Print the N-bonacci sequence

Question

This isn't very widely known, but what we call the Fibonacci sequence, AKA

1, 1, 2, 3, 5, 8, 13, 21, 34...

is actually called the Duonacci sequence. This is because to get the next number, you sum the previous 2 numbers. There is also the Tribonacci sequence,

1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201...

because the next number is the sum of the previous 3 numbers. And the Quadronacci sequence

1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673...

And everybody's favorite, the Pentanacci sequence:

1, 1, 1, 1, 1, 5, 9, 17, 33, 65, 129...

And the Hexanacci sequence, the Septanacci sequence, the Octonacci sequence, and so on and so forth up to the N-Bonacci sequence.

The N-bonacci sequence will always start with N 1s in a row.

The Challenge

You must write a function or program that takes two numbers N and X, and prints out the first X N-Bonacci numbers. N will be a whole number larger than 0, and you can safely assume no N-Bonacci numbers will exceed the default number type in your language. The output can be in any human readable format, and you can take input in any reasonable manner. (Command line arguments, function arguments, STDIN, etc.)

As usual, this is Code-golf, so standard loopholes apply and the shortest answer in bytes wins!

Sample IO

#n,  x,     output
 3,  8  --> 1, 1, 1, 3, 5, 9, 17, 31
 7,  13 --> 1, 1, 1, 1, 1, 1, 1, 7, 13, 25, 49, 97, 193
 1,  20 --> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
 30, 4  --> 1, 1, 1, 1       //Since the first 30 are all 1's
 5,  11 --> 1, 1, 1, 1, 1, 5, 9, 17, 33, 65, 129

Answer 1

Vyxal, 8 bytes

1ẋ{:∑pṫ,

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-2 bytes thanks to emanresu A

Answer 2

JavaScript, 65 bytes

The eval & slice are annoying me!

x=>g=(n,...a)=>n?g(n-1,...a,a[x-1]?eval(a.slice(-x).join`+`):1):a

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Answer 3

Thunno 2, 13 bytes

{1}¹{K°ɱS}K¹ɱ

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Explanation

{1}¹{K°ɱS}K¹ɱ  # Implicit input
{1}            # Push 1 to the stack n times
   ¹{    }     # Repeat x times:
     K°ɱ       #  Take the top n items of the stack
        S      #  And sum this list
          K¹ɱ  # Take the top x items of the stack
               # Implicit output

Answer 4

PHP, 78 bytes

for(list(,$n,$x)=$argv;$i<$x;print${$i++}." ")$s+=$$i=$i<$n?1:$$d+$s-=${$d++};

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-4 Bytes using PHP>=7.1 [,$n,$x] instead of list(,$n,$x)

Answer 5

Jq 1.5, 67 bytes

def C:if length>X then.[:X]else.+=[.[-N:]|add]|C end;[range(N)|1]|C

Assumes input provided by N and X e.g.

def N: 5;
def X: 11;

Expanded

def C:                        # . is current array
    if length>X               # stop when array is as long as X
    then .[:X]                # return first X elements
    else .+=[.[-N:]|add] | C  # recursively add sum of last N elements to array
    end
;
  [range(N)|1]                # initial state
| C

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Answer 6

Mathematica, 59 bytes

((f@#=1)&/@Range@#;f@n_:=Tr[f[n-#]&/@Range@#];f/@Range@#2)&

You'll probably want to Clear@f between function calls. Arguments are n,x, just like the test cases.

Answer 7

Tidy, 36 bytes

{x,n:n^recur(*tile(x,c(1)),sum@c,x)}

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Explanation

{x,n:n^recur(*tile(x,c(1)),sum@c,x)}
{x,n:                              }   lambda taking parameters `x` and `n`
     n^                                take the first `n` terms of...
       recur(                     )        a recursive function
             *tile(x,c(1)),                whose seed is `x` `1`s
                           sum@c,          taking the sum of each window
                                 x         with a window size of `x`

Answer 8

Japt, 18 bytes

@ZsVn)x}gK=Vì1;K¯U

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Explanation:

         K=Vì1        :Start with n 1s in an array K
@      }gK            :Extend K to at least x elements by setting each new element to:
      x               : The sum of
 ZsVn                 : The previous n elements
              ;       :Then
               K¯U    :Return the first n elements of K

Answer 9

MathGolf, 10 bytes

ª*kÆ_Σ▐├p;

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Explanation

ª            push [1]
 *           pop a, b : push(a*b)
  k          read integer from input
   Æ         start block of length 5
    _        duplicate TOS
     Σ       sum(list), digit sum(int)
      ▐      append to end of list
       ├     pop from left of string/array
        p    print with newline
         ;   discard TOS

Answer 10

C# (.NET Core), 130 bytes

(n,x)=>{int j=0,k=0,l=0;var b=new int[x];for(;j<(x<n?x:n);)b[j++]=1;for(j=n;j<x;l=0){for(k=n;k>0;)l+=b[j-k--];b[j++]=l;}return b;}

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Answer 11

Japt, 16 12 bytes

@ZÔ¯V x}hVÆ1

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@ZÔ¯V x}hVÆ1     :Implicit input of integers U=X & V=N
         VÆ1     :Map the range [0,V) returning 1 for each element
        h        :Push the result of the following to the array and repeat until its length equals U
@                :  Pass the array through the following function as Z
 ZÔ              :    Reverse Z
   ¯V            :    Slice to the Vth element
      x          :    Reduce by addition
       }         :  End function

Answer 12

C (gcc), 94 bytes

a(c,C,i){for(int Q[C+c],*q=Q;C--;printf("%d ",*q++))for(*q=Q-q>(i=-c);Q-q<=i&i<0;)*q+=q[i++];}

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Answer 13

krrp, 72 bytes

^nx:\L[take]x,^nl:?nL#!fl@-n1#!r[append]lL[sum]lEl.x[map]^_:1.[range]0n.

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---

Explanation

^nx:              ~ lambda expression in two parameters
 \L               ~  import list module
 [take]x          ~  the first x elements of
  ,^nl:           ~   extend the initial list of n ones
    ?n            ~    if n is non-zero
     L #!fl       ~     keep the first element
      @-n1        ~     recur
       #!r        ~     separate the first element
        [append]l ~      append an additional element,
         L[sum]lE ~      namely the sum of the previous elements
     l            ~    if n is zero, yield l
   . x            ~ extend x elements, yielding in a list of n+x elements
   [map]^_:1.     ~   map a constant function
   [range]0n.     ~   onto a list of n elements

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Answer 14

PostScript, 83 bytes

/f{/x exch def/n exch def[1 1 x{n le{1}{n copy n 1 sub{add}repeat}ifelse}for]==}def

e.g., 5 11 f prints [1 1 1 1 1 5 9 17 33 65 129].

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