Detailed mathematical approach

Problem Summary

The task is to evaluate the prefix sum

$S(n)=\sum_{k=1}^{n} f(k) \pmod{10^9}$

for

$n=3^{37}=450283905890997363.$

The sequence \(f(n)\) is recursive, but the C++, Python, and Java implementations avoid term-by-term generation. They use an equivalent binary description: remove every factor of \(2\) from \(n\), keep the remaining odd part, and reverse its binary digits. This turns the strange recurrence into a family of halving identities that are suitable for memoization.

The built-in checkpoints are

$S(8)=22,\qquad S(100)=3604.$

Mathematical Approach

The decisive observation is that the sequence is much easier to understand in base \(2\) than from its original recursive wording.

Step 1: Express \(f(n)\) through the odd core

Write

$n=2^{\nu_2(n)}u,\qquad \operatorname{odd}(n)=u,$

where \(u\) is odd. If the binary expansion of \(u\) is

$u=(a_r a_{r-1}\dots a_1 1)_2,$

then the implementations use the equivalent closed description

$f(n)=(1 a_1 a_2 \dots a_r)_2.$

In words: strip the trailing zeros from the binary expansion of \(n\), then reverse the remaining odd binary word. Since multiplying by \(2\) only appends one trailing zero before that stripping step, we immediately get

$f(2m)=f(m).$

This is the first self-similarity that makes fast prefix sums possible.

Step 2: Recover the odd-index recurrences

Let

$m=2^s u$

with \(u\) odd, and let \(\ell\) be the bit-length of \(u\). Denote

$R=f(m).$

After reversing the relevant binary words, there exists a power of two

$A=2^{\ell+s}$

such that

$f(2m+1)=A+R,\qquad f(4m+1)=2A+R,\qquad f(4m+3)=3A+R.$

Eliminating \(A\) yields exact identities:

$f(4m+1)=2f(2m+1)-f(m),$

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

These are precisely the relations needed to group odd arguments into pairs and blocks.

Step 3: Introduce two prefix sums

Define the main prefix sum and an auxiliary odd-index prefix sum by

$S(n)=\sum_{k=1}^{n} f(k),\qquad O(n)=\sum_{k=1}^{n} f(2k-1).$

The second quantity is the sum of the first \(n\) odd-indexed terms of the sequence. Because even arguments collapse under the rule \(f(2m)=f(m)\), the main sum satisfies

$\begin{aligned} S(2m) &=\sum_{k=1}^{m} f(2k)+\sum_{k=1}^{m} f(2k-1)\\ &=S(m)+O(m), \end{aligned}$

and therefore

$S(2m+1)=S(2m)+f(2m+1).$

So the main problem reduces to computing \(O(n)\) quickly.

Step 4: Derive a closed recurrence for \(O(n)\)

For an odd upper limit, only one new odd term is added:

$O(2m+1)=O(2m)+f(4m+1)=O(2m)+2f(2m+1)-f(m).$

For an even upper limit, pair the odd arguments as \((4t+1,4t+3)\):

$\begin{aligned} O(2m) &=\sum_{t=0}^{m-1}\bigl(f(4t+1)+f(4t+3)\bigr). \end{aligned}$

Using the identities from Step 2, for every \(t\ge 1\) we have

$f(4t+1)+f(4t+3)=5f(2t+1)-3f(t).$

The case \(t=0\) must be handled separately, because it contributes

$f(1)+f(3)=1+3=4.$

Hence

$\begin{aligned} O(2m) &=4+\sum_{t=1}^{m-1}\bigl(5f(2t+1)-3f(t)\bigr)\\ &=4+5\sum_{t=1}^{m-1}f(2t+1)-3\sum_{t=1}^{m-1}f(t)\\ &=4+5\bigl(O(m)-1\bigr)-3S(m-1)\\ &=-1+5O(m)-3S(m-1). \end{aligned}$

Together with

$S(0)=0,\qquad S(1)=1,\qquad O(0)=0,\qquad O(1)=1,$

these recurrences determine all required prefix sums.

Worked Example: Recover \(S(8)=22\)

The first few sequence values are

$f(1),\dots,f(8)=1,1,3,1,5,3,7,1,$

so a direct sum gives \(22\). The recursive method reaches the same result much more systematically:

$S(1)=1,\qquad O(1)=1.$

Then

$S(2)=S(1)+O(1)=2,$

$O(2)=-1+5O(1)-3S(0)=4,$

$S(4)=S(2)+O(2)=6,$

$O(4)=-1+5O(2)-3S(1)=16,$

$S(8)=S(4)+O(4)=22.$

The same framework also produces the second checkpoint \(S(100)=3604\) without iterating through all one hundred terms one by one.

Step 5: Why the large input becomes manageable

Every recurrence replaces the current argument by either \(n-1\), \(\lfloor n/2\rfloor\), or \(\lfloor n/2\rfloor-1\). An odd step is followed immediately by an even step, so the size drops by roughly a factor of \(2\) after at most one subtraction. That means the state graph is very small compared with the enormous target value \(3^{37}\), and memoization can reuse almost every nontrivial subproblem.

How the Code Works

The C++, Python, and Java implementations all follow the same plan. They store two memo tables: one for \(S(n)\) and one for \(O(n)\). The base entries are the four values listed above.

Whenever a single term \(f(n)\) is needed, the implementation removes all powers of \(2\) from \(n\), reverses the binary digits of the remaining odd integer, and then reduces the result modulo \(10^9\).

To evaluate \(S(n)\), the implementation checks parity. For even \(n\), it applies

$S(n)=S\!\left(\frac{n}{2}\right)+O\!\left(\frac{n}{2}\right);$

for odd \(n\), it uses

$S(n)=S(n-1)+f(n).$

The auxiliary sum is handled analogously: for even \(n\), the code uses

$O(n)=-1+5O\!\left(\frac{n}{2}\right)-3S\!\left(\frac{n}{2}-1\right),$

and for odd \(n\), it adds the next odd contribution through

$O(n)=O(n-1)+2f(n)-f\!\left(\frac{n-1}{2}\right).$

Each result is normalized modulo \(10^9\) before being stored. Because the same arguments recur many times, caching is what turns the mathematical identities into a fast program.

Complexity Analysis

The important point is that fresh states are created only along a halving-style recursion. After at most one subtraction, every branch moves to an argument of about half the previous size. As a result, the number of memoized states grows near-logarithmically in \(n\), not linearly.

At the machine-integer level, each new state performs only constant many modular additions and subtractions, plus one binary reversal whose cost is proportional to the bit-length of the odd core. Therefore the total running time stays far below \(O(n)\), and the memory usage is proportional to the small number of cached states.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=463
2. Binary numeral system: Wikipedia — Binary number
3. Bit-reversal permutation: Wikipedia — Bit-reversal permutation
4. \(p\)-adic valuation and the special case \(\nu_2\): Wikipedia — \(p\)-adic valuation
5. Memoization: Wikipedia — Memoization