Haskell, 45 bytes

n!a=not$any(n!)[x|x<-[n..a-1],mod a x<1]||n>a 

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I define a nice recursive function (!):

n!a checks if any factor of a, in the range [n,a-1], is an n-prime. Then it negates the result. It also makes sure that n>a

Python 2, 39 37 bytes

Thanks to Halvard Hummel for -2 bytes.

f=lambda n,i:n==i or i>i%n>0<f(n+1,i) 

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Python 3, 45 bytes

lambda i,k:(i>k)<all(k%r for r in range(i,k)) 

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How it works

This takes two integers as input, i and k. First checks if k ≥ i. Then generates the range [i, k) and for each integer N in this range, checks if N is coprime to k. If both conditions are fulfilled, then k is an i-prime.

Husk, 6 5 bytes

εf≥⁰Ḋ 

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Thanks to Leo for -1 byte.

Explanation

εf≥⁰Ḋ Inputs are n and k. Ḋ Divisors of k. f Keep those that are ≥⁰ at least n. ε Is the result a one-element list? 

R, 44 37 bytes

function(a,n)a==n|a>n&all(a%%n:(a-1)) 

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-7 bytes thanks to Giuseppe

Returns TRUE if

• a is equal to n or (a==n|)
• a is greater than n and (a>n&) for every number k from n to a-1, a is not evenly divisible by k (all(a%%n:(a-1)))

Returns FALSE otherwise

J, 30 bytes

>:*(-{1=[:+/0=[:|/~]+i.@[)^:>: 

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Takes starting value as the right argument and the value to check at the left argument.

I messed up originally and didn't account for left arguments less than the starting prime. I'm somewhat unhappy with the length of my solution now.

Explanation

Let x be the left argument (the value to check) and y be the right argument (the starting prime).

>:*(-{1=[:+/0=[:|/~]+i.@[)^:>: ^:>: Execute left argument if x >= y i.@[ Create range [0..x] ]+ Add y to it (range now: [y..x+y]) |/~ Form table of residues 0= Equate each element to 0 +/ Sum columns 1= Equate to 1 -{ Take the element at position x-y >:* Multiply by result of x >= y 

Notes

The element at position x-y is the result of primality testing for x (since we added y to the original range).

Multiplying by x >: y ensures that we get a falsey value (0) for x less than y.

JavaScript (ES6), 33 32 30 bytes

Takes input in currying syntax (n)(a). Returns a boolean.

n=>p=(a,k=a)=>k%--a?p(a,k):a<n 

Demo

let f = n=>p=(a,k=a)=>k%--a?p(a,k):a<n for(n = 2; n <= 12; n++) { for(a = n, P = []; P.length < 31; a++) { f(n)(a) && P.push(a); } O.innerText += 'n=' + n + ': ' + P.join(',') + '\n'; }

<pre id=O></pre>

Haskell, 30 bytes

2 bytes saved thanks to H.PWiz's idea which was borrowed from flawr's answer

n!a=[1]==[1|0<-mod a<$>[n..a]] 

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Ok since its been a while, and the only Haskell answer so far is 45 btyes, I decided to post my own answer.

Explanation

This function checks that the only number between n and a that a is divisible by is a itself.

Now the definition only mentions n-primes smaller than a, so why are we checking all these extra numbers? Won't we have problems when a is divisible by some n-composite larger than n?

We won't because if there is an n-composite larger than n it must be divisible by a smaller n-prime by definition. Thus if it divides a so must the smaller n-prime.

If a is smaller than n [n..a] will be [] thus cannot equal [1] causing the check to fail.

Pip, 23 19 14 bytes

b>=a&$&b%(a,b) 

Shortest method is a port of Mr. Xcoder's Python answer. Takes the smallest prime and the number to test as command-line arguments. Try it online!

C, 55 bytes

f(n,a,i){for(i=a;i-->n;)a%i||f(n,i)&&(i=0);return-1<i;} 

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53 bytes if multiple truthy return values are allowed:

f(n,a,i){for(i=a;i-->n;)a%i||f(n,i)&&(i=0);return~i;} 

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dc, _{40 34} 37 bytes

[0p3Q]sf?se[dler%0=f1+dle>u]sudle>u1p 

I would have included a TIO link, but TIO seems to be carrying a faulty distribution of dc seeing as how this works as intended on my system but the Q command functions erroneously on TIO. Instead, here is a link to a bash testing ground with a correctly functioning version of dc:

Demo It!

APL (Dyalog), 24 bytes

{⍵∊o/⍨1=+/¨0=o|⍨⊂o←⍺↓⍳⍵} 

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How?

⍳⍵ - 1 to a

o←⍺↓ - n to a, save to o

o|⍨⊂o - modulo every item in o with every item in o

0= - check where it equals 0 (divides)

+/¨ - sum the number of divisions

1= - if we have only one then the number is only divided by itself

o/⍨ - so we keep these occurences

⍵∊ - is a in that residual array?

Jelly, 8 5 bytes

rḍṛċ< 

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This outputs 0 or 1 for "yes" and a positive integer greater than 1 for "no". This really stretches the "output two distinct values", so I've included a more reasonable 6 byte version:

rḍṛṖi< 

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Outputs 0 for "yes, it is \$n\$-prime" and a positive integer for "no it is not \$n\$-prime". +1 byte to use just 0 and 1.

How they work

rḍṛċ< - Main link. Takes n on the left and a on the right r - Range from n to a inclusive ṛ - Right; Yield a ḍ - For each element in the range, is it divisible by a? < - Is n < a? ċ - Count the occurrences of (n < a) in the divisibility range 

This works because in the range \$[n, a]\$, only one integer can be divisible by \$a\$ if \$a\$ is \$n\$-prime: \$a\$. Therefore, the range, after divisibility, would look like [0, 0, 0, ..., 1]. However, this doesn't check the condition that \$n\$ is the smallest \$n\$-prime, as r will generate a descending range if \$n < a\$. Therefore, we check that \$n < a\$ and count the occurrences of that (1 or 0) in the divisibility range. If \$n > a\$, there will be more than one 0 in the range, and if \$a\$ is not \$n\$-prime, but \$n < a\$, there will be more than 1 in the range.

If \$n = a\$, then r will yield an empty list and ċ will return 0, no matter what the result of < is.

rḍṛṖi< - Main link. Takes n on the left and a on the right r - Range from n to a inclusive ṛ - Right; Yield a ḍ - For each element in the range, is it divisible by a? Ṗ - Remove the final element, 1 < - Is n < a? i - Index of (n < a) or 0 if not found 

JavaScript (Node.js), 27 bytes

i=>f=n=>i==n||i>i%n&&f(n+1) 

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Port of my Python answer, takes input in currying syntax: m(number)(first prime)

JavaScript ES5, 34 Bytes

for(a=i=(p=prompt)();a%--i;);i<p() 

Add++, 20 bytes

L,2Dx@rBcB%B]b*!!A>* 

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L, - Create a lambda function - Example arguments: [5 9] 2D - Copy below; STACK = [5 9 5] x - Repeat; STACK = [5 9 [9 9 9 9 9]] @ - Reverse; STACK = [[9 9 9 9 9] 5 19] r - Range; STACK = [[9 9 9 9 9] [5 6 7 8 9]] Bc - Zip; STACK = [[9 5] [9 6] [9 7] [9 8] [9 9]] B% - Modulo; STACK = [4 3 2 1] B] - Wrap; STACK = [[4 3 2 1]] b* - Product; STACK = [24] !! - Boolean; STACK = [1] A - Arguments; STACK = [1 5 9] > - Greater; STACK = [1 1] * - Product; STACK = [1]