Detailed mathematical approach

Problem Summary

We are given a sequence \(a(n)\) modulo \(M=1000062031\) defined by

$a(0)=1,\qquad a(2m+1)=a(m),\qquad a(2m)=3a(m)+5a\left(2m-2^{\nu_2(m)}\right)\pmod{M}.$

The required value is

$a(N),\qquad N=(1+2^t)^r,\quad t=10^{14}+31,\quad r=62.$

The index \(N\) is far too large for any direct dynamic program or recursion tree that expands all the way up to \(N\). The useful observation is that the recurrence reacts to binary structure, so the problem can be solved by reading the positions of the \(1\)-bits of \(N\) instead of iterating through all smaller integers.

Mathematical Approach

The derivation has two layers. First we understand how the sequence changes when binary digits are appended. Then we apply that rule to the special number \(N=(1+2^t)^r\).

Step 1: Appending a trailing 1 leaves the value unchanged

For every \(m\ge 0\),

$a(2m+1)=a(m).$

In binary, going from \(m\) to \(2m+1\) means shifting left by one position and appending a final \(1\). So a trailing \(1\)-bit does not contribute a multiplicative factor of its own. It only changes the state of the recursion by increasing the number of \(1\)-bits that have already appeared when we continue scanning toward lower bit positions.

This is why the final formula depends on the runs of zeros between successive \(1\)-bits, not on the \(1\)-bits themselves.

Step 2: Appending a trailing 0 multiplies by a state-dependent constant

Suppose \(n\) is odd and has exactly \(i\) set bits. Then \(n-1\) has the same higher bits but only \(i-1\) set bits. Let \(v_i\) denote the factor produced by appending one zero while the current prefix already contains \(i\) ones.

We start with

$v_0=1.$

Now apply the even recurrence to \(2n\):

$a(2n)=3a(n)+5a(n-1).$

Because \(n\) is odd, we also have

$a(n)=a\left(\frac{n-1}{2}\right).$

The number \(n-1\) is obtained from \(\frac{n-1}{2}\) by appending one zero in a state with \(i-1\) ones, hence

$a(n-1)=v_{i-1}a(n).$

Substituting this into the even recurrence gives

$a(2n)=\left(3+5v_{i-1}\right)a(n).$

Therefore the multipliers satisfy

$v_i=5v_{i-1}+3\qquad (i\ge 1).$

The first values are

$v_1=8,\qquad v_2=43,\qquad v_3=218,\qquad v_4=1093.$

Before reduction modulo \(M\), the recurrence has the exact closed form

$v_i=\frac{7\cdot 5^i-3}{4},$

although the implementation only needs the recurrence modulo \(M\).

Step 3: Any positive integer becomes a product over zero gaps

Let the set-bit positions of a positive integer \(n\) be

$p_1>p_2>\cdots>p_s\ge 0.$

Define the zero gaps by

$g_i=\begin{cases} p_i-p_{i+1}-1,&1\le i<s,\\ p_s,&i=s. \end{cases}$

When we scan the binary expansion from the most significant \(1\) downward, the leading \(1\) just starts the process. After the first \(1\), every zero contributes a factor \(v_1\) until the second \(1\) is reached. After the second \(1\), every zero contributes \(v_2\), and so on. Encountering a new \(1\) changes the state from \(i\) seen ones to \(i+1\) seen ones, but does not multiply the value.

Hence the exact evaluation formula is

$a(n)=\prod_{i=1}^{s} v_i^{g_i}\pmod{M}.$

This transforms the original recursion into a finite product determined solely by the locations of the \(1\)-bits.

Step 4: Expand \((1+2^t)^r\) into disjoint binary blocks

Now write

$N=(1+2^t)^r=\sum_{k=0}^{r}\binom{r}{k}2^{kt}.$

If every binomial coefficient satisfies

$\binom{r}{k}<2^t,$

then the binary digits of \(\binom{r}{k}2^{kt}\) stay entirely inside the block of positions \(kt,kt+1,\dots,kt+t-1\). Distinct values of \(k\) occupy disjoint blocks, so their sum produces no carries between blocks.

That is exactly what happens here, because \(t=10^{14}+31\) is vastly larger than the bit-length of every coefficient in row \(r=62\). Therefore the set-bit positions of \(N\) are simply

$\left\{\,kt+b \;:\; 0\le k\le r,\ b\in B_k\,\right\}.$

Here \(B_k\) denotes the set of positions of the \(1\)-bits in \(\binom{r}{k}\).

Once these positions are sorted in descending order, Step 3 gives \(a(N)\) immediately.

Worked Example: \(t=3\) and \(r=2\)

This is the small checkpoint used by the implementation.

First expand the target:

$N=(1+2^3)^2=1+2\cdot 2^3+2^6=1+2^4+2^6=81.$

Its binary form is

$81=1010001_2,$

so the descending set-bit positions are \(6,4,0\). The corresponding gaps are

$g_1=6-4-1=1,\qquad g_2=4-0-1=3,\qquad g_3=0.$

Using

$v_1=8,\qquad v_2=43,\qquad v_3=218,$

we obtain

$a(81)=8^1\cdot 43^3\cdot 218^0\pmod{M}=636056.$

This small case already shows the general pattern: once the set-bit positions are known, the enormous recursion collapses into a short product.

How the Code Works

The C++ implementation contains the full arithmetic. It builds the binomial row for \(r=62\) iteratively, using exact integer division with wide intermediate arithmetic so that every coefficient is computed exactly. For each coefficient, it enumerates the positions of its set bits, shifts them by \(kt\), and stores all resulting positions in one list.

Next, the implementation sorts those positions in descending order, precomputes the multipliers \(v_i\) modulo \(M\), and evaluates

$\prod_{i=1}^{s} v_i^{g_i}\pmod{M}$

by binary modular exponentiation. The Python and Java implementations invoke the same compiled solver, so all three language versions follow the same mathematical logic and produce the same residue.

The program also performs two local consistency checks. It compares the bit-position formula against a direct dynamic-programming computation for all \(n\le 5000\), and it verifies the example \(a\left((1+2^3)^2\right)=636056\). Those checks do not change the algorithm, but they confirm that the product formula matches the original recurrence.

Complexity Analysis

Let

$s=\sum_{k=0}^{r}\operatorname{popcount}\!\left(\binom{r}{k}\right),$

the total number of set bits appearing across the binomial coefficients. Building the row costs \(O(r)\) arithmetic steps. Enumerating the set bits costs \(O(s)\). Sorting the positions costs \(O(s\log s)\), and evaluating the modular powers costs \(O(s\log t)\).

So the total running time is

$O(r+s\log s+s\log t),$

with memory usage \(O(s)\). In the actual problem \(r=62\) is tiny, so the huge size of \(t\) never creates a huge loop; it only appears as the exponent size inside fast modular exponentiation.

Footnotes and References

1. Project Euler problem page: https://projecteuler.net/problem=811
2. Binomial theorem: Wikipedia — Binomial theorem
3. \(p\)-adic valuation, including the \(2\)-adic case \(\nu_2(n)\): Wikipedia — p-adic valuation
4. Hamming weight and set-bit counting: Wikipedia — Hamming weight
5. Modular exponentiation: Wikipedia — Modular exponentiation