Un número secundario es un entero positivo cuyos factores primos (sin multiplicidad) son todos menores o iguales a su raíz cuadrada. 4es un número secundario, porque su único factor primo es 2, que es igual a su raíz cuadrada. Sin embargo, 15no es un número secundario, porque tiene 5como factor primo, que es mayor que su raíz cuadrada ( ~ 3.9). Debido a que todos los números primos se tienen como factores primos, ningún número primo es un número secundario. Los primeros números secundarios son los siguientes:

1, 4, 8, 9, 12, 16, 18, 24, 25, 27, 30, 32, 36, 40, 45, 48, 49, 50, 54, 56

Un número terciario se define de manera similar, excepto que todos los factores primos deben ser menores o iguales que su raíz cúbica. Los primeros números terciarios son los siguientes:

1, 8, 16, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 125, 128, 135, 144, 150, 160, 162

En general, un número n-ary es uno cuyos factores primos son todos menores o iguales a su enésima raíz. Por lo tanto, un número entero positivo $x$ es un nnúmero -ario si cada uno de sus factores primos $p$ satisface $p^n ≤ x$ . Por lo tanto, los números primarios son todos enteros positivos (todos los factores primos menores o iguales a ellos mismos), los números cuaternarios tienen todos sus factores primos menores o iguales que su cuarta raíz, y así sucesivamente.

El reto

Teniendo en cuenta los números enteros ky ncomo entradas, la salida de la kXX nnúmero ary. kpuede ser un índice cero o uno (su elección), y nsiempre será positivo.

Ejemplos

Estos son los primeros 20 elementos en cada secuencia de hasta 10 números arios:

Primary: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
Secondary: 1, 4, 8, 9, 12, 16, 18, 24, 25, 27, 30, 32, 36, 40, 45, 48, 49, 50, 54, 56
Tertiary: 1, 8, 16, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 125, 128, 135, 144, 150, 160, 162
Quarternary: 1, 16, 32, 64, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512
5-ary: 1, 32, 64, 128, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, 1024, 1152
6-ary: 1, 64, 128, 256, 512, 729, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2187, 2304, 2592
7-ary: 1, 128, 256, 512, 1024, 2048, 2187, 2304, 2592, 2916, 3072, 3456, 3888, 4096, 4374, 4608, 5184, 5832, 6144, 6561
8-ary: 1, 256, 512, 1024, 2048, 4096, 6561, 6912, 7776, 8192, 8748, 9216, 10368, 11664, 12288, 13122, 13824, 15552, 16384, 17496
9-ary: 1, 512, 1024, 2048, 4096, 8192, 16384, 19683, 20736, 23328, 24576, 26244, 27648, 31104, 32768, 34992, 36864, 39366, 41472, 46656
10-ary: 1, 1024, 2048, 4096, 8192, 16384, 32768, 59049, 62208, 65536, 69984, 73728, 78732, 82944, 93312, 98304, 104976, 110592, 118098, 124416

Jalea , 12 bytes

Æf*³<‘Ạ
1Ç#Ṫ

Toma n y k (un índice) como argumentos de línea de comandos.

Pruébalo en línea!

Cómo funciona

1Ç#Ṫ     Main link. Left argument: n. Right argument: k

1        Set the return value to 1.
 Ç#      Execute the helper link above for r = 1, 2, 3, ... until k of them return
         a truthy value. Yield the list of all k matches.
   Ṫ     Tail; extract the last match.


Æf*³<‘Ạ  Helper link. Argument: r

Æf       Compute all prime factors of r.
  *³     Elevate them to the n-th power.
    <‘   Compare all powers with r + 1.
      Ạ  All; return 1 if all comparisons were true, 0 if one or more were not.

Brachylog , 21 bytes

:[1]cyt
,1|,.$ph:?^<=

Pruébalo en línea!

Esta respuesta tiene un índice.

Explicación

Input: [N:K]

:[1]cy              Retrieve the first K valid outputs of the predicate below with N as input
      t             Output is the last one

,1                  Output = 1
  |                 Or
   ,.$ph            Take the biggest prime factor of the Output
        :?^<=       Its Nth power is less than the Output

JavaScript (ES7), 95 90 bytes

Razonablemente rápido pero tristemente limitado por el número máximo de recurrencias.

f=(k,n,i=1)=>(F=(i,d)=>i-1?d>1?i%d?F(i,d-1):F(i/d,x):1:--k)(i,x=++i**(1/n)|0)?f(k,n,i):i-1

Cómo funciona

En lugar de factorizar un número entero i y verificar que todos sus factores primos son menores o iguales que x = floor (i ^{1 / n} ) , intentamos validar el último supuesto directamente. Ese es el propósito de la función interna F () :

F = (i, d) =>         // given an integer i and a divisor d:
  i - 1 ?             //   if the initial integer is not yet fully factored
    d > 1 ?           //     if d is still a valid candidate
      i % d ?         //       if d is not a divisor of i
        F(i, d - 1)   //         try again with d-1 
      :               //       else
        F(i / d, x)   //         try to factor i/d
    :                 //     else
      1               //       failed: yield 1
  :                   //   else
    --k               //     success: decrement k

Verificamos si algún número entero d en [2 ... i ^{1 / n} ] divide i . Si no, la suposición no es válida y devolvemos 1 . En caso afirmativo, repetimos el proceso de manera recursiva en i = i / d hasta que falle o el entero inicial se tenga en cuenta por completo ( i == 1 ), en cuyo caso disminuiremos k . A su vez, la función externa f () se llama de forma recursiva hasta k == 0 .

Nota: Debido a errores de redondeo de coma flotante como 125**(1/3) == 4.9999…, por ejemplo , el valor real calculado para x es piso ((i + 1) ^{1 / n} ) .

Manifestación

(Aquí con una versión ES6 de 97 bytes para una mejor compatibilidad).

f=(k,n,i=1)=>(F=(i,d)=>i-1?d>1?i%d?F(i,d-1):F(i/d,x):1:--k)(i,x=Math.pow(++i,1/n)|0)?f(k,n,i):i-1

console.log(f(17, 3)); // 144
console.log(f(13, 4)); // 243
console.log(f(4, 10)); // 4096

JavaScript (ES7), 93 79 bytes

f=(k,n,g=(i,j=2)=>i<2?--k?g(++m):m:j**n>m?g(++m):i%j?g(i,j+1):g(i/j,j))=>g(m=1)

No podía entender la respuesta de Arnauld, así que escribí la mía y convenientemente llegó en dos bytes menos. Editar: guardado 14 bytes con ayuda de @ETHproductions. Sin golf:

function ary(k, n) {
    for (var c = 1;; c++) {
        var m = c;
        for (var i = 2; m > 1 && i ** n <= c; i++)
            while (m % i == 0) m /= i;
        if (m == 1 && --k == 0) return c;
    }
}

Haskell, 86 bytes

m#n=(0:1:filter(\k->last[n|n<-[2..k],all((>0).rem n)[2..n-1],k`rem`n<1]^n<=k)[2..])!!m

Explicación:

( %%%%denota el código de la línea de arriba)

    [n|n<-[2..k],all((>0).rem n)[2..n-1],k`rem`n<1]  -- generates all prime factors of k
    last%%%%^n<=k                                    -- checks whether k is n-ary
    (0:1:filter(\k->%%%%)[2..])!!m                   -- generates all n-ary nubmers and picks the m-th
     m#n=%%%%                                        -- assignment to the function #

Pyth, 13 bytes

e.f.AgL@ZQPZE

Pruébalo en línea.

Funciona realmente igual que la solución Jelly.

e.f.AgL@ZQPZE
                 Implicit: read n into Q.
            E    Read k.
 .f              Find the first k integers >= 1 for which
   .A            all
          P      prime factors of
           Z     the number
     gL          are at most
         Q       the n'th
       @         root of
        Z        the number.
e                Take the last one.

Perl 6, 88 bytes

->\k,\n{sub f(\n,\d){n-1??n%%d??f n/d,d!!f n,d+1!!d};(1,|grep {$_>=f($_,2)**n},2..*)[k]}

My accidental insight is that you don't need to look at every factor of n, just the largest one, which is what the internal function f computes. Unfortunately, it blows the stack with larger inputs.

Robustness can be improved by adding a primality test, using the built-in is-prime method on Ints, at a cost of several more characters.

Husk, 10 bytes

!fS≤ȯ^⁰→pN

Try it online!

Explanation

Took me a while to figure out using → on the empty list returns 1 instead of -Inf for ▲ which leaves 1 in the list (otherwise that would cost 2 bytes to prepend it again):

!fS≤(^⁰→p)N  -- input n as ⁰ and k implicit, for example: 4 3
!f        N  -- filter the natural numbers by the following predicate (example on 16):
  S≤(    )   --   is the following less than the element (16) itself?
        p    --   ..prime factors (in increasing order): [2]
       →     --   ..last element/maximum: 2
     ^⁰      --   ..to the power of n: 16
             --   16 ≤ 16: yes
             -- [1,16,32,64,81..
!            -- get the k'th element: 32

R, 93 bytes

f=function(k,n){x='if'(k,f(k-1,n)+1,1);while(!all(numbers::primeFactors(x)<=x^(1/n)))x=x+1;x}

Zero-indexed.

It is a recursive function that just keeps going until it finds the next number in line. Uses to numbers package to find the prime factors.

MATL, 21 bytes

x~`@YflG^@>~A+t2G<}x@

This solution uses one-based indexing and the inputs are n and k, respectively.

Try it Online!

Explanation

x       % Implicitly grab the first input and delete it
~       % Implicitly grab the second input and make it FALSE (0) (this is the counter)
`       % Beginning of the do-while loop
  @Yf   % Compute the prime factors of the current loop index
  1G^   % Raise them to the power of the first input
  @>    % Determine if each value in the array is greater than the current index
  ~A    % Yield a 0 if any of them were and a 1 if they weren't
  +     % Add this to the counter (increments only for positive cases)
  t2G<  % Determine if we have reached a length specified by the second input
}       % After we exit the loop (finally), ...
x@      % Delete the counter and push the current loop index to the stack
        % Implicitly display the result

Brachylog v2, 16 bytes

{∧.ḋ,1⌉;?^≤}ᶠ⁽t

Try it online!

Credit to Dennis's Jelly solution for getting me thinking in the right direction.

Explanation

Here's a slightly ungolfed version that's easier to parse:

↰₁ᶠ⁽t
∧.ḋ,1⌉;?^≤

Helper predicate (line 2): input is the exponent n, output is unconstrained:

∧           Break implicit unification
 .ḋ         Get the prime factors of the output
   ,1       Append 1 (necessary because 1 has no prime factors and you can't take
            the max of an empty list)
     ⌉      Max (largest prime factor, or 1 if the output is 1)
      ;?    Pair that factor with the input (n)
        ^   Take factor to the power of n
         ≤  Verify that the result is less than or equal to the output

Main predicate (line 1): input is a list containing the index k (1-based) and the exponent n; output is unconstrained:

  ᶠ⁽   Find the first k outputs of
↰₁     calling the helper predicate with n as input
    t  Get the last (i.e. kth) one

APL(NARS), 53 chars, 106 bytes

r←a f w;c
c←0⋄r←1
→3×⍳∼∧/(πr)≤a√r⋄→0×⍳w≤c+←1
r+←1⋄→2

test:

  1 f¨1..20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 
  2 f¨1..20
1 4 8 9 12 16 18 24 25 27 30 32 36 40 45 48 49 50 54 56 
  3 f¨1..20
1 8 16 27 32 36 48 54 64 72 81 96 108 125 128 135 144 150 160 162 
  4 f¨1..20
1 16 32 64 81 96 108 128 144 162 192 216 243 256 288 324 384 432 486 512 
  10 f¨1..20
1 1024 2048 4096 8192 16384 32768 59049 62208 65536 69984 73728 78732 82944 93312 98304 104976 110592 118098 124416 

Python 2, 114 bytes

f=lambda K,n,i=1:K and f(K-all(j**n<=i for j in range(1,i+1)if i%j==0and all(j%k for k in range(2,j))),n,i+1)or~-i

Try it online!

1-indexed function.

Japt, 14 bytes

Takes n as the first input and k (1-indexed) as the second.

@_k e§Vq}aXÄ}g

Try it