Problem 517: A Real Recursion

Mathematical Foundation

Theorem 1 (Combinatorial Interpretation). Let $a = \sqrt{p}$. Then $g_a(n)$ counts the number of ways to express the “walk” from some starting value $x_0 < a$ to $n$ using steps of size $1$ and $a$. Equivalently, if we take $k$ steps of size $a$ and $m$ steps of size $1$, then $g_a(n)$ equals a sum of binomial coefficients:

where $K = \lfloor\sqrt{p}\rfloor$ and $n_k$ depends on the number of large steps taken.

Proof. Starting from a value $x_0 < a$ and reaching $p$ with $k$ jumps of size $a$ and $m$ jumps of size $1$, we need $x_0 + ka + m = p$ with $0 \leq x_0 < a$, i.e., $m = p - ka - x_0$ with $0 \leq x_0 < a$. Since $g_a(x) = 1$ for all $x < a$ (the base case covers a continuous interval), the starting point $x_0$ is implicitly determined. For integer steps, $m + k$ total steps are taken, and we choose the positions of the $k$ large steps among the $m + k$ steps. The number of such arrangements is $\binom{m+k}{k}$.

More precisely, letting $a = \sqrt{p}$, the recursion unfolds to:

where the upper index tracks the number of steps of size $a$, and $\lceil ka \rceil$ accounts for the minimum value consumed by $k$ large steps to stay within the base-case region. Since $a = \sqrt{p}$, the maximum $k$ is $\lfloor\sqrt{p}\rfloor$. The exact value of $n_k$ is $p - 1 - \lfloor k(\sqrt{p} - 1)\rfloor$ or a closely related expression depending on the precise unfolding. $\square$

Lemma 1 (Modular Binomial Computation). For a prime modulus $q$, the binomial coefficient $\binom{n}{k} \bmod q$ can be computed in $O(1)$ time (after $O(n)$ preprocessing of factorials and inverse factorials) via

Proof. Since $q$ is prime, $\mathbb{Z}/q\mathbb{Z}$ is a field, so every nonzero element has a multiplicative inverse. Precompute $i! \bmod q$ and $(i!)^{-1} \bmod q$ for $i = 0, 1, \ldots, n$ using the recurrence $i! = i \cdot (i-1)!$ and Fermat’s little theorem for the inverse. Each binomial coefficient then requires $O(1)$ multiplications. $\square$

Theorem 2 (Prime Sieving in Short Intervals). The primes in $(10^7, 10^7 + 10^4)$ can be found using a segmented sieve of Eratosthenes in $O(\sqrt{10^7} + 10^4)$ time.

Proof. Sieve primes up to $\sqrt{10^7 + 10^4} < 3163$. For each small prime $p$, mark its multiples in the interval $(10^7, 10^7 + 10^4)$. The remaining unmarked entries are prime. $\square$

Editorial

For real a > 1: g_a(x) = 1 for x < a, g_a(x) = g_a(x-1) + g_a(x-a) for x >= a. G(n) = g_{sqrt(n)}(n). Find sum of G(p) mod 10^9+7 for primes p with 10^7 < p < 10^7+10^4. We sieve primes in (10^7, 10^7 + 10^4). We then precompute factorials mod MOD up to 10^7 + 10^4. Finally, compute G(p) for each prime p.

Pseudocode

Sieve primes in (10^7, 10^7 + 10^4)
Precompute factorials mod MOD up to 10^7 + 10^4
Compute G(p) for each prime p
for p in primes

Complexity Analysis

• Time: Segmented sieve: $O(\sqrt{10^7} + 10^4)$. Factorial precomputation: $O(10^7)$. For each of the $\sim$620 primes, the inner sum has $\sim$3162 terms, each $O(1)$: total $O(620 \times 3162) \approx O(2 \times 10^6)$. Overall: $O(10^7)$.
• Space: $O(10^7)$ for factorial arrays.

Answer

#include <bits/stdc++.h>
using namespace std;

const long long MOD = 1000000007;
const int MAXN = 10010001;

long long fac[MAXN], inv_fac[MAXN];

long long power(long long base, long long exp, long long mod) {
    long long result = 1;
    base %= mod;
    while (exp > 0) {
        if (exp & 1) result = result * base % mod;
        base = base * base % mod;
        exp >>= 1;
    }
    return result;
}

void precompute() {
    fac[0] = 1;
    for (int i = 1; i < MAXN; i++)
        fac[i] = fac[i-1] * i % MOD;
    inv_fac[MAXN-1] = power(fac[MAXN-1], MOD-2, MOD);
    for (int i = MAXN-2; i >= 0; i--)
        inv_fac[i] = inv_fac[i+1] * (i+1) % MOD;
}

long long C(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fac[n] % MOD * inv_fac[k] % MOD * inv_fac[n-k] % MOD;
}

int main() {
    precompute();

    // Sieve primes in (10^7, 10^7 + 10^4)
    const int LO = 10000001, HI = 10010000;
    vector<bool> is_prime(HI + 1, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; (long long)i * i <= HI; i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= HI; j += i)
                is_prime[j] = false;
        }
    }

    long long ans = 0;

    for (int p = LO; p < HI; p++) {
        if (!is_prime[p]) continue;

        double a = sqrt((double)p);
        int K = (int)floor(a);  // floor(sqrt(p))

        // G(p) = sum_{k=0}^{K} C(n_k, k)
        // where n_k = p - 1 - floor(k * (a - 1))
        // This counts compositions reaching p from base region [0, a)
        long long gp = 0;
        for (int k = 0; k <= K; k++) {
            // Number of unit steps when k large jumps of size a are taken
            // The total distance covered is p from starting point in [0, a)
            // After k jumps of size a, remaining ~ p - k*a unit steps
            // The jumps can be interleaved: C(unit_steps + k - 1, k) or similar
            // More precisely: n_k = p - 1 - (int)(k * (a - 1))
            int nk = p - 1 - (int)floor(k * (a - 1.0));
            if (nk < k) break;
            gp = (gp + C(nk, k)) % MOD;
        }

        ans = (ans + gp) % MOD;
    }

    cout << ans << endl;
    return 0;
}

"""
Project Euler Problem 517: A Real Recursion

For real a > 1: g_a(x) = 1 for x < a, g_a(x) = g_a(x-1) + g_a(x-a) for x >= a.
G(n) = g_{sqrt(n)}(n).
Find sum of G(p) mod 10^9+7 for primes p with 10^7 < p < 10^7+10^4.
"""
import math

MOD = 1000000007

def solve():
    HI = 10010000
    LO = 10000001

    # Sieve primes up to HI
    is_prime = bytearray([1]) * (HI + 1)
    is_prime[0] = is_prime[1] = 0
    for i in range(2, int(HI**0.5) + 1):
        if is_prime[i]:
            is_prime[i*i::i] = bytearray(len(is_prime[i*i::i]))

    # Precompute factorials and inverse factorials
    MAXN = HI + 1
    fac = [1] * MAXN
    for i in range(1, MAXN):
        fac[i] = fac[i-1] * i % MOD
    inv_fac = [1] * MAXN
    inv_fac[MAXN-1] = pow(fac[MAXN-1], MOD-2, MOD)
    for i in range(MAXN-2, -1, -1):
        inv_fac[i] = inv_fac[i+1] * (i+1) % MOD

    def C(n, k):
        if k < 0 or k > n:
            return 0
        return fac[n] * inv_fac[k] % MOD * inv_fac[n-k] % MOD

    ans = 0
    for p in range(LO, HI):
        if not is_prime[p]:
            continue
        a = math.sqrt(p)
        K = int(math.floor(a))

        gp = 0
        for k in range(K + 1):
            nk = p - 1 - int(math.floor(k * (a - 1.0)))
            if nk < k:
                break
            gp = (gp + C(nk, k)) % MOD

        ans = (ans + gp) % MOD

    print(ans)

if __name__ == "__main__":
    solve()