Detailed mathematical approach

Problem Summary

For fixed integers \(k\), \(t\), and \(r\), define

$\alpha = 2^t + 1,\qquad \beta = 2^k + 1,\qquad M = 10^9 + 62031.$

For each integer \(n\) with \(0 \le n < k\), let \(R_n\) be the least nonnegative residue of \(2^n\alpha^r\) modulo \(\beta\), and let

$D_n = \min(R_n,\beta - R_n).$

The required value is

$F(k,t,r)=\sum_{n=0}^{k-1} 2^{k-n} D_n \pmod{M}.$

The hard part is that \(k\) and \(t\) are enormous, so iterating through all \(n\) directly is impossible. The solution works by expanding \(\alpha^r\), folding powers across the modulus \(2^k+1\), and proving that almost all \(n\)-values inside each interval contribute the same weighted amount.

Mathematical Approach

The implementations use the target value \(r=62\), and the given parameters satisfy \(rt<k\). That inequality is crucial: after expansion, every exponent lies below \(2k\), so any term can cross the \(k\)-boundary at most once.

Step 1: Expand \((2^t+1)^r\) and track the exponents

By the binomial theorem,

$\alpha^r=(1+2^t)^r=\sum_{i=0}^{r}\binom{r}{i}2^{it}.$

Multiplying by \(2^n\) gives

$2^n\alpha^r=\sum_{i=0}^{r}\binom{r}{i}2^{n+it}.$

So the entire problem is controlled by the exponents

$e_i(n)=n+it.$

Because \(0 \le n < k\) and \(rt<k\), every \(e_i(n)\) lies in the half-open range \([0,2k)\). Therefore each term is either already below \(k\), or it exceeds \(k\) once and can be folded back using the congruence \(2^k\equiv -1 \pmod{\beta}\).

Step 2: Split the range of \(n\) into intervals with one fixed folding pattern

For \(s=1,2,\dots,r\), define

$I_s=\left[k-st,\ k-(s-1)t-1\right]\cap [0,k-1].$

There is also a last interval

$I_{r+1}=[0,\ k-rt-1]\cap [0,k-1],$

which may be empty for some parameters.

If \(n\in I_s\), then exactly the terms with \(i\ge s\) satisfy \(e_i(n)\ge k\), while the terms with \(i<s\) stay below \(k\). Hence

$2^{e_i(n)} \equiv \begin{cases} 2^{n+it}, & i<s,\\[4pt] -2^{n+it-k}, & i\ge s, \end{cases}\pmod{\beta}$

and therefore

$R_n \equiv E_s(n):=\sum_{i=0}^{s-1}\binom{r}{i}2^{n+it}-\sum_{i=s}^{r}\binom{r}{i}2^{n+it-k}\pmod{\beta}.$

On the last interval \(I_{r+1}\), nothing folds, so only the first sum remains. This interval decomposition is the key structural simplification: instead of treating each \(n\) separately, we only need to understand one pattern per interval.

Step 3: After weighting by \(2^{k-n}\), the bulk contribution of an interval is constant

Suppose that for some \(n\in I_s\) the centered representative is simply \(D_n=E_s(n)\). Then multiplying by the outer weight gives

$2^{k-n}D_n=\sum_{i=0}^{s-1}\binom{r}{i}2^{k+it}-\sum_{i=s}^{r}\binom{r}{i}2^{it}.$

The right-hand side no longer depends on \(n\). So every non-exceptional \(n\) in the same interval contributes the same weighted amount.

For the last interval, the constant becomes

$2^{k-n}D_n=\sum_{i=0}^{r}\binom{r}{i}2^{k+it},$

again independent of \(n\), whenever the centered residue is taken without switching to \(\beta-R_n\).

This is why the program can add most of the contribution of a huge interval in one bulk step instead of visiting every index inside it.

Step 4: Only the last 62 positions of an interval can be ambiguous

Fix an interval \(I_s\), and let the leading positive exponent be

$e_{\max}=n+(s-1)t$

for \(s\le r\), or \(e_{\max}=n+rt\) on the final interval. Write

$\delta = k-e_{\max}.$

When \(\delta \ge 63\), every other term in the signed sum is at least \(2^{63}\) times smaller than the boundary scale \(2^k\), while the total coefficient mass is bounded by

$\sum_{i=0}^{r}\binom{r}{i}=2^r=2^{62}.$

So the leading term dominates and the entire signed value sits strictly between \(0\) and \(\beta/2\). In that regime there is no need to compare \(R_n\) with \(\beta-R_n\): the nearer value is automatically the unreversed one.

Consequently, only the values with \(\delta\le 62\) need exact centered-remainder logic. Those are exactly the last \(62\) positions of each nonempty interval, which is why the implementations peel off that short suffix and treat the earlier part as bulk.

Step 5: Recover the exact boundary cases from the leading binomial term

For an exceptional \(n\), let \(j=s-1\) for \(s\le r\), and \(j=r\) on the last interval. The dominant term is

$\binom{r}{j}2^{k-\delta}.$

Write the coefficient in base \(2^\delta\):

$\binom{r}{j}=q\,2^\delta+\ell,\qquad 0\le \ell < 2^\delta.$

Then

$\binom{r}{j}2^{k-\delta}=q2^k+\ell 2^{k-\delta}=q(2^k+1)+\ell 2^{k-\delta}-q.$

This identity shows that the exact quotient by \(\beta=2^k+1\) can be read off from the bit shift

$q=\left\lfloor \frac{\binom{r}{j}}{2^\delta}\right\rfloor.$

After subtracting \(q\beta\), the low bits \(\ell\) determine on which side of \(\beta/2\) the remainder lies. If \(\ell<2^{\delta-1}\), the remainder is below half the modulus. If \(\ell>2^{\delta-1}\), the complementary distance \(\beta-R_n\) is smaller. In the tie case \(\ell=2^{\delta-1}\), the lower-order terms decide the outcome. This is exactly what the implementations evaluate for the short exceptional suffix of each interval.

Worked Example: \(F(3,1,1)=42\)

Here

$\alpha=2^1+1=3,\qquad \beta=2^3+1=9.$

Since \(r=1\), we only need the values \(n=0,1,2\):

$\begin{aligned} n=0&:&&2^0\alpha^1=3\equiv 3 \pmod 9,\quad D_0=\min(3,6)=3,\quad 2^{3-0}D_0=24,\\ n=1&:&&2^1\alpha^1=6\equiv 6 \pmod 9,\quad D_1=\min(6,3)=3,\quad 2^{3-1}D_1=12,\\ n=2&:&&2^2\alpha^1=12\equiv 3 \pmod 9,\quad D_2=\min(3,6)=3,\quad 2^{3-2}D_2=6. \end{aligned}$

Therefore

$F(3,1,1)=24+12+6=42,$

which matches the validation value used by the implementations.

How the Code Works

The C++, Python, and Java implementations all follow the same mathematical plan. They first precompute the binomial coefficients \(\binom{r}{i}\), their values modulo \(M\), and the modular powers \(2^{it}\pmod M\). They also compute \(2^k\pmod M\) and the modular inverse of \(2^k\) modulo \(M\), because folded terms naturally introduce factors of \(2^{-k}\) when everything is tracked directly modulo \(M\).

Next, the implementation forms prefix sums for the unfolded part and suffix sums for the folded part. That makes the constant bulk contribution of an interval available in \(O(1)\) modular time once the interval length is known. Only the last \(62\) values of each interval are handled one by one.

For those exceptional positions, the implementation rebuilds the signed folded expression, extracts the quotient \(q\) from the dominant binomial coefficient by a bit shift, subtracts the corresponding multiple of \(\beta\), decides whether the remainder or its complement is closer to zero, and finally multiplies by the outer factor \(2^{k-n}\). The C++ version parallelizes these independent exceptional evaluations; the Python and Java versions evaluate the same cases sequentially.

Complexity Analysis

There are only \(r+1\) interval patterns. Each interval contributes one bulk term and at most \(62\) exceptional positions. Each exceptional evaluation scans the \(r+1\) binomial terms, so the arithmetic core uses \(O(r^2)\) modular operations and \(O(r)\) memory.

For the target problem \(r=62\) is fixed, so the runtime is effectively constant with respect to the sizes of \(k\) and \(t\); it does not grow linearly with either parameter. The only remaining dependence on their magnitudes comes from modular exponentiation, which costs logarithmic time in the exponent values.

Footnotes and References

1. Problem page: Project Euler 889
2. Takagi curve (Blancmange curve): Wikipedia — Takagi curve
3. Binomial theorem: Wikipedia — Binomial theorem
4. Modular arithmetic: Wikipedia — Modular arithmetic
5. Exponentiation by squaring: Wikipedia — Exponentiation by squaring