Calmo numbers: Difference between revisions

=={{header|Lua}}==

{{Trans|ALGOL 68}}

<syntaxhighlight lang="lua">

do --[[ find some "Calmo" numbers: numbers n such that they have 3k divisors

(other than 1 and n) for some k > 0 and the sum of their divisors

taken three at a time is a prime

--]]

local maxNumber = 1000 -- largest number we will consider

-- construct a sieve of (hopefully) enough primes - as we are going to sum

-- the divisors in groups of three, it should be (more than) large enough

local isPrime, maxPrime = {}, maxNumber * 2

for sPos = 1, maxPrime do isPrime[ sPos ] = sPos % 2 == 1 end

isPrime[ 1 ] = false

isPrime[ 2 ] = true

for sPos = 3, math.sqrt( maxPrime ), 2 do

if isPrime[ sPos ] then

for p = sPos * sPos, maxPrime, sPos do isPrime[ p ] = false end

end

end

-- construct tables of the divisor counts and divisor sums and check for

-- the numbers as we do it

-- as we are ignoring 1 and n, the initial counts and sums will be 0

-- but we should ignore primes

local dsum, dcount = {}, {}

for i = 1, maxNumber do

dcount[ i ] = ( isPrime[ i ] and -1 or 0 )

dsum[ i ] = dcount[ i ]

end

for i = 2, maxNumber do

for j = i + i, maxNumber, i do

-- have another proper divisor

if dsum[ j ] >= 0 then

-- this number is still a candidate

dsum[ j ] = dsum[ j ] + i

dcount[ j ] = dcount[ j ] + 1

if dcount[ j ] == 3 then

-- the divisor count is currently 3

-- if the divisor sum isn't prime, ignore it in future

-- if the divisor sum is prime, reset the sum and count

dcount[ j ] = ( not isPrime[ dsum[ j ] ] and -1 or 0 )

dsum[ j ] = dcount[ j ]

end

end

end

end

-- show the numbers, they will have 0 in the divisor count

for i = 2, maxNumber do

if dcount[ i ] == 0 then

io.write( " ", i )

end

end

io.write( "\n" )

end

</syntaxhighlight>

{{out}}

<pre>

165 273 385 399 561 595 665 715 957

</pre>