The sequence of range-exponentiated integers

Question

Consider a triangle where the Nth row (1-indexed) is the array of the first N positive integer powers of N. Here are the first few rows:

N | Triangle

1 | 1
2 | 2 4
3 | 3 9 27
4 | 4 16 64 256
5 | 5 25 125 625 3125
...

Now if we concatenate those powers into a single sequence, we get OEIS A075363:

1, 2, 4, 3, 9, 27, 4, 16, 64, 256, 5, 25, 125, 625, 3125, 6, 36, 216, 1296, 7776, 46656 ...

---

Given an integer N, your task is to return the Nth term of this sequence. You can choose either 0 or 1-indexing.

Test cases

1-indexed:

N  -> Output

1  -> 1
2  -> 2
3  -> 4
5  -> 9
10 -> 256
12 -> 25
15 -> 3125

0-indexed:

N  -> Output

0  -> 1
1  -> 2
2  -> 4
4  -> 9
9  -> 256
11 -> 25
14 -> 3125

Note that these loopholes are forbidden by default. This is code-golf, thus the shortest valid submission in each language wins!

Answer 1

Python 3, 39 bytes

1-indexed

f=lambda x,n=1:n**x*(x<=n)or f(x-n,n+1)

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Answer 2

Husk, 7 bytes

!ṁṠM^ḣN

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1-indexed

Explanation:

      N   Get the list of all natural numbers
 ṁ        Map over each n in that list and then concatenate
  Ṡ  ḣ      Create the range [1,n] then ...
   M^       raise n to the power of each
!         Index into that sequence

Answer 3

Wolfram Language (Mathematica), 32 bytes

r=Range;Flatten[r@#^r@r@#][[#]]&

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Ranges to the power of ranges of ranges...

Answer 4

R, 40 bytes

-2 bytes porting Halvard Hummel's answer

function(N){while(N>T){N=N-T;T=T+1};T^N}

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R, (original answer) 42 bytes

function(N)(rep(1:N,1:N)^sequence(1:N))[N]

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1-indexed.

Answer 5

APL (Dyalog), 15 13 10 bytes

3 bytes saved thanks to @Adám

⊢⊃∘∊⍳*⍳¨∘⍳

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How?

⍳¨∘⍳ - create a range for each number in the range of input

⍳* - raise each number in the range of input to the corresponding powers

∊ - flatten

⊢⊃ - pick the nth element

Answer 6

Jelly, 7 bytes

*R$€Ẏ⁸ị

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-1 thanks to Mr. Xcoder.

1-indexed.

Answer 7

Haskell, 30 28 bytes

Saved 2 bytes thanks to xnor.

([n^m|n<-[1..],m<-[1..n]]!!)

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0-indexed

Answer 8

MATL, 9 bytes

:t!^RXzG)

Indexing is 1-based. Try it online! Or verify all test cases.

Explanation

Consider input 5 as an example.

:     % Implcit input n. Push range [1 2 ... n]
      % STACK: [1 2 3 4 5]
t!^   % Matrix of all pair-wise powers
      % STACK: [1    2    3    4    5;
                1    4    9   16   25;
                1    8   27   64  125;
                1   16   81  256  625;
                1   32  243 1024 3125]
R     % Upper triangular matrix
      % STACK: [1    2    3    4    5;
                0    4    9   16   25;
                0    0   27   64  125;
                0    0    0  256  625;
                0    0    0    0 3125]
Xz    % Nonzeros. Reads values in column-major order
      % STACK: [1; 2; 4; 3; 9; ...; 625; 3125]
G)    % Get n-th entry (1-based). Implcit display
      % STACK: 9

Answer 9

APL (Dyalog), 14 12 bytes

{⍵⌷∊*∘⍳⍨¨⍳⍵}

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Uses 1-indexing

Saved 2 bytes with ↑,/ → ∊, taken from Graham's answer

Note that in the test link, the code requires an extra f←, but this is not counted as per our rules.

Answer 10

Pyth, 8 bytes

@s^RSdSh

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-1 thanks to Steven H..

0-indexed.

Answer 11

05AB1E, 9 bytes

ƒNDLm`}I@

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Explanation

1-indexed.

ƒ           # for N in range [0 ... input]
 N          # push N
  DL        # push range [1 ... N]
    m       # raise N to the power of each in [1 ... N]
     `      # flatten to stack
      }     # end loop
       I@   # get the element at index (input)

Alternative solution over a list instead of a loop

ÝεDLm}˜sè

Answer 12

Perl 6, 29 bytes

{({|($++X**1..$++)}...*)[$_]}

Test it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

  (  # generate the sequence

    {  # code block used to generate each value in the sequence

      |(         # slip the values into the outer sequence

        $++      # post-incremented anonymous state value
        X**      # cross using &infix:«**»
        1 .. $++ # from 1 to post-incremented anonymous state value

      )
    }

    ...          # keep generating values until

    *            # never stop

  )[ $_ ]        # index into the sequence (0-based)
}

Answer 13

Ruby, 34 bytes

->n{w=0;n-=w+=1until n<0;w**=w-~n}

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Answer 14

JavaScript, 30 bytes

-1 byte thanks to Nahuel Fouilleul

f=(x,n=1)=>x>n?f(x-n,n+1):n**x

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Answer 15

Python 2, 57 bytes

lambda N:[n**-~e for n in range(N+2)for e in range(n)][N]

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0-indexed alternative 74 bytes long version.

Answer 16

Jelly,  15  12 bytes

_{Probably not optimal.}

-3 thanks to Erik the Outgolfer's golfing.

=RṁR€T€µT*FḢ

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Answer 17

J, 20 bytes

{1,[:;[:(^&.><\)1+i.

1-indexed

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Answer 18

APL+WIN, 23 bytes

(∊n↑¨⊂[2]n∘.*n←⍳n)[n←⎕]

Explanation:

[n←⎕] prompts for screen input and selects the nth element of the concatenated vector
see below

n←⍳n creates a vector of 1 to n

∘.* outer product with exponentiation as the operator

⊂[2] enclose each row of the resulting array as an element of a nested array

∊n↑¨ take 1 to n elements from the 1 to nth row of the matrix and concatenate into a vector

Answer 19

S.I.L.O.S, 45 bytes

readIO
lblb
a+1
i-a
if i b
i+a
a^i
printInt a

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Answer 20

Perl 5, 30+1 (-p) bytes

$_-=$.++until$.>$_;$_=$.**++$_

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Answer 21

Clojure 51 bytes

0-indexed, for example input 9 returns 256.0.

#(nth(for[i(range)j(range i)](Math/pow i(inc j)))%)

Answer 22

C 76, 62 bytes

y;f(n){for(y=0;n>y*++y/2;);return(int)pow(y-1,n+y*(3-y)/2-1);}

y;f(n){y=(int)(-.5+sqrt(1+8*~-n)/2)+2;n+=y*(3-y)/2-1;return(int)pow(y-1,n);}

I based it on this code

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Answer 23

Pyt, 39 37 bytes

1-indexed

←000`ŕŕ⁺ĐĐř^Đ04Ș↔+⇹Ł-Đ↔3Ș0>łŕ0↔⇹+⁻⦋↔ŕ

Explanation:

Briefly, it calculates each row. If the requested number is in that row, return it; otherwise, go to the next row.