Develop Reference

Javascript Draw pattern

I need some function to return pattern like this :
= * = * =
* = * = *
= * = * =
* = * = *
= * = * =
i try this but i cant get it:
function cetakGambar(angka){
let result = ''
for(let i=0; i<=angka; i++){
for(let j=0; j<=i; j++){
result += '= '
}
for(let k=0; k == i; k++){
result += ' *'
}
result += '\n'
}
return result
}
console.log(cetakGambar(5))
what looping i need to get that pattern

You were on the right track with the two nested loops. Here's an example small modification to solve the problem:
function cetakGambar(angka){
let result = ''
for(let i = 0; i < angka; i++){
for(let j = 0; j < angka; ++j) {
result += ((i + j) % 2 == 0) ? '= ' : '* ';
}
result += '\n';
}
return result
}
For each i a row is generated by looping over j. For each j we append either an = or * , depending on if after adding i and j the result is divisible by two (to create the alternating pattern). After each line a \n (newline) is appended.

Here's an ES6 solution w/o explicit iteration. It is not particularly concise and probably not too useful in this case but fun (IMHO), so I wanted to share it.
It uses Array.from to apply a generator function yielding the next symbol (* or =) for each cell and concats cells with spaces and rows with newlines.
// Utility to apply function <fn> <n> times.
function times(n, fn) {
return Array.from({ length: n }, fn);
}
// Generator function yielding the next symbol to be drawn.
//
function cetakGambar(angka) {
let generator = (function*() {
let [i, s1, s2] = [0, "*", "="];
while (true) {
// Make sure to start a new line with a different symbol for
// even rows in case <angka> is even
if (i++ % angka === 0 && angka % 2 === 0) {
[s1, s2] = [s2, s1];
}
// limit the generator to the number of values we actually need
if (i === angka * angka + 1) return;
yield i % 2 === 0 ? s1 : s2;
}
})();
return times(
angka,
() => times(angka, () => generator.next().value).join(" ") // join symbols w/ spaces ...
).join("\n"); // ... and lines w/ newline
}
console.log(cetakGambar(5));
Codesandbox here: https://codesandbox.io/s/wispy-http-mfxgo?file=/src/index.js

Different approach using one loop
function gen(row,col){
var i = 0;
var out="";
do {
if(i%col==0) out+="\n"
out += (i % 2) == 0 ? ' = ' : ' * ';
i++;
}
while (i < row*col);
return out;
}
console.log(gen(5,5));

Trying to multiply the first two numbers that results to a particular number, is there way i can do this without using nested loops?

Here is the full solution, please is there much better ways to go about this
the question is getting total number of squares in a box but in the inverse way, for example the normal question is 2,9 and the total number of available squares will be 26. but i was given the question in another way such that 26 will be given and im to find all the possible ways m and n can be arranged so the total number of squares will be the given question, which is 26
The code actually works but i need a faster way to do this.
const n = 26;
function count(m, n) {
if (n < m) {
const temp = n;
n = m;
m = temp;
}
return m * (m + 1) * (2 * m + 1) /
6 + (n - m) * m * (m + 1) / 2;
}
const arr = [];
const len = n + 1;
for (let i = 1; i < len; i++) {
if(i<len/2) {
}
for (let b = 1; b < len; b++) {
if (i === 1) {
arr.push(`${i} ${n}`);
break;
} else if (i === n) {
arr.push(`${n} 1`);
break;
} else {
const sum = count(i, b);
if (sum == n) {
arr.push(`${i} ${b}`);
break;
}
}
}
}
const arrLength = arr.length;
console.log(arrLength); //shows the length of the array
for (let g = 0; g < arrLength; g++) {
console.log(arr[g]); // shows each ways it can be arranged to give 26
}

Javascript while loop (Card deck simulation)

I am having an issue with the following code that simulates a card deck.
The deck is created properly (1 array containing 4 arrays (suits) containing 13 elements each (face values)) and when I use the G.test(); function it is correctly pulling 13 random cards but then returns 39x "Empty" (A total of 52).
I hate to ask for help, but I have left the problem overnight and then some and I still cannot find the reason that this is happening. I appreciate any and all insight that can be offered.
var G = {};
G.cards = [[], [], [], []];
G.newCard = function(v) { //currently a useless function, tried a few things
return v;
};
G.deck = {
n: function() { //new deck
var x; var list = [];
list.push(G.newCard("A"));
for (x = 2; x <= 10; x += 1) {
list.push(G.newCard(x.toString()));
}
list.push(G.newCard("J"), G.newCard("Q"), G.newCard("K"));
for (x = 0; x < G.cards.length; x += 1) {
G.cards[x] = list;
}
},
d: function() { //random card - returns suit & value
var s; var c; var v; var drawn = false; var n;
s = random(0, G.cards.length);
c = random(0, G.cards[s].length);
n = 0;
while (!drawn) {
if (G.cards[s].length > 0) {
if (G.cards[s][c]) {
v = G.cards[s].splice(c, 1);
drawn = true;
} else {
c = random(0, G.cards[s].length);
}
} else {
s = (s + 1 >= G.cards.length) ? 0 : s + 1;
n += 1;
console.log(s);
if (n >= G.cards.length) {
console.log(n);
return "Empty";
}
}
}
return {s: s, v: v[0]};
},
}; //G.deck
G.test = function() {
var x; var v;
G.deck.n();
for (x = 0; x < 52; x += 1) {
v = G.deck.d();
console.log(v);
}
};

Replace
for (x = 0; x < G.cards.length; x += 1) {
G.cards[x] = list;
}
with
for (x = 0; x < G.cards.length; x += 1) {
G.cards[x] = list.slice();
}
as this prevents all elements of G.cards[x] binding to the same (single) array instance.
If all elements bind to the same instance, mutating one element equals mutating all elements. list.slice() creates a new copy of list and thus a new array instance to prevent the aforementioned issue.

I won't go through your code, but I built a code that will do what you wanted. I only built this for one deck and not multiple deck play. There are two functions, one will generate the deck, and the other will drawn cards from the deck, bases on how many hands you need and how many cards you wanted for each hand. One a card is drawn, it will not be re-drawn. I might publish a short article for how a card dealing program work or similar in the short future at http://kevinhng86.iblog.website.
function random(min, max){
return Math.floor(Math.random() * (max - min)) + min;
}
function deckGenerate(){
var output = [];
var face = {1: "A", 11: "J", 12: "Q", 13: "K"};
// Heart Space Diamond & Club;
var suit = ["H", "S", "D", "C"];
// Delimiter between card identification and suit identification.
var d = "-";
for(i = 0; i < 4; i++){
output[i] = [];
for(ind = 0; ind < 13; ind++ ){
card = (ind + 1);
output[i][ind] = (card > 10) || (card === 1)? face[card] + d + suit[i] : card.toString() + d + suit[i];
}
}
return output;
}
function randomCard(deck, hand, card){
var output = [];
var randS = 0;
var randC = 0;
if( hand * card > 52 ) throw("Too many card, I built this for one deck only");
for(i = 0; i < hand; i++){
output[i] = [];
for(ind = 0; ind < card; ind++){
randS = random(0, deck.length);
randC = random(0, deck[randS].length);
output[i][ind] = deck[randS][randC];
deck[randS].splice(randC,1);
if(deck[randS].length === 0) deck.splice(randS,1);
}
}
document.write( JSON.stringify(deck, null, 2) );
return output;
}
var deck = deckGenerate()
document.write( JSON.stringify(deck, null, 2) );
document.write("<br><br>");
var randomhands = randomCard(deck, 5, 8);
document.write("<br><br>");
document.write("<br><br>");
document.write( JSON.stringify(randomhands, null, 2) );

Compare Strings Javascript Return %of Likely

I am looking for a JavaScript function that can compare two strings and return the likeliness that they are alike. I have looked at soundex but that's not really great for multi-word strings or non-names. I am looking for a function like:
function compare(strA,strB){
}
compare("Apples","apple") = Some X Percentage.
The function would work with all types of strings, including numbers, multi-word values, and names. Perhaps there's a simple algorithm I could use?
Ultimately none of these served my purpose so I used this:
function compare(c, u) {
var incept = false;
var ca = c.split(",");
u = clean(u);
//ca = correct answer array (Collection of all correct answer)
//caa = a single correct answer word array (collection of words of a single correct answer)
//u = array of user answer words cleaned using custom clean function
for (var z = 0; z < ca.length; z++) {
caa = $.trim(ca[z]).split(" ");
var pc = 0;
for (var x = 0; x < caa.length; x++) {
for (var y = 0; y < u.length; y++) {
if (soundex(u[y]) != null && soundex(caa[x]) != null) {
if (soundex(u[y]) == soundex(caa[x])) {
pc = pc + 1;
}
}
else {
if (u[y].indexOf(caa[x]) > -1) {
pc = pc + 1;
}
}
}
}
if ((pc / caa.length) > 0.5) {
return true;
}
}
return false;
}
// create object listing the SOUNDEX values for each letter
// -1 indicates that the letter is not coded, but is used for coding
// 0 indicates that the letter is omitted for modern census archives
// but acts like -1 for older census archives
// 1 is for BFPV
// 2 is for CGJKQSXZ
// 3 is for DT
// 4 is for L
// 5 is for MN my home state
// 6 is for R
function makesoundex() {
this.a = -1
this.b = 1
this.c = 2
this.d = 3
this.e = -1
this.f = 1
this.g = 2
this.h = 0
this.i = -1
this.j = 2
this.k = 2
this.l = 4
this.m = 5
this.n = 5
this.o = -1
this.p = 1
this.q = 2
this.r = 6
this.s = 2
this.t = 3
this.u = -1
this.v = 1
this.w = 0
this.x = 2
this.y = -1
this.z = 2
}
var sndx = new makesoundex()
// check to see that the input is valid
function isSurname(name) {
if (name == "" || name == null) {
return false
} else {
for (var i = 0; i < name.length; i++) {
var letter = name.charAt(i)
if (!(letter >= 'a' && letter <= 'z' || letter >= 'A' && letter <= 'Z')) {
return false
}
}
}
return true
}
// Collapse out directly adjacent sounds
// 1. Assume that surname.length>=1
// 2. Assume that surname contains only lowercase letters
function collapse(surname) {
if (surname.length == 1) {
return surname
}
var right = collapse(surname.substring(1, surname.length))
if (sndx[surname.charAt(0)] == sndx[right.charAt(0)]) {
return surname.charAt(0) + right.substring(1, right.length)
}
return surname.charAt(0) + right
}
// Collapse out directly adjacent sounds using the new National Archives method
// 1. Assume that surname.length>=1
// 2. Assume that surname contains only lowercase letters
// 3. H and W are completely ignored
function omit(surname) {
if (surname.length == 1) {
return surname
}
var right = omit(surname.substring(1, surname.length))
if (!sndx[right.charAt(0)]) {
return surname.charAt(0) + right.substring(1, right.length)
}
return surname.charAt(0) + right
}
// Output the coded sequence
function output_sequence(seq) {
var output = seq.charAt(0).toUpperCase() // Retain first letter
output += "-" // Separate letter with a dash
var stage2 = seq.substring(1, seq.length)
var count = 0
for (var i = 0; i < stage2.length && count < 3; i++) {
if (sndx[stage2.charAt(i)] > 0) {
output += sndx[stage2.charAt(i)]
count++
}
}
for (; count < 3; count++) {
output += "0"
}
return output
}
// Compute the SOUNDEX code for the surname
function soundex(value) {
if (!isSurname(value)) {
return null
}
var stage1 = collapse(value.toLowerCase())
//form.result.value=output_sequence(stage1);
var stage1 = omit(value.toLowerCase())
var stage2 = collapse(stage1)
return output_sequence(stage2);
}
function clean(u) {
var u = u.replace(/\,/g, "");
u = u.toLowerCase().split(" ");
var cw = ["ARRAY OF WORDS TO BE EXCLUDED FROM COMPARISON"];
var n = [];
for (var y = 0; y < u.length; y++) {
var test = false;
for (var z = 0; z < cw.length; z++) {
if (u[y] != "" && u[y] != cw[z]) {
test = true;
break;
}
}
if (test) {
//Don't use & or $ in comparison
var val = u[y].replace("$", "").replace("&", "");
n.push(val);
}
}
return n;
}

Here's an answer based on Levenshtein distance https://en.wikipedia.org/wiki/Levenshtein_distance
function similarity(s1, s2) {
var longer = s1;
var shorter = s2;
if (s1.length < s2.length) {
longer = s2;
shorter = s1;
}
var longerLength = longer.length;
if (longerLength == 0) {
return 1.0;
}
return (longerLength - editDistance(longer, shorter)) / parseFloat(longerLength);
}
For calculating edit distance
function editDistance(s1, s2) {
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
var costs = new Array();
for (var i = 0; i <= s1.length; i++) {
var lastValue = i;
for (var j = 0; j <= s2.length; j++) {
if (i == 0)
costs[j] = j;
else {
if (j > 0) {
var newValue = costs[j - 1];
if (s1.charAt(i - 1) != s2.charAt(j - 1))
newValue = Math.min(Math.min(newValue, lastValue),
costs[j]) + 1;
costs[j - 1] = lastValue;
lastValue = newValue;
}
}
}
if (i > 0)
costs[s2.length] = lastValue;
}
return costs[s2.length];
}
Usage
similarity('Stack Overflow','Stack Ovrflw')
returns 0.8571428571428571
You can play with it below:
function checkSimilarity(){
var str1 = document.getElementById("lhsInput").value;
var str2 = document.getElementById("rhsInput").value;
document.getElementById("output").innerHTML = similarity(str1, str2);
}
function similarity(s1, s2) {
var longer = s1;
var shorter = s2;
if (s1.length < s2.length) {
longer = s2;
shorter = s1;
}
var longerLength = longer.length;
if (longerLength == 0) {
return 1.0;
}
return (longerLength - editDistance(longer, shorter)) / parseFloat(longerLength);
}
function editDistance(s1, s2) {
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
var costs = new Array();
for (var i = 0; i <= s1.length; i++) {
var lastValue = i;
for (var j = 0; j <= s2.length; j++) {
if (i == 0)
costs[j] = j;
else {
if (j > 0) {
var newValue = costs[j - 1];
if (s1.charAt(i - 1) != s2.charAt(j - 1))
newValue = Math.min(Math.min(newValue, lastValue),
costs[j]) + 1;
costs[j - 1] = lastValue;
lastValue = newValue;
}
}
}
if (i > 0)
costs[s2.length] = lastValue;
}
return costs[s2.length];
}
<div><label for="lhsInput">String 1:</label> <input type="text" id="lhsInput" oninput="checkSimilarity()" /></div>
<div><label for="rhsInput">String 2:</label> <input type="text" id="rhsInput" oninput="checkSimilarity()" /></div>
<div>Match: <span id="output">No Input</span></div>

Using this library for string similarity worked like a charm for me!
Here's the Example -
var similarity = stringSimilarity.compareTwoStrings("Apples","apple"); // => 0.88

Here is a very simple function that does a comparison and returns a percentage based on equivalency. While it has not been tested for all possible scenarios, it may help you get started.
function similar(a,b) {
var equivalency = 0;
var minLength = (a.length > b.length) ? b.length : a.length;
var maxLength = (a.length < b.length) ? b.length : a.length;
for(var i = 0; i < minLength; i++) {
if(a[i] == b[i]) {
equivalency++;
}
}
var weight = equivalency / maxLength;
return (weight * 100) + "%";
}
alert(similar("test","tes")); // 75%
alert(similar("test","test")); // 100%
alert(similar("test","testt")); // 80%
alert(similar("test","tess")); // 75%

To Find degree of similarity between two strings; we can use more than one or two methods but I am mostly inclined towards the usage of 'Dice's Coefficient' . which is better! well in my knowledge than using 'Levenshtein distance'
Using this 'string-similarity' package from npm you will be able to work on what I said above.
some easy usage examples are
var stringSimilarity = require('string-similarity');
var similarity = stringSimilarity.compareTwoStrings('healed', 'sealed');
var matches = stringSimilarity.findBestMatch('healed', ['edward', 'sealed', 'theatre']);
for more please visit the link given above. Thankyou.

Just one I quickly wrote that might be good enough for your purposes:
function Compare(strA,strB){
for(var result = 0, i = strA.length; i--;){
if(typeof strB[i] == 'undefined' || strA[i] == strB[i]);
else if(strA[i].toLowerCase() == strB[i].toLowerCase())
result++;
else
result += 4;
}
return 1 - (result + 4*Math.abs(strA.length - strB.length))/(2*(strA.length+strB.length));
}
This weighs characters that are the same but different case 1 quarter as heavily as characters that are completely different or missing. It returns a number between 0 and 1, 1 meaning the strings are identical. 0 meaning they have no similarities. Examples:
Compare("Apple", "Apple") // 1
Compare("Apples", "Apple") // 0.8181818181818181
Compare("Apples", "apple") // 0.7727272727272727
Compare("a", "A") // 0.75
Compare("Apples", "appppp") // 0.45833333333333337
Compare("a", "b") // 0

How about function similar_text from PHP.js library?
It is based on a PHP function with the same name.
function similar_text (first, second) {
// Calculates the similarity between two strings
// discuss at: http://phpjs.org/functions/similar_text
if (first === null || second === null || typeof first === 'undefined' || typeof second === 'undefined') {
return 0;
}
first += '';
second += '';
var pos1 = 0,
pos2 = 0,
max = 0,
firstLength = first.length,
secondLength = second.length,
p, q, l, sum;
max = 0;
for (p = 0; p < firstLength; p++) {
for (q = 0; q < secondLength; q++) {
for (l = 0;
(p + l < firstLength) && (q + l < secondLength) && (first.charAt(p + l) === second.charAt(q + l)); l++);
if (l > max) {
max = l;
pos1 = p;
pos2 = q;
}
}
}
sum = max;
if (sum) {
if (pos1 && pos2) {
sum += this.similar_text(first.substr(0, pos2), second.substr(0, pos2));
}
if ((pos1 + max < firstLength) && (pos2 + max < secondLength)) {
sum += this.similar_text(first.substr(pos1 + max, firstLength - pos1 - max), second.substr(pos2 + max, secondLength - pos2 - max));
}
}
return sum;
}

fuzzyset - A fuzzy string set for javascript.
fuzzyset is a data structure that performs something akin to fulltext search against data to determine likely mispellings and approximate string matching. Note that this is a javascript port of a python library.

To some extent, I like the ideas of Dice's coefficient embedded in the string-similarity module. But I feel that considering the bigrams only and not taking into account their multiplicities is missing some important data. Below is a version that also handles multiplicities, and I think is a simpler implementation overall. I don't try to use their API, offering only a function which compares two strings after some manipulation (removing non-alphanumeric characters, lower-casing everything, and compressing but not removing whitespace), built atop one which compares them without that manipulation. It would be easy enough to wrap this back in their API, but I see little need.
const stringSimilarity = (a, b) =>
_stringSimilarity (prep (a), prep (b))
const _stringSimilarity = (a, b) => {
const bg1 = bigrams (a)
const bg2 = bigrams (b)
const c1 = count (bg1)
const c2 = count (bg2)
const combined = uniq ([... bg1, ... bg2])
.reduce ((t, k) => t + (Math .min (c1 [k] || 0, c2 [k] || 0)), 0)
return 2 * combined / (bg1 .length + bg2 .length)
}
const prep = (str) => // TODO: unicode support?
str .toLowerCase () .replace (/[^\w\s]/g, ' ') .replace (/\s+/g, ' ')
const bigrams = (str) =>
[...str] .slice (0, -1) .map ((c, i) => c + str [i + 1])
const count = (xs) =>
xs .reduce ((a, x) => ((a [x] = (a [x] || 0) + 1), a), {})
const uniq = (xs) =>
[... new Set (xs)]
console .log (stringSimilarity (
'foobar',
'Foobar'
)) //=> 1
console .log (stringSimilarity (
"healed",
"sealed"
))//=> 0.8
console .log (stringSimilarity (
"Olive-green table for sale, in extremely good condition.",
"For sale: table in very good condition, olive green in colour."
)) //=> 0.7787610619469026
console .log (stringSimilarity (
"Olive-green table for sale, in extremely good condition.",
"For sale: green Subaru Impreza, 210,000 miles"
)) //=> 0.38636363636363635
console .log (stringSimilarity (
"Olive-green table for sale, in extremely good condition.",
"Wanted: mountain bike with at least 21 gears."
)) //=> 0.1702127659574468
console .log (stringSimilarity (
"The rain in Spain falls mainly on the plain.",
"The run in Spun falls munly on the plun.",
)) //=> 0.7560975609756098
console .log (stringSimilarity (
"Fa la la la la, la la la la",
"Fa la la la la, la la",
)) //=> 0.8636363636363636
console .log (stringSimilarity (
"car crash",
"carcrash",
)) //=> 0.8
console .log (stringSimilarity (
"Now is the time for all good men to come to the aid of their party.",
"Huh?",
)) //=> 0
.as-console-wrapper {max-height: 100% !important; top: 0}
Some of the test cases are from string-similarity, others are my own. They show some significant differences from that package, but nothing untoward. The only one I would call out is the difference between "car crash" and "carcrash", which string-similarity sees as identical and I report with a similarity of 0.8. My version finds more similarity in all the olive-green test-cases than does string-similarity, but as these are in any case fairly arbitrary numbers, I'm not sure how much difference it makes; they certainly position them in the same relative order.

string-similarity lib vs Top answer (by #overloard1234) performance comparation you can find below
Based on #Tushar Walzade's advice to use string-similarity library, you can find, that for example
stringSimilatityLib.findBestMatch('KIA','Kia').bestMatch.rating
will return 0.0
So, looks like better to compare it in lowerCase.
Better base usage (for arrays) :
findBestMatch(str, strArr) {
const lowerCaseArr = strArr.map(element => element.toLowerCase());//creating lower case array
const match = stringSimilatityLib.findBestMatch(str.toLowerCase(), lowerCaseArr).bestMatch; //trying to find bestMatch
if (match.rating > 0) {
const foundIndex = lowerCaseArr.findIndex(x => x === match.target); //finding the index of found best case
return strArr[foundIndex]; //returning initial value from array
}
return null;
},
Performance
Also, i compared top answer here (made by #overloard1234) and string-similarity lib (v4.0.4).
The results you can find here : https://jsbench.me/szkzojoskq/1
Result : string-similarity is ~ twice faster
Just for fun : v2.0 of string-similarity library slower, than latest 4.0.4 about 2.2 times. So update it, if you are still using < 3.0 :)

const str1 = " pARTH PARmar r ";
const str2 = " parmar r par ";
function calculateSimilarity(str1 = "", str2 = "") {
let longer = str1.trim();
let shorter = str2.trim();
let a1 = longer.toLowerCase().split(" ");
let b1 = shorter.toLowerCase().split(" ");
let result = a1.every((aa, i) => aa[0] === b1[i][0]);
if (longer.length < shorter.length) [longer,shorter] = [shorter,longer];
var arr = [];
let count = 0;
for(var i = 0;i<longer.length;i++){
if(shorter && shorter.includes(longer[i])) {
shorter = shorter.replace(longer[i],"")
count++
};
}
return {
score : (count*100)/longer.length,
result
}
}
console.log(calculateSimilarity(str1, str2));

I used #overlord1234 function, but corrected ь: '', cuz English words don't have this letter, and next need return a[char] ?? char instead of return a[char] || char

Generating Fibonacci Sequence

var x = 0;
var y = 1;
var z;
fib[0] = 0;
fib[1] = 1;
for (i = 2; i <= 10; i++) {
alert(x + y);
fib[i] = x + y;
x = y;
z = y;
}
I'm trying to get to generate a simple Fibonacci Sequence but there no output.
Can anybody let me know what's wrong?

You have never declared fib to be an array. Use var fib = []; to solve this.
Also, you're never modifying the y variable, neither using it.
The code below makes more sense, plus, it doesn't create unused variables:
var i;
var fib = [0, 1]; // Initialize array!
for (i = 2; i <= 10; i++) {
// Next fibonacci number = previous + one before previous
// Translated to JavaScript:
fib[i] = fib[i - 2] + fib[i - 1];
console.log(fib[i]);
}

According to the Interview Cake question, the sequence goes 0,1,1,2,3,5,8,13,21. If this is the case, this solution works and is recursive without the use of arrays.
function fibonacci(n) {
return n < 1 ? 0
: n <= 2 ? 1
: fibonacci(n - 1) + fibonacci(n - 2)
}
console.log(fibonacci(4))
Think of it like this.
fibonacci(4) .--------> 2 + 1 = 3
| / |
'--> fibonacci(3) + fibonacci(2)
| ^
| '----------- 2 = 1 + 1 <----------.
1st step -> | ^ |
| | |
'----> fibonacci(2) -' + fibonacci(1)-'
Take note, this solution is not very efficient though.

Yet another answer would be to use es6 generator functions.
function* fib() {
var current = a = b = 1;
yield 1;
while (true) {
current = b;
yield current;
b = a + b;
a = current;
}
}
sequence = fib();
sequence.next(); // 1
sequence.next(); // 1
sequence.next(); // 2
// ...

Here's a simple function to iterate the Fibonacci sequence into an array using arguments in the for function more than the body of the loop:
fib = function(numMax){
for(var fibArray = [0,1], i=0,j=1,k=0; k<numMax;i=j,j=x,k++ ){
x=i+j;
fibArray.push(x);
}
console.log(fibArray);
}
fib(10)
[ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ]

You should've declared the fib variable to be an array in the first place (such as var fib = [] or var fib = new Array()) and I think you're a bit confused about the algorithm.
If you use an array to store the fibonacci sequence, you do not need the other auxiliar variables (x,y,z) :
var fib = [0, 1];
for(var i=fib.length; i<10; i++) {
fib[i] = fib[i-2] + fib[i-1];
}
console.log(fib);
Click for the demo
You should consider the recursive method too (note that this is an optimised version) :
function fib(n, undefined){
if(fib.cache[n] === undefined){
fib.cache[n] = fib(n-1) + fib(n-2);
}
return fib.cache[n];
}
fib.cache = [0, 1, 1];
and then, after you call the fibonacci function, you have all the sequence in the fib.cache field :
fib(1000);
console.log(fib.cache);

The golden ration "phi" ^ n / sqrt(5) is asymptotic to the fibonacci of n, if we round that value up, we indeed get the fibonacci value.
function fib(n) {
let phi = (1 + Math.sqrt(5))/2;
let asymp = Math.pow(phi, n) / Math.sqrt(5);
return Math.round(asymp);
}
fib(1000); // 4.346655768693734e+208 in just a few milliseconds
This runs faster on large numbers compared to the recursion based solutions.

You're not assigning a value to z, so what do you expect y=z; to do? Likewise you're never actually reading from the array. It looks like you're trying a combination of two different approaches here... try getting rid of the array entirely, and just use:
// Initialization of x and y as before
for (i = 2; i <= 10; i++)
{
alert(x + y);
z = x + y;
x = y;
y = z;
}
EDIT: The OP changed the code after I'd added this answer. Originally the last line of the loop was y = z; - and that makes sense if you've initialized z as per my code.
If the array is required later, then obviously that needs to be populated still - but otherwise, the code I've given should be fine.

Another easy way to achieve this:
function fibonacciGenerator(n) {
// declare the array starting with the first 2 values of the fibonacci sequence
// starting at array index 1, and push current index + previous index to the array
for (var fibonacci = [0, 1], i = 2; i < n; i++)
fibonacci.push(fibonacci[i-1] + fibonacci[i - 2])
return fibonacci
}
console.log( fibonacciGenerator(10) )

function fib(n) {
if (n <= 1) {
return n;
} else {
return fib(n - 1) + fib(n - 2);
}
}
fib(10); // returns 55

fibonacci 1,000 ... 10,000 ... 100,000
Some answers run into issues when trying to calculate large fibonacci numbers. Others are approximating numbers using phi. This answer will show you how to calculate a precise series of large fibonacci numbers without running into limitations set by JavaScript's floating point implementation.
Below, we generate the first 1,000 fibonacci numbers in a few milliseconds. Later, we'll do 100,000!
const { fromInt, toString, add } =
Bignum
const bigfib = function* (n = 0)
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n >= 0) {
yield toString (a)
_ = a
a = b
b = add (b, _)
n = n - 1
}
}
console.time ('bigfib')
const seq = Array.from (bigfib (1000))
console.timeEnd ('bigfib')
// 25 ms
console.log (seq.length)
// 1001
console.log (seq)
// [ 0, 1, 1, 2, 3, ... 995 more elements ]
Let's see the 1,000th fibonacci number
console.log (seq [1000])
// 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
10,000
This solution scales quite nicely. We can calculate the first 10,000 fibonacci numbers in under 2 seconds. At this point in the sequence, the numbers are over 2,000 digits long – way beyond the capacity of JavaScript's floating point numbers. Still, our result includes precise values without making approximations.
console.time ('bigfib')
const seq = Array.from (bigfib (10000))
console.timeEnd ('bigfib')
// 1877 ms
console.log (seq.length)
// 10001
console.log (seq [10000] .length)
// 2090
console.log (seq [10000])
// 3364476487 ... 2070 more digits ... 9947366875
Of course all of that magic takes place in Bignum, which we will share now. To get an intuition for how we will design Bignum, recall how you added big numbers using pen and paper as a child...
1259601512351095520986368
+ 50695640938240596831104
---------------------------
?
You add each column, right to left, and when a column overflows into the double digits, remembering to carry the 1 over to the next column...
... <-001
1259601512351095520986368
+ 50695640938240596831104
---------------------------
... <-472
Above, we can see that if we had two 10-digit numbers, it would take approximately 30 simple additions (3 per column) to compute the answer. This is how we will design Bignum to work
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s, Number) .reverse ()
, toString: (b) =>
Array.from (b) .reverse () .join ('')
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
We'll run a quick test to verify our example above
const x =
fromString ('1259601512351095520986368')
const y =
fromString ('50695640938240596831104')
console.log (toString (add (x,y)))
// 1310297153289336117817472
And now a complete program demonstration. Expand it to calculate the precise 10,000th fibonacci number in your own browser! Note, the result is the same as the answer provided by wolfram alpha
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s, Number) .reverse ()
, toString: (b) =>
Array.from (b) .reverse () .join ('')
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
const { fromInt, toString, add } =
Bignum
const bigfib = function* (n = 0)
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n >= 0) {
yield toString (a)
_ = a
a = b
b = add (b, _)
n = n - 1
}
}
console.time ('bigfib')
const seq = Array.from (bigfib (10000))
console.timeEnd ('bigfib')
// 1877 ms
console.log (seq.length)
// 10001
console.log (seq [10000] .length)
// 2090
console.log (seq [10000])
// 3364476487 ... 2070 more digits ... 9947366875
100,000
I was just curious how far this little script could go. It seems like the only limitation is just time and memory. Below, we calculate the first 100,000 fibonacci numbers without approximation. Numbers at this point in the sequence are over 20,000 digits long, wow! It takes 3.18 minutes to complete but the result still matches the answer from wolfram alpha
console.time ('bigfib')
const seq = Array.from (bigfib (100000))
console.timeEnd ('bigfib')
// 191078 ms
console.log (seq .length)
// 100001
console.log (seq [100000] .length)
// 20899
console.log (seq [100000])
// 2597406934 ... 20879 more digits ... 3428746875
BigInt
JavaScript now has native support for BigInt. This allows for calculating huge integers very quickly -
function* fib (n)
{ let a = 0n
let b = 1n
let _
while (n >= 0) {
yield a.toString()
_ = a
a = b
b = b + _
n = n - 1
}
}
console.time("fib(1000)")
const result = Array.from(fib(1000))
console.timeEnd("fib(1000)")
document.body.textContent = JSON.stringify(result, null, 2)
body {
font-family: monospace;
white-space: pre;
}

I like the fact that there are so many ways to create a fibonacci sequence in JS. I will try to reproduce a few of them. The goal is to output a sequence to console (like {n: 6, fiboNum: 8})
Good ol' closure
// The IIFE form is purposefully omitted. See below.
const fiboGenClosure = () => {
let [a, b] = [0, 1];
let n = 0;
return (fiboNum = a) => {
[a, b] = [b, a + b];
return {
n: n++,
fiboNum: fiboNum
};
};
}
// Gets the sequence until given nth number. Always returns a new copy of the main function, so it is possible to generate multiple independent sequences.
const generateFiboClosure = n => {
const newSequence = fiboGenClosure();
for (let i = 0; i <= n; i++) {
console.log(newSequence());
}
}
generateFiboClosure(21);
Fancy ES6 generator
Similar to the closure pattern above, using the advantages of generator function and for..of loop.
// The 'n' argument is a substitute for index.
function* fiboGen(n = 0) {
let [a, b] = [0, 1];
while (true) {
yield [a, n++];
[a, b] = [b, a + b];
}
}
// Also gives a new sequence every time is invoked.
const generateFibonacci = n => {
const iterator = fiboGen();
for (let [value, index] of iterator) {
console.log({
n: index,
fiboNum: value
});
if (index >= n) break;
}
}
generateFibonacci(21);
Tail call recursion
This one is a little tricky, because, now in late 2018, TC optimization is still an issue. But honestly – if you don't use any smart tricks to allow the default JS engine to use a really big numbers, it will get dizzy and claims that the next fibonacci number is "Infinity" by iteration 1477. The stack would probably overflow somewhere around iteration 10 000 (vastly depends on browser, memory etc…). Could be probably padded by try… catch block or check if "Infinity" was reached.
const fibonacciRTC = (n, i = 0, a = 0, b = 1) => {
console.log({
n: i,
fibonacci: a
});
if (n === 0) return;
return fibonacciRTC(--n, ++i, b, a + b);
}
fibonacciRTC(21)
It can be written as a one-liner, if we throe away the console.log thing and simply return a number:
const fibonacciRTC2 = (n, a = 0, b = 1) => n === 0 ? a : fibonacciRTC2(n - 1, b, a + b);
console.log(fibonacciRTC2(21))
Important note!
As I found out reading this mathIsFun article, the fibonacci sequence is valid for negative numbers as well! I tried to implement that in the recursive tail call form above like that:
const fibonacciRTC3 = (n, a = 0, b = 1, sign = n >= 0 ? 1 : -1) => {
if (n === 0) return a * sign;
return fibonacciRTC3(n - sign, b, a + b, sign);
}
console.log(fibonacciRTC3(8)); // 21
console.log(fibonacciRTC3(-8)); // -21

There is also a generalization of Binet's formula for negative integers:
static float phi = (1.0f + sqrt(5.0f)) / 2.0f;
int generalized_binet_fib(int n) {
return round( (pow(phi, n) - cos(n * M_PI) * pow(phi, -n)) / sqrt(5.0f) );
}
...
for(int i = -10; i < 10; ++i)
printf("%i ", generalized_binet_fib(i));

A quick way to get ~75
ty #geeves for the catch, I replaced Math.floor for Math.round which seems to get it up to 76 where floating point issues come into play :/ ...
either way, I wouldn't want to be using recursion up and until that point.
/**
* Binet Fibonacci number formula for determining
* sequence values
* #param {int} pos - the position in sequence to lookup
* #returns {int} the Fibonacci value of sequence #pos
*/
var test = [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025,20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041,1548008755920,2504730781961,4052739537881,6557470319842,10610209857723,17167680177565,27777890035288,44945570212853,72723460248141,117669030460994,190392490709135,308061521170129,498454011879264,806515533049393,1304969544928657,2111485077978050,3416454622906707,5527939700884757,8944394323791464,14472334024676221,23416728348467685,37889062373143906,61305790721611591,99194853094755497,160500643816367088,259695496911122585,420196140727489673,679891637638612258,1100087778366101931,1779979416004714189,2880067194370816120,4660046610375530309,7540113804746346429,12200160415121876738,19740274219868223167,31940434634990099905,51680708854858323072,83621143489848422977,135301852344706746049,218922995834555169026];
var fib = function (pos) {
return Math.round((Math.pow( 1 + Math.sqrt(5), pos)
- Math.pow( 1 - Math.sqrt(5), pos))
/ (Math.pow(2, pos) * Math.sqrt(5)));
};
/* This is only for the test */
var max = test.length,
i = 0,
frag = document.createDocumentFragment(),
_div = document.createElement('div'),
_text = document.createTextNode(''),
div,
text,
err,
num;
for ( ; i < max; i++) {
div = _div.cloneNode();
text = _text.cloneNode();
num = fib(i);
if (num !== test[i]) {
err = i + ' == ' + test[i] + '; got ' + num;
div.style.color = 'red';
}
text.nodeValue = i + ': ' + num;
div.appendChild(text);
frag.appendChild(div);
}
document.body.appendChild(frag);

You can get some cache to speedup the algorithm...
var tools = {
fibonacci : function(n) {
var cache = {};
// optional seed cache
cache[2] = 1;
cache[3] = 2;
cache[4] = 3;
cache[5] = 5;
cache[6] = 8;
return execute(n);
function execute(n) {
// special cases 0 or 1
if (n < 2) return n;
var a = n - 1;
var b = n - 2;
if(!cache[a]) cache[a] = execute(a);
if(!cache[b]) cache[b] = execute(b);
return cache[a] + cache[b];
}
}
};

If using ES2015
const fib = (n, prev = 0, current = 1) => n
? fib(--n, current, prev + current)
: prev + current
console.log( fib(10) )

If you need to build a list of fibonacci numbers easily you can use array destructuring assignment to ease your pain:
function fibonacci(n) {
let fibList = [];
let [a, b] = [0, 1]; // array destructuring to ease your pain
while (a < n) {
fibList.push(a);
[a, b] = [b, a + b]; // less pain, more gain
}
return fibList;
}
console.log(fibonacci(10)); // prints [0, 1, 1, 2, 3, 5, 8]

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>fibonacci series</title>
<script type="text/javascript">
function generateseries(){
var fno = document.getElementById("firstno").value;
var sno = document.getElementById("secondno").value;
var a = parseInt(fno);
var result = new Array();
result[0] = a;
var b = ++fno;
var c = b;
while (b <= sno) {
result.push(c);
document.getElementById("maindiv").innerHTML = "Fibonacci Series between "+fno+ " and " +sno+ " is " +result;
c = a + b;
a = b;
b = c;
}
}
function numeric(evt){
var theEvent = evt || window.event;
var key = theEvent.keyCode || theEvent.which;
key = String.fromCharCode(key);
var regex = /[0-9]|\./;
if (!regex.test(key)) {
theEvent.returnValue = false;
if (theEvent.preventDefault)
theEvent.preventDefault();
}
}
</script>
<h1 align="center">Fibonacci Series</h1>
</head>
<body>
<div id="resultdiv" align="center">
<input type="text" name="firstno" id="firstno" onkeypress="numeric(event)"><br>
<input type="text" name="secondno" id="secondno" onkeypress="numeric(event)"><br>
<input type="button" id="result" value="Result" onclick="generateseries();">
<div id="maindiv"></div>
</div>
</body>
</html>

I know this is a bit of an old question, but I realized that many of the answers here are utilizing for loops rather than while loops.
Sometimes, while loops are faster than for loops, so I figured I'd contribute some code that runs the Fibonacci sequence in a while loop as well! Use whatever you find suitable to your needs.
function fib(length) {
var fibArr = [],
i = 0,
j = 1;
fibArr.push(i);
fibArr.push(j);
while (fibArr.length <= length) {
fibArr.push(fibArr[j] + fibArr[i]);
j++;
i++;
}
return fibArr;
};
fib(15);

sparkida, found an issue with your method. If you check position 10, it returns 54 and causes all subsequent values to be incorrect. You can see this appearing here: http://jsfiddle.net/createanaccount/cdrgyzdz/5/
(function() {
function fib(n) {
var root5 = Math.sqrt(5);
var val1 = (1 + root5) / 2;
var val2 = 1 - val1;
var value = (Math.pow(val1, n) - Math.pow(val2, n)) / root5;
return Math.floor(value + 0.5);
}
for (var i = 0; i < 100; i++) {
document.getElementById("sequence").innerHTML += (0 < i ? ", " : "") + fib(i);
}
}());
<div id="sequence">
</div>

Here are examples how to write fibonacci using recursion, generator and reduce.
'use strict'
//------------- using recursion ------------
function fibonacciRecursion(n) {
return (n < 2) ? n : fibonacciRecursion(n - 2) + fibonacciRecursion(n - 1)
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciRecursion(i))
}
//-------------- using generator -----------------
function* fibonacciGenerator() {
let a = 1,
b = 0
while (true) {
yield b;
[a, b] = [b, a + b]
}
}
// usage
const gen = fibonacciGenerator()
for (let i = 0; i < 10; i++) {
console.log(gen.next().value)
}
//------------- using reduce ---------------------
function fibonacciReduce(n) {
return new Array(n).fill(0)
.reduce((prev, curr) => ([prev[0], prev[1]] = [prev[1], prev[0] + prev[1]], prev), [0, 1])[0]
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciReduce(i))
}

I just would like to contribute with a tail call optimized version by ES6. It's quite simple;
var fibonacci = (n, f = 0, s = 1) => n === 0 ? f : fibonacci(--n, s, f + s);
console.log(fibonacci(12));

There is no need for slow loops, generators or recursive functions (with or without caching). Here is a fast one-liner using Array and reduce.
ECMAScript 6:
var fibonacci=(n)=>Array(n).fill().reduce((a,b,c)=>a.concat(c<2?c:a[c-1]+a[c-2]),[])
ECMAScript 5:
function fibonacci(n){
return Array.apply(null,{length:n}).reduce(function(a,b,c){return a.concat((c<2)?c:a[c-1]+a[c-2]);},[]);
}
Tested in Chrome 59 (Windows 10):
fibonacci(10); // 0 ms -> (10) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
JavaScript can handle numbers up to 1476 before reaching Infinity.
fibonacci(1476); // 11ms -> (1476) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...]

Another implementation, while recursive is very fast and uses single inline function. It hits the javascript 64-bit number precision limit, starting 80th sequence (as do all other algorithms):
For example if you want the 78th term (78 goes in the last parenthesis):
(function (n,i,p,r){p=(p||0)+r||1;i=i?i+1:1;return i<=n?arguments.callee(n,i,r,p):r}(78));
will return: 8944394323791464
This is backwards compatible all the way to ECMASCRIPT4 - I tested it with IE7 and it works!

This script will take a number as parameter, that you want your Fibonacci sequence to go.
function calculateFib(num) {
var fibArray = [];
var counter = 0;
if (fibArray.length == 0) {
fibArray.push(
counter
);
counter++
};
fibArray.push(fibArray[fibArray.length - 1] + counter);
do {
var lastIndex = fibArray[fibArray.length - 1];
var snLastIndex = fibArray[fibArray.length - 2];
if (lastIndex + snLastIndex < num) {
fibArray.push(lastIndex + snLastIndex);
}
} while (lastIndex + snLastIndex < num);
return fibArray;
};

This is what I came up with
//fibonacci numbers
//0,1,1,2,3,5,8,13,21,34,55,89
//print out the first ten fibonacci numbers
'use strict';
function printFobonacciNumbers(n) {
var firstNumber = 0,
secondNumber = 1,
fibNumbers = [];
if (n <= 0) {
return fibNumbers;
}
if (n === 1) {
return fibNumbers.push(firstNumber);
}
//if we are here,we should have at least two numbers in the array
fibNumbers[0] = firstNumber;
fibNumbers[1] = secondNumber;
for (var i = 2; i <= n; i++) {
fibNumbers[i] = fibNumbers[(i - 1)] + fibNumbers[(i - 2)];
}
return fibNumbers;
}
var result = printFobonacciNumbers(10);
if (result) {
for (var i = 0; i < result.length; i++) {
console.log(result[i]);
}
}

Beginner, not too elegant, but shows the basic steps and deductions in JavaScript
/* Array Four Million Numbers */
var j = [];
var x = [1,2];
var even = [];
for (var i = 1;i<4000001;i++){
j.push(i);
}
// Array Even Million
i = 1;
while (i<4000001){
var k = j[i] + j[i-1];
j[i + 1] = k;
if (k < 4000001){
x.push(k);
}
i++;
}
var total = 0;
for (w in x){
if (x[w] %2 === 0){
even.push(x[w]);
}
}
for (num in even){
total += even[num];
}
console.log(x);
console.log(even);
console.log(total);

My 2 cents:
function fibonacci(num) {
return Array.apply(null, Array(num)).reduce(function(acc, curr, idx) {
return idx > 2 ? acc.concat(acc[idx-1] + acc[idx-2]) : acc;
}, [0, 1, 1]);
}
console.log(fibonacci(10));

I would like to add some more code as an answer :), Its never too late to code :P
function fibonacciRecursive(a, b, counter, len) {
if (counter <= len) {
console.log(a);
fibonacciRecursive(b, a + b, counter + 1, len);
}
}
fibonacciRecursive(0, 1, 1, 20);
Result
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181

function fibo(count) {
//when count is 0, just return
if (!count) return;
//Push 0 as the first element into an array
var fibArr = [0];
//when count is 1, just print and return
if (count === 1) {
console.log(fibArr);
return;
}
//Now push 1 as the next element to the same array
fibArr.push(1);
//Start the iteration from 2 to the count
for(var i = 2, len = count; i < len; i++) {
//Addition of previous and one before previous
fibArr.push(fibArr[i-1] + fibArr[i-2]);
}
//outputs the final fibonacci series
console.log(fibArr);
}
Whatever count we need, we can give it to above fibo method and get the fibonacci series upto the count.
fibo(20); //output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

Fibonacci (one-liner)
function fibonacci(n) {
return (n <= 1) ? n : fibonacci(n - 1) + fibonacci(n - 2);
}
Fibonacci (recursive)
function fibonacci(number) {
// n <= 1
if (number <= 0) {
return n;
} else {
// f(n) = f(n-1) + f(n-2)
return fibonacci(number - 1) + fibonacci(number - 2);
}
};
console.log('f(14) = ' + fibonacci(14)); // 377
Fibonacci (iterative)
function fibonacci(number) {
// n < 2
if (number <= 0) {
return number ;
} else {
var n = 2; // n = 2
var fn_1 = 0; // f(n-2), if n=2
var fn_2 = 1; // f(n-1), if n=2
// n >= 2
while (n <= number) {
var aa = fn_2; // f(n-1)
var fn = fn_1 + fn_2; // f(n)
// Preparation for next loop
fn_1 = aa;
fn_2 = fn;
n++;
}
return fn_2;
}
};
console.log('f(14) = ' + fibonacci(14)); // 377
Fibonacci (with Tail Call Optimization)
function fibonacci(number) {
if (number <= 1) {
return number;
}
function recursion(length, originalLength, previous, next) {
if (length === originalLength)
return previous + next;
return recursion(length + 1, originalLength, next, previous + next);
}
return recursion(1, number - 1, 0, 1);
}
console.log(`f(14) = ${fibonacci(14)}`); // 377