Module:Factorization

local p = {}

function p.factor(frame) -- Consider calling the parser function #expr --  to simplify a potential mathematical expression? number = tonumber(frame.args[1]) if number == nil then return ' Error: input not recognized as a number ' end

productSymbol = frame.args['product'] or '·' bold = frame.args['bold'] and true big = frame.args['big'] and true serif = frame.args['serif'] and true primeLink = frame.args['prime'] and true

number = math.floor(number) if number 1000000000 or number == math.huge then return ' Error: ' .. number .. ' out of range ' end

result = "" currentNumber = number power = 0 divisor = 2

-- Attempt factoring by the value of the divisor --  divisor increments by 2, except first iteration (2 to 3) while divisor <= math.sqrt(currentNumber) do       power = 0 while currentNumber % divisor == 0 do           currentNumber = currentNumber / divisor power = power + 1 end

-- Concat result and increment divisor -- when divisor is 2, go to 3. All other times, add 2 result = result .. powerformat(divisor, power, productSymbol) divisor = divisor + (divisor == 2 and 1 or 2) end

if currentNumber ~= 1 then result = result .. currentNumber .. ' ' .. productSymbol .. ' '   end

if currentNumber == number and primeLink then return 'prime' end

result = string.sub(result,1,-4)

return format(result) end

function powerformat(divisor, power, productSymbol) if power < 1     then return '' elseif power == 1 then return divisor .. ' ' .. productSymbol .. ' '   else return divisor .. '' .. power .. ' ' .. productSymbol .. ' '   end end

function format(numString) if bold then numString = ..numString.. end

ret = (serif or big) and '' or '') .. numString .. ((serif or big) and ' ' or '')

return ret end

return p