syntacticsugar

=begin

Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P^4 + P^7 = 22 + 70 = 92 = P^8.
However, their difference, 70 − 22 = 48, is not pentagonal.

Find a pair of pentagonal numbers P^j and P^k,

syntacticsugar

How to Use

Open up Terminal.app in your /Applications/Utilities directory, then type in these commands, one after each other:

1. Create a temporary directory for cycript:

   mkdir cycript && cd cycript
   

2. Pull the latest cycript from cycript.org:

syntacticsugar

import Either
 
--main = asText (filterrr (\x -> x > 2) [1, 2, 3, 4])
main = asText (map squirtWithError [-2, -1, 0, 16, 36])
 
 
-- types
 
data Option a = None | Some a

syntacticsugar

=begin
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?


NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
=end

# "and" COUNTS!

syntacticsugar

class String
  def palindrome?
    self == self.reverse
  end
end

class Integer
  def palindrome?(b=10)
    self.to_s(b).palindrome?
  end

syntacticsugar

# proccc

def one proccc, x
  proccc.(x)
end

def two proccc, x
  proccc.(proccc.(x))
end

syntacticsugar

class Fixnum
  def ordinal
    %w[first second third fourth fifth sixth seventh eighth ninth tenth eleventh twelfth][self - 1]
  end

  # Use #to_word rather than #to_s, so it doesn't break any other printing of numbers.
  def to_word
    %w[one two three four five six seven eight nine ten eleven twelve][self - 1]
  end
end

syntacticsugar

$array = []

module Collatz
  def self.how_many n
    if n == 1
      ($array.push n).length
    elsif n.even?
      $array.push n
      n = n / 2
      how_many n

syntacticsugar

(define menu
  (lambda (x y)
    (conde
     [(== x 'tea) (== y 'biscuit)]
     [(== x 'coffee) (== y 'cookie)])))

(define conso
  (lambda (x y o)
    (== `(,x . ,y) o)))

syntacticsugar

(ns hs-js-clj.core)

(def foo {:bar 1})

(defn fooify [n] (str "foo" n))

(defn commaify [[x y]]
  (str x ", " y))