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

• Morgan Thrapp: Man, I had this idea a while ago and never got around to writing it up.
• Umbrella: Wouldn't 3-bonacci be 1, 1, 2, 4, 7 as the third position would be 0 + 1 + 1? ... and so one with the others?
• DJMcMayhem: @umbrella No, the tribonacci starts with 3 1s. See my edit to clarify this point.
• Deadcode: Well then, that rules out using this approach. What a shame.

Answer 1

Boolfuck, 6 bytes

,,[;+]

You can safely assume no N-Bonacci numbers will exceed the default number type in your language.

The default number type in Boolfuck is a bit. Assuming this also extends to the input numbers N and X, and given that N>0, there are only two possible inputs - 10 (which outputs nothing) and 11 (which outputs 1).

, reads a bit into the current memory location. N is ignored as it must be 1. If X is 0, the loop body (surrounded by []) is skipped. If X is 1, it is output and then flipped to 0 so the loop does not repeat.

• Stack Exchange Broke The Law: @StanStrum From before or after this answer?
• Unmitigated: @StanStrum In fact, this answer is the example linked to in the initial version of that loophole.
• The Fifth Marshal: Indeed, abusing native number types is one of the loopholes that doesn't apply retroactively. I personally disagree with this policy and think it should, but I accept that consensus is against me.

Answer 2

Python 2, 79 bytes

n,x=input()
i,f=0,[]
while i<x:v=[sum(f[i-n:]),1][i<n];f.append(v);print v;i+=1

Try it online

• Cyoce: Try replacing the last line with exec"v=[sum(f[i-n:]),1][i<n];f+=[v];print v;i+=1;"*x

Answer 3

Haskell, 56 bytes

g l=sum l:g(sum l:init l)
n#x|i<-1<$[1..n]=take x$i++g i

Usage example: 3 # 8-> [1,1,1,3,5,9,17,31].

How it works

i<-1<$[1..n] -- bind i to n copies of 1
take x -- take the first x elements of
 i++g i -- the list starting with i followed by (g i), which is
sum l: -- the sum of it's argument followed by
 g(sum l:init l) -- a recursive call to itself with the the first element
 -- of the argument list replaced by the sum

• proud haskeller: Shouldn't that be tail l instead of init l?
• nimi: @proudhaskeller: it doesn't matter. we keep the last n elements int the list. There's no difference between removing from the end and adding to the front and the other way around, i.e. removing from the front and adding to the end, because the initial list is made up of only 1s.
• proud haskeller: Oh, I get it. That's a nifty way to replace ++[] by :!
• nimi: @proudhaskeller: yes, exactly!

Answer 4

Pyth, 13

<Qu+Gs>QGEm1Q

Test Suite

Takes input newline separated, with n first.

Explanation:

<Qu+Gs>QGEm1Q ## implicit: Q = eval(input)
 u Em1Q ## read a line of input, and reduce that many times starting with
 ## Q 1s in a list, with a lambda G,H
 ## where G is the old value and H is the new one
 +G ## append to the old value
 s>QG ## the sum of the last Q values of the old value
<Q ## discard the last Q values of this list

• DJMcMayhem: Wow, that was fast. I barely had time to close my browser before you'd already posted this!

Answer 5

Javascript ES6/ES2015, (削除) 107 (削除ここまで) (削除) 97 (削除ここまで) (削除) 85 (削除ここまで) 80 Bytes

Thanks to @user81655, @Neil and @ETHproductions for save some bytes

---

(i,n)=>eval("for(l=Array(i).fill(1);n-->i;)l.push(eval(l.slice(-i).join`+`));l")

try it online

---

Test cases:

console.log(f(3, 8))// 1, 1, 1, 3, 5, 9, 17, 31
console.log(f(7, 13))// 1, 1, 1, 1, 1, 1, 1, 7, 13, 25, 49, 97, 193
console.log(f(5, 11))// 1, 1, 1, 1, 1, 5, 9, 17, 33, 65, 129

• user81655: Nice. A couple of golfing tips: for is always better than while, x.split('') -> [...x], ~~a -> +a, n-=1 -> n--, if you enclose the entire function body in an eval you don't need to write return. Also, even shorter than [...'1'.repeat(i)] is Array(i).fill(1) and you can remove the ~~ from a and b. And you're allowed to remove the f=.
• user81655: This is what it looks like with my tips (85 bytes): (i,n)=>eval("for(l=Array(i).fill(1);n-->i;)l.push(l.slice(-i).reduce((a,b)=>a+b));l"). I changed the order of statements, combined the n-- into n-i and removed l from the arguments to save a few extra bytes.
• Neil: @user81655 I don't get the eval saving; (i,n)=>{for(l=Array(i).fill(1);n-->i;)l.push(l.slice(-i).reduce((a,b)=>a+b));return l} is still 85 bytes.
• user81655: @Neil Looks like 86 bytes to me...
• ETHproductions: l.slice(-i).reduce((a,b)=>a+b) => eval(l.slice(-i).join`+`)

Answer 6

ES6, 66 bytes

(i,n)=>[...Array(n)].map((_,j,a)=>a[j]=j<i?1:j-i?s+=s-a[j+~i]:s=i)

Sadly map won't let you access the result array in the callback.

• Shaggy: Save a byte by currying the parameters.

Answer 7

Jelly, 12 bytes

ḣ3S;
b1Ç4¡Uḣ

Try it online!

How it works

b1Ç4¡Uḣ Main link. Left input: n. Right input: x.
b1 Convert n to base 1.
 ¡ Call...
 Ç the helper link...
 4 x times.
 U Reverse the resulting array.
 ḣ Take its first x elements.
ḣ3S; Helper link. Argument: A (list)
ḣ3 Take the first n elements of A.
 S Compute their sum.
 ; Prepend the sum to A.

Answer 8

Python 2, 55 bytes

def f(x,n):l=[1]*n;exec"print l[0];l=l[1:]+[sum(l)];"*x

Tracks a length-n window of the sequence in the list l, updated by appending the sum and removing the first element. Prints the first element each iteration for x iterations.

A different approach of storing all the elements and summing the last n values gave the same length (55).

def f(x,n):l=[1]*n;exec"l+=sum(l[-n:]),;"*x;print l[:x]

Answer 9

Haskell, 47 bytes

q(h:t)=h:q(t++[h+sum t])
n?x=take x$q1ドル<$[1..n]

Try it online!

<$ might have been introduced into Prelude after this challenge was posted.

---

Haskell, 53 bytes

n%i|i>n=sum$map(n%)[i-n..i-1]|0<1=1
n?x=map(n%)[1..x]

Try it online!

Defines the binary function ?, used like 3?8 == [1,1,1,3,5,9,17,31].

The auxiliary function % recursively finds the ith element of the n-bonacci sequence by summing the previous n values. Then, the function ? tabulates the first x values of %.

• Conor O'Brien: Old answer, but do you mean "The auxiliary function %"?
• Ørjan Johansen: Switching the guards will turn i<=n into i>n.
• xnor: @ØrjanJohansen I noticed that too when editing, though looking back the whole method seems bad, so I might just re-do the whole golf.

Answer 10

Husk, 9 bytes

↑§¡ȯΣ↑_B1

Try it online!

Starts from the Base-1 representation of N (simply a list of N ones) and ¡teratively sums (Σ) the last (↑_) N elements and appends the result to the list. Finally, takes (↑) the first X numbers in this list and returns them.

Answer 11

C#, 109 bytes

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

Try it online!

---

(n, x) => { 
 int i, j, total; 
 var result = new int[x];
 for (i = -1; ++i < x; ) { //calculate each term one at a time
 total = 0; //value of the current term
 for (j = i - n; j > -1 & j < i;) //sum the last n terms
 {
 total += result[j++]; 
 }
 result[i] = i < n ? 1 : total; //first n terms are always 1 so ignore the total if i < n
 } 
 return result; 
}

Answer 12

Haskell, 46 bytes

g _ 0=[]
g(h:t)n=h:g(t++[sum$h:t])(n-1)

g.g[1]

Attempt This Online!

This is based on xnor's answer but it employs 1 clever trick.

The \$n\$-bonacci sequence always start with \$n\$ 1s. A more general version of this might start with just any \$n\$ values. And this more general version is what we implement with g. g takes a list of \$n\$ integers and a value \$m\$ and gives us a list of the first \$m\$ terms of this generalized \$n\$-bonacci sequence.

The trick is then that g[1] is a cheap way to generate \$n\$ 1s. Since the sequence starting with [1] is just an endless stream of 1s. So we use g in two ways, and even though g might be slightly longer because it implements something a little more general, it saves bytes because it serves two purposes.

• xnor: Very nice, I see you re-used g as a helper to initialize the list of n ones, which also avoids using <$.

Answer 13

Java, 82 + 58 = 140 bytes

Function to find the ith n-bonacci number (82 bytes):

int f(int i,int n){if(i<=n)return 1;int s=0,q=0;while(q++<n)s+=f(i-q,n);return s;}

Function to print first k n-bonacci number (58 bytes):

(k,n)->{for(int i=0;i<k;i++){System.out.println(f(i,n));}}

Answer 14

C++11, 360 bytes

Hi I just like this question. I know c++ is a very hard language to win this competition. But I'll thrown a dime any way.

#include<vector>
#include<numeric>
#include<iostream>
using namespace std;typedef vector<int>v;void p(v& i) {for(auto&v:i)cout<<v<<" ";cout<<endl;}v b(int s,int n){v r(n<s?n:s,1);r.reserve(n);for(auto i=r.begin();r.size()<n;i++){r.push_back(accumulate(i,i+s,0));}return r;}int main(int c, char** a){if(c<3)return 1;v s=b(atoi(a[1]),atoi(a[2]));p(s);return 0;}

I'll leave this as the readable explanation of the code above.

#include <vector>
#include <numeric>
#include <iostream>
using namespace std;
typedef vector<int> vi;
void p(const vi& in) {
 for (auto& v : in )
 cout << v << " ";
 cout << endl;
}
vi bonacci(int se, int n) {
 vi s(n < se? n : se, 1);
 s.reserve(n);
 for (auto it = s.begin(); s.size() < n; it++){
 s.push_back(accumulate(it, it + se, 0));
 }
 return s;
}
int main (int c, char** v) {
 if (c < 3) return 1;
 vi s = bonacci(atoi(v[1]), atoi(v[2]));
 p(s);
 return 0;
}

• wizzwizz4: Welcome to Programming Puzzles and Code Golf. This is a good answer, however I have noticed that you have lots of whitespace, and variable and function names that are longer than 1 character long. As it stands, this is a good readable version of your code, but you should add a golfed version. When you do, I will give you an upvote, but until it is golfed I will not.
• hetepeperfan: @wizzwizz4 Hi, added a golfed version of the code above. I left the ungolfed code around to let people see how I did it. Besides I like to read a function bonacci that returns vi which still sounds like vibonacci. I do feel I should not make the main function shorter because the standardard mandates using int main(int, char**) as entry point of the program. Further I believe all variables are max 1 character long and all non significant whitespaces are removed.
• wizzwizz4: This is not code-"comply with the standards". This is code-golf. We manipulate and take advantage of our languages. If any variables are ints, remove the int. If any functions are called foo, call them f. Be brutal; ignore the standard and exploit the compiler. That is how you golf.
• wizzwizz4: Puns and nice code belong in the ungolfed code only. But feel free to keep them there; actually, it is recommended to. But be really, really mean to the compiler when you golf your code. Get it as small as possible no matter what. (Oh, and here's the +1 I promised!)
• DJMcMayhem: @wizzwizz4 Is removing "int" valid? I thought implied int won't run.

Answer 15

R, 60 bytes

function(n,m){for(i in 1:m)T[i]=`if`(i>n,sum(T[i-1:n]),1);T}

Try it online!

Answer 16

Uiua (version 0.0.7), 34 bytes

→(;;)⍥(⊂∶/+↙∶→,⇌.)∶→(-,,)[⍥.-1∶1].

Answer 17

APL, 21

{⍵↑⍺{⍵,+/⍺↑⌽⍵}⍣⍵+⍺/1}

This is a function that takes n as its left argument and x as its right argument.

Explanation:

{⍵↑⍺{⍵,+/⍺↑⌽⍵}⍣⍵+⍺/1}
 ⍺/1 ⍝ begin state: X ones 
 + ⍝ identity function (to separate it from the ⍵)
 ⍺{ }⍣⍵ ⍝ apply this function N times to it with X as left argument
 ⍵, ⍝ result of the previous iteration, followed by...
 +/ ⍝ the sum of
 ⍺↑ ⍝ the first X of
 ⌽ ⍝ the reverse of
 ⍵ ⍝ the previous iteration
 ⍵↑ ⍝ take the first X numbers of the result

Test cases:

 ↑⍕ ̈ {⍵↑⍺{⍵,+/⍺↑⌽⍵}⍣⍵+⍺/1} / ̈ (3 8)(7 13)(1 20)(30 4)(5 11)
 1 1 1 3 5 9 17 31 
 1 1 1 1 1 1 1 7 13 25 49 97 193 
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
 1 1 1 1 
 1 1 1 1 1 5 9 17 33 65 129 

Answer 18

Python 3, 59

Saved 20 bytes thanks to FryAmTheEggman.

Not a great solution, but it'll work for now.

def r(n,x):f=[1]*n;exec('f+=[sum(f[-n:])];'*x);return f[:x]

Also, here are test cases:

assert r(3, 8) == [1, 1, 1, 3, 5, 9, 17, 31]
assert r(7, 13) == [1, 1, 1, 1, 1, 1, 1, 7, 13, 25, 49, 97, 193]
assert r(30, 4) == [1, 1, 1, 1]

Answer 19

C, 132 bytes

The recursive approach is shorter by a couple of bytes.

k,n;f(i,s,j){for(j=s=0;j<i&j++<n;)s+=f(i-j);return i<n?1:s;}main(_,v)int**v;{for(n=atoi(v[1]);k++<atoi(v[2]);)printf("%d ",f(k-1));}

Ungolfed

k,n; /* loop index, n */
f(i,s,j) /* recursive function */
{
 for(j=s=0;j<i && j++<n;) /* sum previous n n-bonacci values */
 s+=f(i-j);
 return i<n?1:s; /* return either sum or n, depending on what index we're at */
}
main(_,v) int **v;
{
 for(n=atoi(v[1]);k++<atoi(v[2]);) /* print out n-bonacci numbers */
 printf("%d ", f(k-1));
}

Answer 20

Brain-Flak, (削除) 144 (削除ここまで) (削除) 124 (削除ここまで) 122 bytes

-20 bytes thanks to Nitroden

This is my first Brain-Flak answer, and I'm sure it can be improved. Any help is appreciated.

(([{}]<>)<{({}(()))}{}>)<>{({}[()]<<>({<({}<({}<>)<>>())>[()]}{})({}<><({({}<>)<>}<>)>)<>>)}{}<>{({}<{}>())}{}{({}<>)<>}<>

Try it online!

Answer 21

Pari/GP, 46 bytes

The generating function of the sequence is:

$$\frac{(n-1)x^n}{x^{n+1}-2x+1}-\frac{1}{x-1}$$

(n,m)->Vec(n--/(x-(2-1/x)/x^n)-1/(x-1)+O(x^m))

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Answer 22

Vyxal , 8 bytes

1ẋ?‡W∑*Ḟ

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1ẋ # [1] * input
 ‡W∑ # Function summing its inputs
 ? * # Set the arity to the input
 Ḟ # Generate an infinite list from the function and the vector of 1s.

Answer 23

Julia, 78 bytes

f(n,x)=(z=ones(Int,n);while endof(z)<x push!(z,sum(z[end-n+1:end]))end;z[1:x])

This is a function that accepts two integers and returns an integer array. The approach is simple: Generate an array of ones of length n, then grow the array by adding the sum of the previous n elements until the array has length x.

Ungolfed:

function f(n, x)
 z = ones(Int, n)
 while endof(z) < x
 push!(z, sum(z[end-n+1:end]))
 end
 return z[1:x]
end

Answer 24

Perl 6, (削除) 52~72 (削除ここまで) 47~67 bytes

sub a($n,$x){EVAL("1,"x$n~"+*"x$n~"...*")[^$x]}

Requires the module MONKEY-SEE-NO-EVAL, because of the following error:

===SORRY!=== Error while compiling -e
EVAL is a very dangerous function!!! (use MONKEY-SEE-NO-EVAL to override,
but only if you're VERY sure your data contains no injection attacks)
at -e:1

$ perl6 -MMONKEY-SEE-NO-EVAL -e'a(3,8).say;sub a($n,$x){EVAL("1,"x$n~"+*"x$n~"...*")[^$x]}'
(1 1 1 3 5 9 17 31)

• Andreas Louv: Anyone know of a way to turn off strict mode, etc?
• Batman: I think if you use a pre-christmas 2015 Perl 6 release, it doesn't enforce monkey-see-no-eval.

Answer 25

MATL, 22 (削除) 26 (削除ここまで) bytes

1tiXIX"i:XK"tPI:)sh]K)

This uses current release (10.2.1) of the language/compiler.

Try it online!

A few extra bytes :-( due to a bug in the G function (paste input; now corrected for next release)

Explanation

1tiXIX" % input N. Copy to clipboard I. Build row array of N ones
i:XK % input X. Build row array [1,2,...X]. Copy to clipboard I
" % for loop: repeat X times. Consumes array [1,2,...X]
 t % duplicate (initially array of N ones)
 PI:) % flip array and take first N elements
 sh % compute sum and append to array
] % end
K) % take the first X elements of array. Implicitly display

Answer 26

Perl 6, 38 bytes

->\N,\X{({@_[*-N..*].sum||1}...*)[^X]} # 38 bytes

-> \N, \X {
 (
 {
 @_[
 *-N .. * # previous N values
 ].sum # added together
 || # if that produces 0 or an error
 1 # return 1
 } ... * # produce an infinite list of such values
 )[^X] # return the first X values produced
}

Usage:

# give it a lexical name
my &n-bonacci = >\N,\X{...}
for ( (3,8), (7,13), (1,20), (30,4), (5,11), ) {
 say n-bonacci |@_
}

(1 1 1 3 5 9 17 31)
(1 1 1 1 1 1 1 7 13 25 49 97 193)
(1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1)
(1 1 1 1)
(1 1 1 1 1 5 9 17 33 65 129)

Answer 27

J, 31 bytes

]{.(],[:+/[{.])^:(-@[`]`(1#~[))

Ungolfed:

] {. (] , [: +/ [ {. ])^:(-@[`]`(1 #~ [))

explanation

Fun times with the power verb in its gerund form:

(-@[`]`(1 #~ [)) NB. gerund pre-processing

Breakdown in detail:

• ] {. ... Take the first <right arg> elements from all this stuff to the right that does the work...
• <left> ^: <right> apply the verb <left> repeatedly <right> times... where <right> is specified by the middle gerund in (-@[](1 #~ [), ie, ], ie, the right arg passed into the function itself. So what is <left>? ...
• (] , [: +/ [ {. ]) The left argument to this entire phrase is first transformed by the first gerund, ie, -@[. That means the left argument to this phrase is the negative of the left argument to the overall function. This is needed so that the phrase [ {. ] takes the last elements from the return list we're building up. Those are then summed: +/. And finally appended to that same return list: ] ,.
• So how is the return list initialized? That's what the third pre-processing gerund accomplishes: (1 #~ [) -- repeat 1 "left arg" number of times.

Try it online!

Answer 28

Ruby, 41 bytes

->a,b{r=[1]*a;b.times{z,*r=r<<r.sum;p z}}

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Answer 29

R, 68 bytes

function(n,x){a=1+!1:n;for(i in n+1:x){a[i]=sum(a[(i-n:1)])};a[1:x]}

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Answer 30

K (ngn/k), (削除) 26 (削除ここまで) 24 bytes

{(y-x){y,+/x#y}[-x]/x#1}

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