buffer