推理流程:

首先以n=1000进行打表可以得到以下数据
#define For(i, a, b) for (int i = (a); i <= (b); i += 1)
#define pb push_back
#define all(x) x.begin(), x.end()
#define Sz(x) (int)(x.size()) 
void db(int n) { 
    vector<int> a;
    a.pb(0);
    For(i, 1, n) {
        a.pb(i * 2 - 2);
    }
    vector<int> pre(n + 1);
    For (i, 1, n) {
        pre[i] = pre[i - 1] + a[i];
    }
    vector<int> b;
    For(i, 1, n) {
        For(j, i + 1, n) {
            // cout << i << ' ' << j << '\n';
            b.pb(pre[j] - pre[i - 1]);
        }
    }
    sort(all(b));
    b.erase(unique(all(b)), b.end());
    if (b.empty()) {
        cout << '\n';
        return;
    }
    For(i, 0, Sz(b) - 1) {
        cout << b[i] << '\n';
    }
    cout << '\n';
}

2
6
10
12
14 
... 数据很大 会明显发现1~1000种恰好少以4 8 16.. 二次幂 但是不包括2,其余偶数全部在,
  那么大概guess出想要找到第k小,就可以去除4 8 16这些二的幂次 进行二分即可

Code

vector<i64> ans;
 
void init() { 
    for (i64 i = 2; i <= 63; i++) {
        i128 x = 1LL << i;
        ans.push_back(x);
    }
}
 
std::ostream &operator<<(std::ostream &os, i128 n) {
    if (n == 0) {
        return os << 0;
    }
    std::string s;
    while (n > 0) {
        s += char('0' + n % 10);
        n /= 10;
    }
    std::reverse(s.begin(), s.end());
    return os << s;
}
 
void Solve() {
    i64 k;
    cin >> k; 
    auto check = [&](i128 x) -> i128 {
        if (x < 2) return 0;
        i128 cnt = x / 2;  
        i128 it = upper_bound(all(ans), x) - ans.begin();  
        return cnt - it; 
    } ; 
    i128 l = 0, r = 4e18; 
    while (l < r) {
        i128 mid = l + (r - l) / 2;
        if (check(mid) >= k) {
            r = mid;
        } else {
            l = mid + 1;
        }
    } 
    if (l & 1) {
        l++;
    } 
    cout << l << '\n';
}