I did by finding Euler Circuit of a complete graph of odd degree, and then adjusting it by removing the last edge and adding path of required length.

What I did was...
Three case :

1) $n \mod 3 == 0$

6, 10, 15, 6*p1, 10*p2, 15*p3, 6*p4, 10*p5, ... so on

2) $n \mod 3 == 1$

21, 14, 10, 15, 6*p1, 10*p2, 15*p3, 6*p4, 10*p5, ... so on

3) $n \mod 3 == 2$

33, 77, 14, 10, 15, 6*p1, 10*p2, 15*p3, 6*p4, 10*p5, ... so on

Where p1,p2,p3,... are primes greater than 11...

(as I used primes till 11 in these case to make sequence N%3==0)

One interesting property which I found was that "No three consecutive odd numbers have a common divisor" starting from 3. So, I tried multiplying 3 to the first 2 elements, then multiplied 5 to the second and third elements, then multiplied 7 to the third and fourth elements and so on... (completing the cycle with a prime number for the last and the first elements just in case if the first element and the second last element don't have a common divisor). This passed subtask-1, but didn't pass subtask-2. Maybe it was because of generating numbers > 10^9. Is there a modification I could do to this approach to get it ACed?

I am a bit disappointed that the property of coprimes weren't considered in the editorial. In fact, it is not necessary at all to generate prime numbers.

If we consider each positive integer as a multiset of its prime factors, then the problem could be reduced to:

Find N ordered multisets of integers that for every 2 consecutive sets, the intersection is not empty, but the each intersection of 2 consecutive intersections is an empty set, i.e. coprime to each other.

Therefore if we find a list of consecutive coprimes, the result would be product of 2 consecutive elements of that list.

Implementation in Python:

from math import gcd
from itertools import cycle, islice
from sys import stdin

CANDIDATES = cycle(range(2, 30000))

a, b, new = 30011, 30013, {}
coprimes, result = [a, b], [a * b]
for _ in range(49998):
    for i in CANDIDATES:
        if new.get(b * i, True) and gcd(a, i) == 1 == gcd(b, i):
            coprimes.append(i)
            a, b = b, i
            break
    new[a * b] = False
    result.append(a * b)

next(stdin)
for N in map(int, stdin):
    N -= 1
    print(coprimes[N] * 30011, end=' ')
    print(*islice(result, N))

Since the first 2 coprimes are hardcoded to be primes, we can reuse the list for every N in the given constraint, thus the overall complexity is drastically reduced.

What i did was ... 1)found out out some 4000 consecutive prime numbers using seive

2)appended them to a new array under the condition that the product of two consecutive numbers does not cross 10^9.This helped me pass the first sub task.

3)For the next part I started my series from 5 and accepted primes at different intervals starting from 2 and incrementing interval by 1 and also a different starting point for every interval.

4)this helped me generate all(50000 in this case) possible combinations of products of primes and all i had to do was consider the first 550 prime numbers, and also including the previous

The time complexity was O(d^3),d=550. but since the inner loops depended on the interval it worked even faster.

my solution does not use generating primes at all. we forget some facts about numbers with small difference and their common factors.Here is my solution

https://www.codechef.com/viewsolution/22244894

Urzbeckuvayy...can u please explain ur logic a bit?

in java only 12500 primes can be stored for 50000 Bytes

https://www.codechef.com/viewsolution/22395844

what is wrong in this solution?.

Can you elaborate how you got this pattern @l_returns?

in the setter's solution can anyone explain how and int x = 4 and if (i == n - 1) {ind = 5;} how this two constant are fixed ? I tried some other constants and some of them give right answer and some of them dont't; @eartemov