Detailed mathematical approach

Problem Summary

Problem 918 defines an integer sequence by

$a_1=1,\qquad a_{2k}=2a_k,\qquad a_{2k+1}=a_k-3a_{k+1}.$

The quantity to compute is the prefix sum

$S(n)=\sum_{i=1}^{n} a_i$

at the large input \(n=10^{12}\).

A direct approach would try to generate every term up to \(a_{10^{12}}\), but the index is far too large for that. The recurrence also shows why a one-term viewpoint is insufficient: the odd rule depends on both \(a_k\) and \(a_{k+1}\). The real task is to find a state that stays closed when the index is halved, then express the whole prefix sum through that smaller state.

Mathematical Approach

The solution has two decisive ingredients. First, the correct recursive object is the adjacent pair \((a_n,a_{n+1})\), not a single sequence value. Second, once the prefix sum is grouped into even-odd blocks, it telescopes to a formula involving only one sequence term at roughly half the original index.

Why a Single Term Does Not Form a Closed State

If we want \(a_{2k}\), the recurrence gives it immediately from \(a_k\). But for an odd index we have

$a_{2k+1}=a_k-3a_{k+1},$

so the value at index \(2k+1\) depends on two neighboring terms at the smaller scale. That means any recursive algorithm based only on \(a_n\) would keep needing one extra value. The natural way to close the recursion is therefore to carry those two values together.

The Adjacent-Pair Recursion

Define the state

$P(n)=(a_n,a_{n+1}).$

The base case is immediate:

$P(1)=(a_1,a_2)=(1,2).$

Now write \(P(k)=(x,y)=(a_k,a_{k+1})\). Then the recurrence gives two parity-dependent transitions:

$P(2k)=(a_{2k},a_{2k+1})=(2a_k,\ a_k-3a_{k+1})=(2x,\ x-3y),$

$P(2k+1)=(a_{2k+1},a_{2k+2})=(a_k-3a_{k+1},\ 2a_{k+1})=(x-3y,\ 2y).$

This pair is the exact invariant used by the implementations: once \((a_k,a_{k+1})\) is known, either \((a_{2k},a_{2k+1})\) or \((a_{2k+1},a_{2k+2})\) follows in constant time.

Halving the Index All the Way Down

To compute \(P(n)\), the formulas above require only \(P(\lfloor n/2\rfloor)\). Each recursive call therefore makes exactly one smaller call:

$n,\ \left\lfloor \frac{n}{2} \right\rfloor,\ \left\lfloor \frac{n}{4} \right\rfloor,\ \dots,\ 1.$

When the recursion unwinds, the parity of each intermediate index chooses which linear transformation to apply. The depth is the number of binary digits of \(n\), so recovering \(a_n\) or \(P(n)\) takes only \(O(\log n)\) arithmetic steps.

Collapsing Even-Odd Blocks in the Prefix Sum

The prefix sum becomes tractable when terms are grouped as \((a_{2j},a_{2j+1})\). For every \(j\ge 1\),

$a_{2j}+a_{2j+1}=2a_j+(a_j-3a_{j+1})=3(a_j-a_{j+1}).$

This identity is the key simplification. A block at large indices becomes a simple difference of adjacent terms at smaller indices, and differences of that form telescope when summed over \(j\).

Closed Forms for Odd and Even Prefix Lengths

Let \(m\ge 1\). For an odd prefix length, isolate the first term and pair the rest:

$S(2m-1)=a_1+\sum_{j=1}^{m-1}(a_{2j}+a_{2j+1}).$

Substituting the block identity yields

$S(2m-1)=1+3\sum_{j=1}^{m-1}(a_j-a_{j+1}).$

The sum telescopes immediately:

$\sum_{j=1}^{m-1}(a_j-a_{j+1})=a_1-a_m=1-a_m,$

so

$S(2m-1)=1+3(1-a_m)=4-3a_m.$

For an even prefix, add the final even term \(a_{2m}=2a_m\):

$S(2m)=S(2m-1)+a_{2m}=4-3a_m+2a_m=4-a_m.$

That reduces the whole problem to one sequence value:

$S(2m-1)=4-3a_m,\qquad S(2m)=4-a_m.$

Because \(10^{12}=2\cdot 5\cdot 10^{11}\), the target answer comes from a single term:

$S(10^{12})=4-a_{5\cdot 10^{11}}.$

Worked Example: Recovering \(S(9)\) and \(S(10)\)

The first ten terms are

$a_1=1,\quad a_2=2,\quad a_3=-5,\quad a_4=4,\quad a_5=17,\quad a_6=-10,\quad a_7=-17,\quad a_8=8,\quad a_9=-47,\quad a_{10}=34.$

To recover \(a_5\), follow the halving chain \(5\to 2\to 1\):

$P(1)=(1,2),\qquad P(2)=(2,-5),\qquad P(5)=(17,-10).$

So \(a_5=17\). The closed forms then give

$S(9)=4-3a_5=4-51=-47,$

$S(10)=4-a_5=4-17=-13.$

Direct summation confirms the same values:

$1+2-5+4+17-10-17+8-47=-47,$

$1+2-5+4+17-10-17+8-47+34=-13.$

This example already contains the full method used for the large input: compute one carefully chosen term by recursive halving, then substitute it into the appropriate prefix identity.

How the Code Works

The C++, Python, and Java implementations all follow the derivation above. None of them expands the sequence up to the target index. Instead, they solve only the single recursive branch forced by repeated halving.

Recursive Reconstruction of \((a_n,a_{n+1})\)

The core routine receives an index \(n\) and returns the adjacent pair \((a_n,a_{n+1})\). The base case is \((1,2)\). For larger \(n\), the implementation first computes the pair at \(\lfloor n/2\rfloor\), then rebuilds the answer with one of the two formulas above depending on whether \(n\) is even or odd.

Because there is only one recursive branch, no table or memoization is necessary. Each level contributes only a handful of additions, subtractions, and doublings.

Turning One Sequence Term into the Prefix Sum

After the implementation can recover \(a_m\), the prefix sum wrapper is just a parity test on \(n\). If \(n\) is even, it returns \(4-a_{n/2}\). If \(n\) is odd, it returns \(4-3a_{(n+1)/2}\). For the Project Euler input \(10^{12}\), the computation therefore stops at the single term \(a_{5\cdot 10^{11}}\).

What the Implementations Verify

The C++ implementation also includes a small built-in validation step: it checks the first ten sequence values against explicit known data and compares the fast prefix formula with direct summation for small indices before printing the large answer. The Python and Java implementations keep only the compact fast path.

Complexity Analysis

Each recursive level halves the index, so computing \((a_n,a_{n+1})\) takes \(O(\log n)\) arithmetic operations. Converting that result into \(S(n)\) adds only \(O(1)\) work.

The recursive form uses \(O(\log n)\) extra memory for the call stack. An iterative reconstruction along the same halving chain could reduce the extra space to \(O(1)\), but for this problem the recursive version is already short and direct.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=918
2. Recurrence relation: Wikipedia - Recurrence relation
3. Integer sequence: Wikipedia - Integer sequence
4. Telescoping series: Wikipedia - Telescoping series
5. Divide-and-conquer algorithm: Wikipedia - Divide-and-conquer algorithm
6. Binary number: Wikipedia - Binary number