**π‘ Problem Formulation:** We are tasked with determining whether a given number `n`

can be expressed as the sum of `k`

consecutive natural numbers. For example, if `n = 9`

and `k = 2`

, the output should be `True`

because 9 can be represented as 4+5. The challenge lies in finding an efficient algorithm to identify such pairs for any given `n`

and `k`

.

## Method 1: Iteration Over Range

This method involves checking every range of `k`

consecutive numbers up to `n`

to see if their sum equals `n`

. It is straightforward but not the most efficient for large values of `n`

and `k`

. The function `can_sum_to_n_k_times(n, k)`

will iterate from 1 to `n-k+1`

and sum each range of `k`

numbers to determine if the sum equals `n`

.

Here’s an example:

def can_sum_to_n_k_times(n, k): for start in range(1, n-k+2): if sum(range(start, start+k)) == n: return True return False print(can_sum_to_n_k_times(9, 2))

**Output:** True

This code snippet defines a function that loops through a range starting from 1, computing the sum of `k`

consecutive numbers, and compares it to `n`

. If a match is found, it returns `True`

. Otherwise, it concludes with `False`

after the loop. This method is easy to understand but escalates in complexity as `n`

and `k`

grow.

## Method 2: Mathematical Formula

Using a mathematical approach, we can determine if `n`

can be shown as a sum of `k`

consecutive numbers by utilizing the formula for the sum of an arithmetic series. For a range to sum up to `n`

, the starting point of the series has to be a positive integer, which can be deduced from the rearranged formula. The function `is_sum_possible(n, k)`

performs this calculation.

Here’s an example:

def is_sum_possible(n, k): # Formula to find the starting point: x = (n/k) - (k-1)/2 x = (n/k) - (k-1)/2 return x.is_integer() and x > 0 print(is_sum_possible(9, 2))

**Output:** True

This method uses the arithmetic series formula to find the starting point `x`

of the sequence. If `x`

is a positive integer, `n`

can be expressed as the sum of `k`

consecutive numbers. It’s significantly more efficient than the first method as it performs the calculation in constant time rather than relying on an iterative sum.

## Method 3: Optimized Iteration Based on Midpoint

A variation of the iterative approach can be employed, which relies on the fact that for a series to sum to `n`

, `n/k`

will nearly always be around the midpoint of the series. This optimized version only checks sums around this midpoint, significantly reducing the number of iterations compared to the first method. The function `optimized_can_sum(n, k)`

implements this optimization.

Here’s an example:

def optimized_can_sum(n, k): midpoint = n // k lower_bound = max(1, midpoint - k // 2) upper_bound = lower_bound + k for start in range(lower_bound, upper_bound): if sum(range(start, start + k)) == n: return True return False print(optimized_can_sum(9, 2))

**Output:** True

This method reduces the number of iterations by only looking at a range of numbers surrounding what would be the midpoint of the series that adds up to `n`

. It maintains simplicity while being more efficient than a full iteration through all possible ranges.

## Method 4: Binary Search

Binary search can be applied to find the starting number where the sum of `k`

consecutive numbers is equal to `n`

. This method works by continuously halving the search space until it finds the correct starting point or determines that no such sequence exists. The function `binary_search_sum(n, k)`

demonstrates implementing a binary search algorithm for this purpose.

Here’s an example:

def binary_search_sum(n, k): left, right = 1, n while left <= right: mid = (left + right) // 2 current_sum = k * (2*mid + k - 1) // 2 if current_sum == n: return True elif current_sum < n: left = mid + 1 else: right = mid - 1 return False print(binary_search_sum(9, 2))

**Output:** True

The binary search algorithm seeks to minimize the range in which the starting number could lie. By comparing the sum of the series with `n`

, it adjusts the search range until the solution is found or is proven to not exist. This algorithm has a logarithmic time complexity, proving much faster than a simple iterative approach.

## Bonus One-Liner Method 5: Symmetry Property

With the understanding that if `n`

can be divided by `k`

with a remainder of `k/2`

, the sum of `k`

consecutive numbers centering around `n/k`

will equal `n`

. This method leverages this symmetry property through a simple one-liner conditional check. The function `is_symmetric_sum(n, k)`

encapsulates this property.

Here’s an example:

is_symmetric_sum = lambda n, k: n % k == k // 2 print(is_symmetric_sum(9, 2))

**Output:** True

This lambda function quickly checks if `n`

modulo `k`

is equal to `k`

divided by 2. The use of this symmetry property offers a very concise and efficient solution, applicable in cases where the conditions are met.

## Summary/Discussion

**Method 1: Iteration Over Range.**Easy to understand. Not efficient for large numbers.**Method 2: Mathematical Formula.**Very efficient with constant time complexity. Requires understanding of arithmetic series.**Method 3: Optimized Iteration Based on Midpoint.**More efficient than full iteration. Still less efficient than mathematical solutions.**Method 4: Binary Search.**Efficient with logarithmic time complexity. The best for a range of practical scenarios.**Method 5: Symmetry Property.**Extremely succinct and fast. Applicable only when specific divisibility conditions are met.