Files
JavaScript/Maths/FriendlyNumbers.js
Roland Hummel 86d333ee94 feat: Test running overhaul, switch to Prettier & reformat everything (#1407)
* chore: Switch to Node 20 + Vitest

* chore: migrate to vitest mock functions

* chore: code style (switch to prettier)

* test: re-enable long-running test

Seems the switch to Node 20 and Vitest has vastly improved the code's and / or the test's runtime!

see #1193

* chore: code style

* chore: fix failing tests

* Updated Documentation in README.md

* Update contribution guidelines to state usage of Prettier

* fix: set prettier printWidth back to 80

* chore: apply updated code style automatically

* fix: set prettier line endings to lf again

* chore: apply updated code style automatically

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Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
2023-10-04 02:38:19 +05:30

39 lines
1.2 KiB
JavaScript

/*
'In number theory, friendly numbers are two or more natural numbers with a common abundancy index, the
ratio between the sum of divisors of a number and the number itself.'
Source: https://en.wikipedia.org/wiki/Friendly_number
See also: https://mathworld.wolfram.com/FriendlyNumber.html#:~:text=The%20numbers%20known%20to%20be,numbers%20have%20a%20positive%20density.
*/
export const FriendlyNumbers = (firstNumber, secondNumber) => {
// input: two integers
// output: true if the two integers are friendly numbers, false if they are not friendly numbers
// First, check that the parameters are valid
if (
!Number.isInteger(firstNumber) ||
!Number.isInteger(secondNumber) ||
firstNumber === 0 ||
secondNumber === 0 ||
firstNumber === secondNumber
) {
throw new Error('The two parameters must be distinct, non-null integers')
}
return abundancyIndex(firstNumber) === abundancyIndex(secondNumber)
}
function abundancyIndex(number) {
return sumDivisors(number) / number
}
function sumDivisors(number) {
let runningSumDivisors = number
for (let i = 0; i < number / 2; i++) {
if (Number.isInteger(number / i)) {
runningSumDivisors += i
}
}
return runningSumDivisors
}