Generate Lucky Numbers

Question

Story:

Lucy asked George what his Lucky Number was. After some contemplation, George replied that he had several Lucky Numbers. After some brief confusion, Lucy asked George what his first n Lucky Numbers are. George then asked you, his buddy, to write him a program to do the work for him.

The Challenge:

You will write a program/function that will receive from standard input/function argument a string or integer n. The program/function will then return/output the first n Lucky Numbers. Lucky numbers are defined via a sieve as follows.

Start with the positive integers:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, ...

Now remove every second number:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, ...

The second remaining number is 3, so remove every third number:

1, 3, 7, 9, 13, 15, 19, 21, 25, ...

Now the next remaining number is 7, so remove every seventh number:

1, 3, 7, 9, 13, 15, 21, 25, ...

Next, remove every ninth number and so on. The resulting sequence are the lucky numbers.

Winning:

As usual for codegolf, fewest bytes wins.

As usual, submissions using standard loopholes are disqualified.

Answer 1

Python 2, 79

n=input()
L=range(1,2**n)
for r in L:r+=r<2;map(L.remove,L[r-1::r])
print L[:n]

The magic of iterating over a list as the loop modifies it!

The list L starts with all the integers 1 to a sufficiently high value. The code iterates over each element r of L, taking the sublist of every r'th element, and removing each of those values. As a result, the removed values aren't iterated over. At the end, print the first n elements.

The expression map(A.remove,B) is a trick I've been waiting a long time to use. It calls A.remove for each element of B, which causes all the elements of B to removed from A. Effectively, it takes the list difference, though it requires B to be a sublist of A. It requires Python 2, since Python 3 wouldn't actually evaluate the map.

The first loop needs to be special-cased to convert r from 1 to 2, as r+=r<2.

The sufficiently high upper bound of 2**n makes the program very slow for large values of n. Using n*n+1 suffices, but costs a character. Note that n*n doesn't work for n=1.

Answer 2

Haskell, 71 69 bytes

s(n:k)p=n:s[m|(i,m)<-zip[p..]k,i`mod`n>0](p+1)
f n=take n$1:s[3,5..]3

Defines a function f. The expression 1:s[3,5..]3 evaluates to the infinite list of lucky numbers, and f simply takes the first n of them by take n.

f 20
[1,3,7,9,13,15,21,25,31,33,37,43,49,51,63,67,69,73,75,79]

I could shave off 5 bytes from the sieve using a parallel list comprehension

s(n:k)p=n:s[m|m<-k|i<-[p..],i`mod`n>0](p+1)

but that would require passing the humongous compiler flag -XParallelListComp to GHC to enable the extension.

Explanation of the sieve

s(n:k)p=               -- Sieving a list with head n and tail k with accumulator p is
 n:                    -- the head n, followed by
  s[m|                 -- the result of sieving the list of numbers m
    (i,m)<-zip[p..]k,  -- where (i,m) is drawn from [(p,k_0),(p+1,k_1),(p+2,k_2),..] and
    i`mod`n>0]         -- i does not divide n,
   (p+1)               -- using p+1 as the accumulator

The basic idea is that s(n:k)p produces the (p-1)th lucky number n, drops every nth number from the infinite tail k (offset by p to account for the numbers produced earlier), and recurses to that list with accumulator (p+1). In f, we initialize the process with the odd numbers starting from 3, and tack 1 to the front, obtaining exactly the lucky numbers.

Answer 3

Python 2, 71 69 67

At first, I thought this would be a great challenge for Python's array slicing. However, I encountered a stumbling block when I discovered that slices with a step other than 1 can only have another slice of identical length assigned to them. But after googling "python remove slice", my faith was restored: I found a funky del statement that does the trick perfectly.

n=input()
l=range(n*n+9)
for v in l:del l[v&~1::v or 2]
print l[:n]

Old version

n=input()
l=range(1,n*n+9)
for v in l:del l[v-1%v::v+1/v]
print l[:n]

-2 bytes thanks to Sp3000.

Answer 4

CJam - 25

Lri{1$W%{1$\(1e>/+}/)+}/p

Try it online

Explanation:

This implementation doesn't remove numbers successively from an array, but calculates each number based on how many would have been removed before it.
For each index i (from 0 to n-1) and each previous lucky number l, in reverse order, we increment i by i/(l-1), except for l=1 we use 1 instead of 0, and also add 1 at the end.
E.g. for i=4 we have the first 4 numbers, [1 3 7 9], and calculate:
4 + 4/(9-1) = 4
4 + 4/(7-1) = 4
4 + 4/(3-1) = 6
6 + 6/1 = 12
12 + 1 = 13

L              empty array - the first 0 lucky numbers :)
ri             read and convert to integer (n)
{…}/           for each number (i) from 0 to n-1
    1$         copy the previous array
    W%         reverse the order
    {…}/       for each array element (l)
        1$     copy i
        \(     swap with l and decrement l
        1e>    use 1 if l=1
        /+     divide and add to i
    )+         increment and add the new lucky number to the array
p              pretty print

Answer 5

><>, 121 114 111 bytes

i2+:&:*1\
:})?v:2+>l{
nao2\r~1
)?vv>1+:&:&=?;:[{::nao}]$}l1-[01+}:l3-$%$l1-@@-{$[{~l1
3.\ ff+
!?:<]-1v
~]{43. >

I have only a few words to say...

... "Argh my brain hurts."

---

Explanation

><> is a 2D esoteric programming language and is definitely not suited for this task, due to its lack of arrays. In fact, the only data type in ><> is strange mix of int/float/char, and everything happens on a stack of stacks.

Here's the rundown:

Line 1:            i2+:&:*1\

i2+:&              Read a char as input (n) and add 2, copying n+2 into the register
:*                 Duplicate and multiply, giving (n+2)^2 on the stack
1\                 Push 1 and go to the next line

Line 2:            >l{:})?v:2+

l{:})?v            Go to the next line if the stack's length is greater than (n+2)^2
:2+                Otherwise duplicate the top of the stack and add 2 to it

Line 3:            \r~1nao2

r~                 Reverse the stack and pop; stack contains the first (n+2)^2 odd integers
1nao               Print 1 (special case)
2\                 Push 2 (let's call this "i" for "iterations") and go to the next line

Line 4:            >1+:&:&=?;:[{::nao}]$}l1-[01+}:l3-$%$l1-@@-{$[{~l1)?vv

1+                 Increment i
:&:&=?;            If i is equal to n+2 (+2 because we started at 2), halt
:[{::nao}]$}       Print the i-th element down (the next lucky number) and also
                   copy it to the top of the stack, while moving i to the bottom
l1-[               Move everything but i to a new stack
0                  Push 0 (let's call this "r" for recursion depth)

Sieve loop:

1+                 Increment r
}:l3-$%$l1-@@-{$[  Move everything up to the last element to be sieved out to a new stack
{~                 Remove said last element
1)?vv              If the length is 1, go to line 6 (sieving complete)
                   Otherwise go to line 5, which repeats this sieve loop by teleporting

Line 6:            :?!v1-]

:?!v1-]            Keep unrolling and decrementing r until r is 0

Line 7:            >~]{43.             

~]                 Pop r and unroll once more (to the stack where i waits)
43.                Loop, performing everything from line 4 all over again

Here's a mock example which demonstrates roughly how the sieving works (here k is the lucky number which we sieve with):

[[15 13 11 9 7 5 3 1 k=3 r=0]]     -- move -->
[[15 13] [11 9 7 5 3 1 k=3 r=1]]   -- pop  -->
[[15 13] [9 7 5 3 1 k=3 r=1]]      -- move -->
[[15 13] [9 7] [5 3 1 k=3 r=2]]    -- pop  -->
[[15 13] [9 7] [3 1 k=3 r=2]]      -- move -->
[[15 13] [9 7] [3 1] [k=3 r=3]]    -- pop  -->
[[15 13] [9 7] [3 1] [r=3]]        (now we unroll)

Answer 6

Pyth: 23 22 bytes

<u-G%@GhH+0GQ%2r1^hQ2Q

Try it online: Pyth Compiler/Executor

Explanation:

<u-G%@GhH+0GQ%2r1^hQ2Q    Q = input()
             %2r1^hQ2     create the list [1, 2, ..., (Q+1)^2-1][::2]
 u          Q%2r1^hQ2     G = [1, 2, ..., (Q+1)^2-1][::2]
                           modify G for each H in [0, 1, 2, ..., Q]:
  -G%:GhH+0G                  G = G - ([0] + G)[::G[H+1]]
                               (minus is remove in Pyth)
<                    Q    print the first Q elements of the resulting list

The reduce actually calculates more than Q lucky numbers (the remove command is called Q+1 times, Q-1 should be enough).

Answer 7

R, 58 bytes

n=scan();s=r=1:n^2;for(j in 1:n)r=r[-max(2,r[j])*s];r[1:n]

With line breaks:

n=scan()              #user input
s=r=1:n^2             #declare r and s simultaneously, both large enough to sieve
for(j in 1:n)
  r=r[-max(2,r[j])*s] #iteratively remove elements by position in vector
r[1:n]                #print

Previous version, 62 bytes

function(n){
  s=r=1:n^2             #declare r and s simultaneously, both large enough to sieve
  for(j in 1:n)
    r=r[-max(2,r[j])*s] #iteratively remove elements by position in vector
  r[1:n]                #print
}

Previous version, 78 bytes

n=as.numeric(readline())   #ask for user input and convert it to numeric
r=1:n^2                    #create a large enough vector to sieve
for(j in 1:n){             #loop
  r=r[-max(2,r[j])*1:n^2]  #iteratively remove elements by position in vector
}
r[1:n]                     #print

Answer 8

CJam, 32 30 bytes

3ri:N#,N{0\__I1e>)=%-+}fI(;N<p

Takes input from STDIN.

Code explanation:

3ri:N#,                          "Read the input in N and get first 3^N whole numbers";
       N{0\__I1e>)=%-+}fI        "Run the code block N times, storing index in I";
         0\__                    "Put 0 before the array and take 2 copies";
             I1e>)=              "Take min(2, I + 1) th index from the copy";
                   %             "Take every array[ min (2, I + 1)] element from the array";
                    -+           "Remove it from the list and prepend 0 to the list";
                         (;N<p   "Print number index 1 to N";

Try it online here

Answer 9

Python 2, 105 101 bytes

n=input()
L=range(-1,n*n+9,2)
i=2
while L[i:]:L=sorted(set(L)-set(L[L[i]::L[i]]));i+=1
print L[1:n+1]

Just a straightforward implementation.

Pyth, 39 36 35 32 bytes

J%2r1h^Q2VJI>JhN=J-J%@JhN+2J;<JQ

Similar to the approach above, but things are 0-indexed rather than 1-indexed. Try it online.

Thanks to @Jakube for pointing out a byte saving.

Answer 10

Perl, 86 81 78

86:

@a=(1..($k=<>)**2);for$n(@a){$i=1;@a=map{$i++%($n+($n<2))?$_:()}@a;$k-=$k&&print"$n "}

UPDATE: obviously, grep{...} is better than map{...?$_:()} 81:

@a=(1..($k=<>)**2);for$n(@a){$i=1;@a=grep{$i++%($n+($n<2))}@a;$k-=$k&&print"$n "}

UPDATE: OK, actually a one-liner now. I can stop. (?) 78:

@a=(1..($k=<>)**2);for$n(@a){$k-=$i=$k&&print"$n ";@a=grep{$i++%($n+=$n<2)}@a}

Answer 11

Mathematica, 80 bytes

(For[l=Range[#^2];i=1,(m=l[[i++]]~Max~2)<=Length@l,l=l~Drop~{m,-1,m}];l[[;;#]])&

Straight-forward implementation of the definition. As some other answers, starts with a range from 1 to n2 and then keeps filtering.

Answer 12

J, 60 52 bytes

   ({.}.@((>:@{.,]#~0<({~{.)|i.@#)@]^:[2,1+2*i.@*:@>:)) 8
1 3 7 9 13 15 21 25

Explanation (from right to left):

2,1+2*i.@*:@>:  generates the list 2 1 3 5 7 9... with (n+1)^2 odd numbers
^:[             repeats n times the following
@]                using the list
0<({~{.)|i.@#     is the remainder of the indexes of the lists elements with the first element positive (i.e. index divisible by first element)
]#~               keep those elements from the list
>:@{.,            concatenate a first element with the value of the current one +1
}.@             drop first element
{.              take the first n element

2,1+2*i.@*:@>: seems way too long but I can only shorten it by 1 byte swapping *: with ! making the list grow exponentially.

Answer 13

JavaScript (ES6) 96 99

Edit Counting down in first loop - thanks @DocMax

F=n=>(i=>{
  for(o=[1];--i;)o[i]=i-~i;
  for(;++i<n;)o=o.filter((x,j)=>++j%o[i]);
})(n*n)||o.slice(0,n)

Ungolfed

F=n=>{
  for (i = n*n, o = [1]; --i;)
    o[i] = i+i+1;
  for (; ++i < n; )
    o = o.filter((x, j) => (j+1) % o[i])
  return o.slice(0,n)
}

Test In Firefox / FireBug console

F(57)

Output

[1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, 51, 63, 67, 69, 73, 75, 79, 87, 93, 99, 105, 111, 115, 127, 129, 133, 135, 141, 151, 159, 163, 169, 171, 189, 193, 195, 201, 205, 211, 219, 223, 231, 235, 237, 241, 259, 261, 267, 273, 283, 285, 289, 297, 303]

Answer 14

Octave, 139 83 72

function r=l(n)r=1:2:n^2;for(i=2:n)h=r(i);r(h:h:end)=[];end;r=r(1:n);end

Ungolfed:

function r=l(n)
  r=1:2:n^2;
  for(i=2:n)
    h=r(i);
    r(h:h:end)=[];
  end
r=r(1:n);  # reduce it to only N lucky numbers
end

Answer 15

Matlab, 104 bytes

function x=f(n)
k=1;N=n;x=0;while nnz(x)<n
x=1:N;m=1;while m-nnz(x)
m=x(x>m);x(m:m:end)=[];end
N=N+2;end

With thanks to @flawr for very appropriate comments and suggestions.

Example from Matlab command prompt:

>> f(40)
ans =
  Columns 1 through 22
     1     3     7     9    13    15    21    25    31    33    37    43    49    51    63    67    69    73    75    79    87    93
  Columns 23 through 40
    99   105   111   115   127   129   133   135   141   151   159   163   169   171   189   193   195   201

Answer 16

Julia, 66 bytes

x->(l=[1:x^2;];[deleteat!(l,(i+=i<2):i:length(l)) for i=l];l[1:x])

Try it online!

Answer 17

Bash+coreutils, 136

I'd hoped to golf this down more, but oh well. Not every day that you pipe into a recursive function in a shell script:

f(){
mapfile -tn$2 a
(($1>$2))&&{
tr \  \\n<<<${a[@]}
sed $[${a[-1]}-$2]~${a[-1]}d
}|f $1 $[$2+1]||echo ${a[@]}
}
yes|sed -n 1~2=|f $1 2

Output:

$ ./lucky.sh 23
1 3 7 9 13 15 21 25 31 33 37 43 49 51 63 67 69 73 75 79 87 93 99
$ 

---

Bash+coreutils, 104

Shorter using a more straightforward implementation:

a=(`seq 1 2 $[3+$1**2]`)
for((;i++<$1;));{
a=($(tr \  \\n<<<${a[@]}|sed 0~${a[i]}d))
}
echo ${a[@]:0:$1}

Answer 18

MATLAB, 62 characters

n=input('');o=1:2:n^2;for i=2:n;o(o(i):o(i):end)=[];end;o(1:n)

I misinterpreted the challenge at first - my revised version is now actually shorter.

Answer 19

05AB1E, 17 16 bytes

∞IFD®LKн©ô€¨˜}I£

Try it online.

Explanation:

∞             # Push an infinite positive list: [1,2,3,...]
 IF           # Loop the input amount of times:
   D          #  Duplicate the infinite list
    ®L        #  Push a list in the range [1,`®`]
              #  (`®` is -1 by default, so the list is [1,0,-1] the first iteration)
      K       #  Remove all those values from the duplicated list
       н      #  Pop the list and leave the first (smallest) value
              #  (this is basically the smallest value above `®`)
        ©     #  Store it as new `®` (without popping)
         ô    #  Split the infinite list into parts of size `®`
          ۬  #  Remove the last item of each inner list
            ˜ #  Flatten it back to a single list
  }I£         # After the loop: only leave the first input amount of values
              # (after which the list is output implicitly as result)

Answer 20

Go, 326

package main
import"fmt"
func s(k, p int,in chan int)chan int{
    o := make(chan int)
    go func(){
        for p>0{
            o<-<-in;p--
        }
        for{
            <-in
            for i:=1;i<k;i++{o<-<-in}
        }
    }()
    return o
}
func main(){
    n := 20
    fmt.Print(1)
    c := make(chan int)
    go func(c chan int){
        for i:=3;;i+=2{c<-i}
    }(c)
    for i:=1;i<n;i++{
        v := <-c
        fmt.Print(" ", v)
        c = s(v, v-i-2, c)
    }
}

Straight forward implementation using goroutine and pipes to make sieves.

Answer 21

Racket 196 bytes

Produces lucky numbers upto n:

(λ(n)(let loop((l(filter odd?(range 1 n)))(i 1))(define x(list-ref l i))(set! l(for/list((j(length l))
#:unless(= 0(modulo(add1 j)x)))(list-ref l j)))(if(>= i(sub1(length l)))l(loop l(add1 i)))))

Ungolfed version:

(define f
 (λ(n)
    (let loop ((l (filter odd? (range 1 n))) (i 1))
      (define x (list-ref l i))
      (set! l (for/list ((j (length l)) #:unless (= 0 (modulo (add1 j) x)))
                (list-ref l j)))
      (if (>= i (sub1 (length l)))
          l
          (loop l (add1 i)))))
  )

Testing:

(f 100)

Output:

'(1 3 7 9 13 15 21 25 31 33 37 43 49 51 63 67 69 73 75 79 87 93 99)