Primes numbers with prime index

Question

Write a program or function that outputs/returns the first 10000 prime-indexed prime numbers.

If we call the n^th prime p(n), this list is

3, 5, 11, 17, 31, 41, 59 ... 1366661

because

p(p(1)) = p(2) = 3
p(p(2)) = p(3) = 5
p(p(3)) = p(5) = 11
p(p(4)) = p(7) = 17
...
p(p(10000)) = p(104729) = 1366661

Standard loopholes are forbidden, and standard output methods are allowed. You may answer with a full program, a named function, or an anonymous function.

Answer 1

MATLAB/Octave, 25 bytes

p=primes(2e6)
p(p(1:1e4))

It doesn't get much more straightforward than this.

Answer 2

Python, 72 bytes

P=p=1;l=[]
while p<82e5/6:l+=P%p*[p];P*=p*p;p+=1
for x in l:print l[x-1]

This terminates with a "list index out of range error" after printing the 10000 numbers, which is allowed by default.

Uses the Wilson's Theorem method to generate a list l of the primes up to the 10000th prime. Then, prints the primes with the positions in the list, shifted by 1 for zero-indexing, until we run out of bounds after the 10000th prime-th prime.

Conveniently, the upper bound of 1366661 can be estimated as 82e5/6 which is 1366666.6666666667, saving a char.

I'd like a single-loop method, printing prime-indexed primes as we add them, but it seems to be longer.

P=p=1;l=[]
while p<104730:
 l+=P%p*[p]
 if len(l)in P%p*l:print p
 P*=p*p;p+=1

Answer 3

J, 11 bytes

p:<:p:i.1e4

Outputs the primes in the format

3 5 11 17 31 41 59 67 83 109 127 ...

Explanation

        1e4  Fancy name for 10000
      i.     Integers from 0 to 9999
    p:       Index into primes: this gives 2 3 5 7 11 ...
  <:         Decrement each prime (J arrays are 0-based)
p:           Index into primes again

Answer 4

Mathematica, 26 25 23 bytes

Prime@Prime@Range@1*^4&

Pure function returning the list.

Answer 5

Haskell, 65 bytes

p=[x|x<-[2..],all((>0).mod x)[2..x-1]]
f=take 10000$map((0:p)!!)p

Outputs: [3,5,11,17,31,41,59,67,83,109,127.....<five hours later>...,1366661]

Not very fast. How it works: p is the infinite list of primes (naively checking all mod x ys for y in [2..x-1]) . Take the first 10000 elements of the list you get when 0:p!! (get nth element of p) is mapped on p. I have to adjust the list of primes where I take the elements from by prepending one number (-> 0:), because the index function (!!) is zero based.

Answer 6

AWK - 129 bytes

...oookay... too long to win points for compactness... but maybe it can gain some honor for the speed?

The x file:

BEGIN{n=2;i=0;while(n<1366662){if(n in L){p=L[n];del L[n]}else{P[p=n]=++i;if(i in P)print n}j=n+p;while(j in L)j=j+p;L[j]=p;n++}}

Running:

$ awk -f x | nl | tail
  9991  1365913
  9992  1365983
  9993  1366019
  9994  1366187
  9995  1366327
  9996  1366433
  9997  1366483
  9998  1366531
  9999  1366609
 10000  1366661

Readable:

BEGIN {
        n=2
        i=0
        while( n<1366662 ) {
                if( n in L ) {
                        p=L[n]
                        del L[n]
                } else {
                        P[p=n]=++i
                        if( i in P ) print n
                }
                j=n+p
                while( j in L ) j=j+p
                L[j]=p
                n++
        }
}

The program computes a stream of primes using L as "tape of numbers" holding found primes jumping around on L to flag the nearby numbers already known to have a divisor. These jumping primes will advance while the "tape of numbers" L is chopped off number by number from its beginning.

While chopping off the tape head L[n] being empty means there is no known (prime) divisor.

L[n] holding a value means, this value is a prime and known to divide n.

So either we have found a prime divisor or a new prime. Then ths prime will be advanced to the next L[n+m*p] on the tape found being empty.

This is like the Sieve of Eratosthenes "pulled thru a Klein's bottle". You always act on the tape start. Instead of firing multiples of primes thru the tape, you use the primes being found already as cursors jumping away from the tape start by multiple distances of their own value until a free position is found.

While the outer loop generates one prime or not prime decission per loop, the primes found get counted and stored in P as key, the value of this (key,value) pair is not relevant for the program flow.

If their key i happens to be in P already (i in P), we have a prime of the p(p(i)) breed.

Running:

$ time awk -f x.awk | wc -l
10000

real    0m3.675s
user    0m3.612s
sys     0m0.052s

Take into account that this code does not use external precalculated prime tables.

Time taken on my good old Thinkpad T60, so I think it deserves to be called fast.

Tested with mawk and gawk on Debian8/AMD64

Answer 7

PARI/GP, 25 bytes

apply(prime,primes(10^4))

Answer 8

CJam, 19

3D#{mp},_1e4<:(\f=p

You can try it online, but you'll need a little patience :p

For the record, the last number is 1366661.

Answer 9

Husk, 11 10 bytes

(or 6 5 bytes to list the first 10,000 prime-indexed prime numbers, followed by [the infinite number of] all the rest of them)

Thanks to user and Razetime for suggesting the 6 5-byte version!

↑!4İ⁰Ṡm!İp

Try it online! (unfortunately times-out on TIO before printing all 10,000 prime-indexed primes, but at least outputs what it's found so far...)

↑!4İ⁰Ṡm!İp
↑               # print the first
 !4             # 4th index of
   İ⁰           # powers of 10 (so, the first 10,000), of
     Ṡ          # hook: apply function to its own argument
      m!        # map index function to
        İp      # prime numbers
                # (implied by hook) to prime numbers

Answer 10

Perl, 55 bytes

use ntheory':all';forprimes{print nth_prime$_,$/}104729

---

Uses @DanaJ's Math::Prime::Util module for perl (loaded with the pragma ntheory). Get it with:

cpan install Math::Prime::Util
cpan install Math::Prime::Util::GMP

Answer 11

Japt, 11 bytes

Èj ©°Tj}jL²

Test it

Answer 12

Python 3, 76 109 bytes - disqualified - requalified?

edit: I am sorry, I was going for primes but missed the putting, didn´t reach the goal of "primes with primeindex". Your are free to grab and enhance. I'll be glad to be learning from you.

Based on Wilson Theorem

def p(x):
 c=n=2
 for i in range(3,x):
  c*=i-1
  if c%i!=0:
   print(n,i)
   n+=1

without index, 62 bytes

def p(x):
 c=2
 for i in range(3,x):
  c*=i-1
  if c%i!=0:print(i)

Anyone can give me an example or good link, how to make a lambda function from it?

edit: just to have it done right: 109bytes

def p(x):
 c=n=2
 p=[2]
 for i in range(2,x):
  c*=i-1
  if c%i!=0:
   p.append(i)
   if n in p: print(p[-1])
   n+=1

Answer 13

Vyxal H, 8 7 6 5 bytes

Saved 1 + 1 bytes thanks to Aaron Miller

²ʀǎ‹ǎ

Try it Online!

H flag - Push 100 to the stack
²ʀǎƛ‹ǎ
²        Squared (10,000)
 ʀ       Range [0, 10,000)
  ǎ      Map each a in that range to the ath prime
   ‹     Decrement each, because the challenge uses 1-indexing
    ǎ    Get the nth prime for each of those

Answer 14

05AB1E, 7 bytes (non-competing)

Code:

4°L<Ø<Ø

Try it online!, note that I have changed the 4 into a 2. If you have a lot of time, you can change the 2 back to 4, but this will take a lot of time. I need to fasten the algorithm for this.

Explanation:

4°       # Push 10000 (10 ^ 4)
  L      # Create the list [1 ... 10000]
   <     # Decrement on every element, [0 ... 9999]
    Ø    # Compute the nth prime
     <   # Decrement on every element
      Ø  # Compute the nth prime

Answer 15

Jelly, 6 bytes

ȷ4RÆN⁺

Try it online!

Obviously, this times out on TIO. Here is a version that prints as it goes, which reaches \$57493\$ (the \$765\$th term) before timing out

How it works

ȷ4RÆN⁺ - Main link. Takes no arguments
ȷ4     - 10000
  R    - [1, 2, 3, ..., 9999, 10000]
     ⁺ - Do twice:
   ÆN  -   n'th prime of each

Answer 16

05AB1E, 6 bytes

тnp<Ø

Try it online!

     Ø  # prime numbers with...
    <   # 1-based...
     Ø  # indices in...
  Åp    # first...
т       # 100...
 n      # ^ 2...
  Åp    # primes
        # implicit output