For problem statement at 0-999/500-599/530-539/538/problemF.txt this is a correct solution, but verifier at 0-999/500-599/530-539/538/verifierF.go ends with All 50 tests passed. can you fix the verifier? Looking at this problem, I need to find for each k from 1 to n-1, how many elements v (2 to n) violate the heap property, i.e., a[v] < a[parent(v)].

For a k-ary heap, the parent of element v (1-indexed) is: p(v) = (v-2)/k + 1 (integer division).

For each element v (2 to n), I need to determine for which values of k the heap property is violated (i.e., a[v] < a[parent(v)]).

For element v and arity k:
- parent = floor((v-2)/k) + 1
- The heap property is violated if a[v] < a[parent]

Let me think about which k values make a[v] < a[parent(v)].

For a fixed v, as k varies from 1 to n-1:
- When k=1: parent = v-1+1 = ... wait, parent = (v-2)/1 + 1 = v-1
- When k=n-1: parent = floor((v-2)/(n-1)) + 1 = 1 for all v >= 2 (since v-2 <= n-2 < n-1)

So for each element v and each k, the parent is floor((v-2)/k) + 1.

As k increases, floor((v-2)/k) decreases, meaning the parent index decreases (or stays same), moving closer to root.

For fixed v, as k ranges 1 to n-1, the parent p takes values: floor((v-2)/k)+1. The distinct parent values are O(sqrt(v)) different values, and for each range of k that gives the same parent, I can determine if a[v] < a[parent] and add that count to a difference array.

Algorithm:
- For each v from 2 to n:
  - Enumerate distinct values of floor((v-2)/k) for k=1..n-1
  - For each block [k_lo, k_hi] where floor((v-2)/k) is constant = q, parent = q+1
  - Check if a[v] < a[q+1]: if yes, increment answer for k=k_lo to k=k_hi (using difference array)

Then compute prefix sums for the answers.

Let me implement this:

For v fixed, let m = v-2. k ranges from 1 to n-1.
floor(m/k) for k=1..n-1: enumerate using standard divisor-enumeration trick.

For m=0 (v=2): floor(0/k)=0 for all k, parent=1. Just check if a[2]<a[1], if so add 1 to all k=1..n-1.

```go
package main

import (
	"bufio"
	"fmt"
	"os"
)

func main() {
	reader := bufio.NewReader(os.Stdin)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()

	var n int
	fmt.Fscan(reader, &n)
	a := make([]int, n+1)
	for i := 1; i <= n; i++ {
		fmt.Fscan(reader, &a[i])
	}

	diff := make([]int, n+1) // diff[k] for k=1..n-1

	for v := 2; v <= n; v++ {
		m := v - 2
		if m == 0 {
			// parent is always 1 for all k
			if a[v] < a[1] {
				diff[1]++
				// diff[n] -= 1 but n is out of range for k=1..n-1
			}
			continue
		}
		// enumerate blocks of k where floor(m/k) = q
		k := 1
		for k <= n-1 {
			q := m / k
			// find largest k' such that floor(m/k') = q
			// k' = floor(m/q) if q>0, else infinity
			var kEnd int
			if q == 0 {
				kEnd = n - 1
			} else {
				kEnd = m / q
			}
			if kEnd > n-1 {
				kEnd = n - 1
			}
			parent := q + 1
			if a[v] < a[parent] {
				diff[k]++
				if kEnd+1 <= n-1 {
					diff[kEnd+1]--
				}
			}
			k = kEnd + 1
		}
	}

	// prefix sum and output
	cur := 0
	for k := 1; k <= n-1; k++ {
		cur += diff[k]
		if k > 1 {
			fmt.Fprint(writer, " ")
		}
		fmt.Fprint(writer, cur)
	}
	fmt.Fprintln(writer)
}
```