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The N-bonacci sequence, originally invented by @DJMcMayhem in this question, is a sequence generated by starting with the integers 0 and 1, and then adding the previous N numbers to generate the next number. The special N-bonacci sequence is an N-bonacci sequence beginning with a pair of numbers other than 0 and 1, which will be named X and Y. If N is greater than the number of terms already in the sequence, simply add all available terms.

So for example the normal fibonacci sequence has an N of 2 (takes the previous two items), and an X and Y of 0 and 1, or 1 and 1, depending on who you ask.

Your Task:

You are to write a program or function that checks whether an inputted integer (A) is part of the special N-bonacci sequence generated by the next three integers (using the second input as N, and the third and fourth as X and Y). Ensure that you handle the special case of N=1.

Input:

Four non-negative integers, A, N, X, and Y.

Output:

A truthy/falsy value that indicates whether A is part of the N-bonacci sequence generated by the N, X, and Y inputs.

Test Cases:

Input:    Output:
13,2,0,1->truthy
12,3,1,4->falsy
4,5,0,1-->truthy
8,1,8,9-->truthy
9,1,8,9-->truthy

12,5,0,1->falsy  [0,1]>[0,1,1]>[0,1,1,2]>[0,1,1,2,4]>[0,1,1,2,4,8]>[0,1,1,2,4,8,16]>etc.

Scoring:

This is , so the lowest score in bytes wins.

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  • 1
    \$\begingroup\$ N==1 is such a weird case. \$\endgroup\$ – Magic Octopus Urn Jun 16 '17 at 14:43
  • \$\begingroup\$ Yep, but weird cases are what makes this fun :) \$\endgroup\$ – Gryphon Jun 16 '17 at 14:44
  • \$\begingroup\$ If you do indeed want answers to handle the case N=1, you might want to call it out in the question, since many answers (including all current answers, I think) will have a failure condition that assumes a strictly increasing series. Also, can X and Y be negative? That will probably also invalidate all existing answers. \$\endgroup\$ – apsillers Jun 16 '17 at 19:43
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    \$\begingroup\$ I think all existing answers fail to handle the non-increasing case where both X and Y are zero. Is it necessary to handle that case as well? \$\endgroup\$ – apsillers Jun 16 '17 at 23:11
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    \$\begingroup\$ I think you should add the truthy cases 8,1,8,9 and 9,1,8,9 to ensure that N=1 case handling detects the non-repeated X value as well as the Y value. (If you want to handle 0,0 cases you should add that as well.) \$\endgroup\$ – apsillers Jun 16 '17 at 23:19

10 Answers 10

Active Oldest Votes
5
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Jelly, 12 bytes

ḣ⁴S;µṀ<⁵µ¿⁵e

A full program taking [X,Y], N, A.

Try it online!

How?

ḣ⁴S;µṀ<⁵µ¿⁵e - Main link (monadic): [X,Y]
    µ   µ¿   - while:
     Ṁ       -   maximum value of the list
       ⁵     -   5th command line argument (3rd input) = A
      <      -   less than?
             - ...do:
 ⁴           -   4th command line argument (2nd input) = N
ḣ            -   head (get the first N (or less) items from the list)
  S          -   sum
   ;         -   concatenate (add the result to the front of the list)
          ⁵  - 5th command line argument (3rd input) = A
           e - exists in the resulting list?
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3
  • \$\begingroup\$ Excellent. Seems to work for me, anyway. +1 \$\endgroup\$ – Gryphon Jun 16 '17 at 14:10
  • \$\begingroup\$ To instead see the reversed N-bonacci sequence up to a value greater than or equal to A just remove the ⁵e from the end; much easier to tell it will work then (noting that the order of the first two terms is of no consequence). \$\endgroup\$ – Jonathan Allan Jun 16 '17 at 14:19
  • \$\begingroup\$ Tried a bunch of test cases, so unless someone finds one it fails, it's good with me. \$\endgroup\$ – Gryphon Jun 16 '17 at 14:20
5
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05AB1E, 18 bytes

[DR²£O©‚˜³®>‹#]³QZ

Try it online!

Uses: [X,Y], N, A

I feel like some unintended functionality made that harder than it needed to be.

There's no greater-than-or-equal-to, never noticed that before.

And #³ didn't work, and required a ], for +1 bytes #]³.

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4
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Python 2, 59 56 bytes

a,n,l=input()
while l[0]<a:l=[sum(l[:n])]+l
print a in l

Try it online!

Takes input as A,N,[X,Y]

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2
  • \$\begingroup\$ Here is a wrapper for all test cases if you like it. \$\endgroup\$ – Leaky Nun Jun 16 '17 at 15:14
  • \$\begingroup\$ Here are 2 bytes golfed off. \$\endgroup\$ – Leaky Nun Jun 16 '17 at 15:15
3
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Perl 6, 47 bytes

->\A,\N,\X,\Y{A∈(X,Y,{[+] @_.tail(N)}...*>A)}

test it

Expanded:

->
  \A,
  \N,
  \X, \Y
{
    A          # is ｢A｣

  ∈            # an element of

    (          # this Sequence

      X, Y,        # seed values of sequence

      {            # generate the rest of the Seq using this code block

        [+]        # reduce by addition

          @_       # of all previously generated values
          .tail(N) # only use the last ｢N｣ of them
      }

      ...          # keep generating values until

      * > A        # it is greater than ｢A｣

    )
}
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2
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Python 2, 50 bytes

a,n,l=input()
while[a]>l:l=[sum(l[:n])]+l
a in l>x

Takes input as A,N,[Y,X]. Outputs via exit code.

Try it online!

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1
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R, 69 60 bytes

function(a,n,l){while(l<a)l=c(sum(l[1:n],na.rm=T),l)
a%in%l}

Try it online!

Returns an anonymous function, taking a,n and a vector l=c(y,x). Constructs the N-bonacci sequence backwards (i.e., smaller index is further in the sequence), since while(l<a) only checks the first element of l.

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1
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Common Lisp, 164 bytes

(defun f(a n x y &aux(l(list y x)))(if(= n 1)(or(= a x)(= a y))(loop(if(<= a(car l))(return(member a l))(setf l(cons(reduce'+ l)(if(<(length l)n)l(butlast l))))))))

This function returns NIL for false, non-NIL for true (according to the definition of generalized boolean of Common Lisp).

(defun f(a n x y &aux (l (list y x)))    ; initialize a list l for the N values
  (if (= n 1)                            ; special case for N = 1
      (or (= a x) (= a y))               ;    true only if A = X or A = Y
      (loop
        (if (<= a (car l))               ; when the last number generated is greater than A
            (return (member a l))        ; return true if A is in the list
            (setf l (cons (reduce '+ l)  ; otherwise compute the sum of l
                          (if (< (length l) n)   ; and push it to l (truncating the list at 
                              l                  ; end if it has already size = N)
                              (butlast l))))))))
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3
  • \$\begingroup\$ Does you special-case handling for N=1 detect an A of, e.g., both 1 and/or 2 when X=1 Y=2? My Lisp-reading skills are not great, but it looks like you might only compare A to one of the two initial values. \$\endgroup\$ – apsillers Jun 16 '17 at 23:16
  • \$\begingroup\$ @apsillers , when N=1 I compare A only with X and not with Y. should I compare it with both returning true if it is equal to one of them? Maybe the sequence is not well defined for this case? \$\endgroup\$ – Renzo Jun 16 '17 at 23:20
  • \$\begingroup\$ Ok, now I see the question has been changed, I have updated my answer. \$\endgroup\$ – Renzo Jun 16 '17 at 23:24
0
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k, 29 bytes

{x=*(*x>){(x=#y)_y,+/y}[y]/z}

Try it online! 1 is truthy, 0 is falsey. Input is [A;N;X,Y].

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3
  • \$\begingroup\$ I ran it on all examples that I saw. 1 is truthy, 0 is falsey. \$\endgroup\$ – zgrep Jun 16 '17 at 14:10
  • \$\begingroup\$ @Gryphon I moved the input over to the footer instead of the body, but I'm not certain what you want me to change. It both is and was the same function. \$\endgroup\$ – zgrep Jun 16 '17 at 14:21
  • \$\begingroup\$ Oh, I see now. I thought you weren't taking any input, but you were taking it in the code. Makes much more sense now. I don't know k, so I assumed you'd missinterpreted the question, as all it would do was output 1 0 1 1 \$\endgroup\$ – Gryphon Jun 16 '17 at 14:23
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PHP>=7.1, 103 Bytes

for([,$n,$d,$s,$e]=$argv,$f=[$s,$e];$x<$n;)$x=$f[]=array_sum(array_slice($f,-$d));echo+in_array($n,$f);

Testcases

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0
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Mathematica, 94 bytes

(s={#3,#4};t=1;While[t<#2-1,s~AppendTo~Tr@s;t++];!LinearRecurrence[1~Table~#2,s,#^2]~FreeQ~#)&

input format

[A,N,X,Y]

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