05AB1E, 13 bytes

I think it has been 389 days since I last posted something here haha. There is definitely some golfing potential left in this program.

Code

Uses the 05AB1E-encoding.

∞æ¹ùʒPy.¿Q}²£ 

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Explanation

It is worth noting that for two numbers \$n, m \in \mathbb{Z}^+\$ that:

$$ \tag{1} \label{1} \gcd(n, m) \cdot \text{lcm}(n, m) = n \cdot m $$

This means that for two numbers \$n, m \in \mathbb{Z}^+\$ where the \$\gcd(n, m) = 1\$, we can conclude that the \$\text{lcm}(n, m) = n \cdot m\$.

Furthermore, the \$\gcd\$ function is a multiplicative function, which means that if \$n_1\$ and \$n_2\$ are relatively prime, then:

$$ \gcd(n_1 \cdot n_2, m) = \gcd(n_1, m) \cdot \gcd(n_2, m) $$

From this, we obtain the fact that:

$$ \tag{2} \label{2} \gcd(a, bc) = 1 \iff \gcd(a, b) = 1 \wedge \gcd(a, c) = 1 $$

Let us denote a \$k\$-tuple of positive integers as \$S = \{x_1, x_2, \dots, x_k\}\$. A set \$S\$ is pairwise coprime, if and only if:

$$ \tag{3} \label{3} \forall a, b \in S \wedge a \not = b \rightarrow \gcd(a, b) = 1 $$

Using Equations \$\eqref{1}, \eqref{2}\$ and \$\eqref{3}\$, we can conclude that a set \$S = \{x_1, x_2, \dots, x_k\}\$ is pairwise coprime, if and only if:

$$ \text{lcm}(x_1, x_2, \dots, x_k) = \prod_{x \in S} x $$

Code Explanation

 ∞æ¹ùʒPy.¿Q}²£ ∞æ # Powerset of the infinite list [1, ..., ∞]. ¹ù # Keep only lists of length k. ʒ } # Filter. Keep lists where the P # product of the list Q # is equal to y.¿ # the least common multiple of the list ²£ # Retrieve the first n elements. 

Husk, 9 bytes

↑fËoε⌋`ṖN 

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Explanation

A straightforward solution, not the most exciting.

↑fËoε⌋`ṖN Implicit inputs, say k=3, n=2. N Natural numbers: [1,2,3,4,.. `Ṗ All k-element subsets: [[1,2,3],[2,3,4],[1,3,4],.. ` flips the arguments of Ṗ since it expects the number first. f Keep those that satisfy this: Ë All pairs x,y (not necessarily adjacent) satisfy this: ⌋ their gcd oε is at most 1. Result is all pairwise coprime subsets: [[1,2,3],[1,3,4],.. ↑ Take the first n: [[1,2,3],[1,3,4]] 

Jelly, 16 bytes

‘×ÆNœcŒcg/€$ÐṂḣ⁸ 

A dyadic Link accepting n on the left and k on the right.

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There must be a better way than this inefficient monstrosity! It'll time out for quite small inputs since it inspects all k-tuples of the natural numbers up to the (n+1)*k-th prime! (The +1 is only needed to handle n=0.)

Wolfram Language (Mathematica), 106 bytes

(s=Range[#2#];If[#==1,List/@s,SortBy[Select[s~(S=Subsets)~{#},Union[GCD@@@#~S~{2}]=={1}&],Last][[;;#2]]])& 

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Python 3, 153 bytes

lambda n,k,R=range:[[*t,r]for r in R(n+k+2)for t in combinations(R(1,r),k-1)if all(sum(x%i<1for x in[*t,r])<2for i in R(2,r))][:n] from itertools import* 

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A function that takes n, k as arguments and returns out the list of n co-prime k-tuples.

The tuple are generated with increasing maximum, so it's guaranteed that every co-prime tuple will eventually be printed as n increases.

Charcoal, 58 bytes

ＮθＮη≔⁰ζ⊞υ⟦⟧Ｗ‹ＬΦυ⁼Ｌκθη«≦⊕ζＦΦυ⬤κ⬤…²ζ∨﹪μξ﹪ζξ⊞υ⁺⟦ζ⟧κ»Ｉ…Φυ⁼Ｌιθη 

Try it online! Link is to verbose version of code. Explanation:

ＮθＮη 

Input k and n.

≔⁰ζ⊞υ⟦⟧ 

Start the master list with a 0-tuple whose largest number is 0.

Ｗ‹ＬΦυ⁼Ｌκθη« 

Repeat until we have at least k n-tuples.

≦⊕ζ 

Increment the candidate number.

ＦΦυ⬤κ⬤…²ζ∨﹪μξ﹪ζξ 

Filter out all of the existing tuples where at least one member has a common factor with the candidate.

⊞υ⁺⟦ζ⟧κ 

Prepend the candidate to each remaining tuple and push all the resulting tuples back to the master list.

»Ｉ…Φυ⁼Ｌιθη 

Print the first n k-tuples.

JavaScript (ES6), 143 bytes

Takes input as (k)(n).

(k,x=0)=>F=n=>n?(g=a=>x>>i?x>>i++&1?a.some(x=>(C=(a,b)=>b?C(b,a%b):a>1)(x,i))?[]:g([...a,i]):g(a):b=a)(i=[],x++).length-k?F(n):[b,...F(n-1)]:[] 

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Commented

( k, // outer function taking k x = 0 // x = bit mask of integers to include in the tuple ) => // F = n => // F = recursive function taking n n ? // if n is not equal to 0: ( g = a => // g is a recursive function taking a[]: x >> i ? // if x is greater than or equal to 2**i: x >> i++ & 1 ? // if the i-th bit is set in x: a.some(x => // for each value x in a[]: ( C = (a, b) => // C tests whether a and b are coprime: b ? // if b is not equal to 0: C(b, a % b) // recursive call with (b, a mod b) : // else: a > 1 // true if *not* coprime )(x, i) // initial call to C with (x, i) ) ? // end of some(); if truthy: [] // abort by returning an empty array : // else: g([...a, i]) // append i to a[] and call g again : // else: g(a) // just call g with a[] unchanged : // else: b = a // done: return a[] and save it in b[] )(i = [], x++) // initial call to g with a = [], i = 0; increment x .length - k ? // if the length of the result is not equal to k: F(n) // just call F with n unchanged : // else: [b, ...F(n - 1)] // append b[] to the final result and decrement n : // else: [] // stop recursion