For Python3 Platform

The efficient way to solve this problem lies in understanding the concept of binomial expansion in mathematics

from math import comb

def solve(n):
    res = []
    
    for r in range(n//2+1):
        res.append(comb(n, r) % pow(10, 9))
    
    if(n % 2 == 0):
        res += res[:-1][::-1]
    else:
        res += res[::-1]
    
    return res

t = int(input())
for i in range(t):
    n = int(input())
    
    result = solve(n)
    
    print(*result)

great post

An nCr table, also known as a combination table, is a systematic way to list combinations of items, especially useful in probability and statistics. It typically shows the number of ways to choose r items from a set of n items without regard to the order of selection. Such tables are invaluable in fields like combinatorics, where precise calculations of permutations and combinations are required. They serve as quick references, aiding in computations for various scenarios, from binomial coefficients to complex probability distributions. For those interested in exploring practical applications or enhancing their understanding of these concepts, resources like Studio Trataka's Qutub Copper Candle Stand T-Light, available at https://studiotrataka.com/product/qutub-copper-candle-stand-t-light/, can add a touch of elegance to your study environment while you delve into the intricacies of combinatorial mathematics.

Python math.comb is so fast that the following brute-force one-liner works:

def solve(n):
    return [math.comb(n, i)%10**9 for i in range(n+1)]

You can watch Raymond Hettinger's amazing talk "Numerical Marvels Inside Python" to understand why math.comb is so fast, including with very very large n (group theory ahead!).

def solve(n): ncr = [] for r in range (n//2 + 1): ncr.append(comb(n, r)%(10**9)) if n % 2== 0: return ncr + ncr[:-1][::-1] else: return ncr + ncr[::-1]