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A super-prime is a prime number whose index in the list of all primes is also prime. The sequence looks like this:
3, 5, 11, 17, 31, 41, 59, 67, 83, 109, 127, 157, 179, 191, ...
This is sequence A006450 in the OEIS.
Given a positive integer, determine whether it is a super-prime.
2: false 3: true 4: false 5: true 7: false 11: true 13: false 17: true 709: true 851: false 991: true
This is code-golf, so the shortest answer in each language wins.
ÆRÆNċ
ÆRÆNċ Main link. Argument: n
ÆR Prime range; yield the array of all primes up to n.
ÆN N-th prime; for each p in the result, yield the p-th prime.
ċ Count the occurrences of n.
Thanks to user202729 for saving 3 bytes.
PrimeQ/@(#&&PrimePi@#)&
This makes use of the fact that Mathematica leaves most nonsensical expressions unevaluated (in this case, the logical And of two numbers) and Map can be applied to any expression, not just lists. So we compute the And of the input and its prime index, which just remains like that, and then we Map the primality test over this expression which turns the two operands of the And into booleans, such that the And can then be evaluated.
ÆRi³ÆP
Uses the same technique as my Japt answer: Generate the primes up to n, get the index of n in that list, and check that for primality. If n itself is not prime, the index is 0, which is also not prime, so 0 is returned anyway.
õ fj bU Ä j
This is actually very straight-forward, unlike my original submission:
õ fj bU Ä j
Uõ fj bU +1 j Ungolfed
Implicit: U = input integer
Uõ Generate the range [1..U].
fj Take only the items that are prime.
bU Take the (0-indexed) index of U in this list (-1 if it doesn't exist).
+1 j Add 1 and check for primality.
This is true iff U is at a prime index in the infinite list of primes.
Implicit: output result of last expression
^(?=((?=((\2x*?|^)((?!(xx+)\5+$)))xx)x)*(x*))\6(?4)xx
Try it online! - Perl
Try it online! - PCRE
Takes its input in unary, as a string of x characters whose length represents the number. Based on Prime counting function.
^ # tail = N = input value
(?=
# Calculate π(N) = the number of primes <= N, by counting
# from the largest to the smallest prime.
( # J = 0
(?=
# \2 starts at zero, and on each subsequent iteration, contains the
# difference N-P-(J-1) where P is the previously found prime, and J
# is the running total of our prime count.
(
(
\2 # Start from the previous value of \2. This
# will make tail = P-1, where P is the
# previously found prime.
x*? # Advance as little as necessary to make the
# following match, and add this to \2, while
# subtracting it from tail.
| # or
^ # Don't advance at all on the first iteration
# (when \2 is is still unset), so as to assert
# that N is not composite.
)
((?!(xx+)\5+$)) # Define and use subroutine (?4): Assert tail
# is prime. Note that this needs to be inside
# group \2 for it to work in PCRE1 and older
# versions of PCRE2, which atomicize groups
# that have nested backreferences.
)
xx # Assert tail is prime by eliminating the
# false positives 0, 1
)
x # J += 1; tail -= 1
)* # Iterate zero or more times, until there are no smaller primes
# remaining.
# At this point, head = π(N), and tail = N - π(N)
(x*) # \6 = tail = N - head = tool to make tail = head
)
\6 # tail = π(N)
(?4)xx # Assert tail is prime
Alternative 55 54 53 bytes:
^(?=((?=((\2x*?|^)((?!(xx+)\5+$|x?$))))x)*(x*))\6(?4)
Try it online! - Perl
Try it online! - PCRE
This eliminates the false negatives \$0\$ and \$1\$ in the definition of the primality subroutine instead of patching its second and third individual uses.
^(?=(\3*?(?!(xx+)\2+$)(x))*x)(?<-3>x){2,}(?<!\3|^\4+(x+x))
This uses the .NET feature of balanced groups to do the prime counting. Since .NET regex has no subroutines, this has three two copies of the primality test.
^ # tail = N = input value
# Calculate π(N) = the number of primes <= N, by counting
# from the largest to the smallest prime.
(?= # Atomic lookahead - lock in this match once it completes
(
\3*? # Advance as little as necessary to make the following
# match, but don't advance at all on the first iteration
# (when \3 is is still unset), so as to assert that N is
# not composite.
(?!(xx+)\2+$) # Assert tail is not composite
(x) # Eliminate the false primality positive of 0, and advance
# forward so that the next prime can be found (if we didn't
# do this, the regex engine would exit the loop due to a
# zero-width match); \3 = 1, to signal the first iteration
# has been done, and push a capture onto the Group 3 stack.
)* # Loop as many times as can be done until its stops
# matching.
x # Eliminate the false primality positive of 1
)
(?<-3>x){2,} # Pop all Group 3 captures off the stack, asserting that
# the count is ≥ 2, and doing head += 1 for each one. Since
# this isn't done atomically, we need to subsequently
# verify that all captures were popped, due to backtracking
# if the following assertion fails.
(?<!\3|^\4+(x+x)) # Assert that the Group 3 capture stack is empty, and that
# head is not composite (and since it's already been
# forced to be ≥ 2, this asserts it to be prime).
^(?=((?=((\2x*?|^)(?!(xx+)\4+$))xx)x)*(x*))\5(?!(xx+)\6+$)xx
Try it online! - Perl
Try it online! - Java
Try it online! - PCRE
Try it online! - .NET
This is a straight port of the Perl/PCRE version, to regex engines that lack subroutine calls but support nested backreferences. The subroutine call has been replaced with a copy of the routine.
regex), 61 bytes^(?=((?=((\6x*?|^)((?!(xx+)\5+$)))xx)(?=(\2))x)*(x*))\7(?4)xx
Try it online! - Perl
Try it online! - PCRE
Try it online! - Python import regex
This is a port of the Perl/PCRE version. \2 is copied into \6 using the lookahead (?=(\2)), since Python (even with import regex) lacks nested backreferences.
^(?=((?=((\6x*?|^)((?!(xx+)\5+$)))xx)(?=(\2))x)*(x*))\7\g<4>xx
Try it online! - PCRE
Try it online! - Ruby
This is a straight port of the Pythonregex version, to Ruby's subroutine call syntax.
regex / Ruby / .NET), 68 bytes^(?=((?=((\5x*?|^)(?!(xx+)\4+$))xx)(?=(\2))x)*(x*))\6(?!(xx+)\7+$)xx
Try it online! - Perl
Try it online! - Java
Try it online! - PCRE
Try it online! - Python import regex
Try it online! - Ruby
Try it online! - .NET
To be portable to all 6 regex engines, this uses neither nested backreferences nor subroutine calls.
p=lambda m:(m>1)*all(m%x for x in range(2,m))
f=lambda n:p(n)*p(len([*filter(p,range(n+1))]))
Returns 0/1, at most 4 bytes longer if it has to be True/False.
f, you can reformat your code like this and exclude it from the byte count.
\$\endgroup\$
Jul 7, 2017 at 21:57
n=input()
s={2};p=i=1
while i<n:
p*=i;i+=1
if p*p%i:s|={i}
print{n,len(s)}<=s
ÆCÆPaÆP
ÆC counts the number of primes less than or equal to the input (so, if the input is the nth prime, it returns n). Then ÆP tests this index for primality. Finally, a does a logical AND between this result and ÆP (primality test) of the original input.
p x=2==sum[1|0<-mod x<$>[1..x]]
f x=p$sum[1|y<-[1..x],p y,p x]
Try it online! Usage: f 991 yields True.
ṗṗ
One of the cases when the behaviours of functions returning non booleans is really helpful.
user's solution.
ṗ£İp
⁰ due to implicit input, plus, putting it on a seperate line just declares it as a new function, which doesn't affect the first line
\$\endgroup\$
ÝØ<Øså
ÝØ<Øså
Ý # Push range from 0 to input
Ø # Push nth prime number (vectorized over the array)
< # Decrement each element by one (vectorized)
Ø # Push nth prime number again
s # swap top items of stack (gets input)
å # Is the input in the list?
ÅP can be used instead of ÝØ (not that it saves any bytes, but hey, it's a built-in that does the trick).
\$\endgroup\$
&P_QP_smP_dS
&P_QP_smP_dS
Implicit input
mP_dS Primality of all numbers from 1 to N
s Sum of terms (equal to number of primes ≤ N)
P_ Are both that number
&P_Q and N prime?
P=Prime;!P@P@Range@#~FreeQ~#&
-6 bytes from @MartinEnder
sI~p>@hs
s - is_prime(input)
I~p>@hs - if ^:
~p> - first_n_primes(input)
@ - ^.index(input)
h - ^+1
s - is_prime(^)
def a(n):
r=[2];x=2
while x<n:x+=1;r+=[x]*all(x%i for i in r)
return{n,len(r)}<=set(r)
Constructs r, the list of primes <= n; if n is prime, then n is the len(r)'th prime. So n is a super prime iff n in r and len(r) in r.
math.primes, Saved 13 bytes thanks to @Bubbler! Saved 4 bytes thanks to @chunes!
[ dup nprimes index ?1+ prime? ]
[ dup nprimes index ?1+ prime? ]
dup nprimes ! First n primes (generates extra primes)
index ! Find the index of the input in that
! when the index is not found, an f is outputted
! If the index exists,
?1+ ! If it's a number, increment
! If it's f, output 0 (which is not prime)
prime? ! Is that prime?
[ dup primes-upto index ?1+ prime? ]. ?1+ increments an integer or consumes an f and gives 0.
\$\endgroup\$
nprimes instead of primes-upto. It's less efficient of course, but small price to pay.
\$\endgroup\$
g, 4 bytes¡†"æ
Try it Online! - test cases only
This is the same as the 5 byte answer below except the minimum is taken by the g flag rather than a g element.
¡†"æg
Try it Online!
Try it Online! - test cases only
This makes use of Prime counting function, based on Dennis's 05AB1E solution to that challenge. This is actually much faster than the other two 5 byte solutions presented below it.
¡ # Factorial
† # Number of distinct prime factors – when applied to the factorial, this
# gives the index of the greatest prime ≤ the input number (1-indexed).
" # Wrap the top two items (the input number, and the above result) in a list.
æ # Is the number prime? (Vectorized)
g # Minimum - effectively applies a boolean AND to the list
Alternative 5 bytes:
Þpḟ›æ
Try it Online!
Try it Online! - test cases only
Þp # An infinite list of primes
ḟ # Index of input number in that list (0-indexed; -1 if not found)
› # Add 1
æ # Is the number prime?
Another alternative 5 bytes:
Þp‹ǎc
Try it Online!
Try it Online! - test cases only
Port of Datboi's 05AB1E solution. This is much slower than the primary 5 byte solution above, but much faster than its 05AB1E counterpart.
Þp # An infinite list of primes
‹ # Subtract 1 from every item on the list
ǎ # Nth prime - vectorized to the entire list
c # Contains - Is the input number a member of that list?
D, \$ 8 \log_{256}(96) \approx \$ 6.58 bytesFAFLZPNP
Port of Deadcode's Vyxal answer.
It's all uppercase letters!
FAFLZPNP # Implicit input
F # Factorial
AF # Unique prime factors
L # Length
ZP # Pair
N # Is prime
P # Product
# Implicit output
{$/=&is-prime;all($_∈*,$/)([grep $/,1..$_])}
grep {.is-prime&&is-prime ++$},^∞
Six years of golfing experience let me cut down on this solution considerably.
~µ:||\_x0]{p=p-µq|~q=a|_xµp]q=q+1
~ | IF .... THEN (do nothing)
: the number 'a' (read from cmd line)
µ | is Prime
\_x0 ELSE (non-primes) quit, printing 0
] END IF
{ DO
In this next bit, q is raised by 1 every loop, and tested for primality.
p keeps track of how may primes we've seen (but does so negatively)
µq| test q for primality (-1 if so, 0 if not)
p=p- and subtract that result from p (at the start of QBIC: q = 1, p = 0)
~q=a| IF q == a
_xµp QUIT, and print the prime-test over p (note that -3 is as prime as 3 is)
] END IF
q=q+1 Reaise q and run again.
f=filter
p x=2==(length$f(\a->mod(x)a==0)[1..x])
s=map(\(_,x)->x)$f(\(x,_)->p x)$zip[1..]$f(p)[2..]
r x=x`elem`(take x s)
(\(_,x)->x) is snd, (\(x,_)->p x) is (p.fst). Both fst and snd are in Prelude, so no need for imports.
\$\endgroup\$
r x=elem x$take x s. However, in this case you can go pointfree (introducing backticks again) and omit the function name: elem<*>(`take`s).
\$\endgroup\$
x=#(input@@)a=function{p=1;k=$1==2;f=2;while(f<$1)do{p=p*$1%f;k=1;f=f+1};return p*k}r=1;i=2;while(i<x)do{if(a@i)then{r=r+1}i=i+1}print@((a@x*a@r)>0)
Saved 2 bytes thanks to Tom Carpenter.
A very naive implementation using built-in functions:
isprime(x)&isprime(nzz(primes(x)))
(p=@isprime)(x)&p(nnz(primes(x)))
\$\endgroup\$
Jul 8, 2017 at 17:28
return 1 if x is a super-prime, 0 otherwise.
without using an isprime-kind function.
x->a=[0,1];for i=3:x push!(a,0∉i%(2:i-1))end;a[sum(a)]&a[x]
P_hx.fP_Z
Uses a fairly different approach to the existing Pyth solution. Returns True for super-primes, False for non-super-primes, and [] (falsy) for composite numbers.
Tired of all the extremely obscure golfing languages in this thread? Here's the practical language solution you're looking for! Let me know if you see a way to shave off a few bytes.
require'prime';n=gets.to_i;p n.prime?&&(Prime.take(n).index(n)+1).prime?
.Ø>‚pß
Try it online! - 05AB1E
Try it online! - 05AB1E - test cases only
Try it online! - 05AB1E (legacy) - test cases only
Equal in length to Datboi's solution, but much faster on numbers that are not super-prime.
.Ø # Get 0-indexed number of the greatest prime ≤ input number
> # Add 1
‚ # Wrap the top two items (the input number, and the above result) in a list
p # Is the number prime? (Vectorized)
ß # Minimum - effectively applies a boolean AND to the list
Alternative 6 bytes:
!fg‚pß
Try it online!
Try it online! - test cases only
This makes use of Dennis's Prime counting function solution, which is equal in length to (but slower than) the ones using built-in prime-counting functions.
#define s for(i=e;--i&&e%i;)p+=i <3
p,r,i;m(e){s;for(p=r=i==e%2;--e;)s;e=p+1;s;return r<i<r;}
(added whitespace for readability)
i==e%2 \$\equiv\$ i==1&er<i<r \$\equiv\$ 1<i<r \$\equiv\$ i<2&r \$\equiv\$ i/2<r \$\equiv\$ i<r+r \$\equiv\$ i<2*r