A166681 -id:A166681 - cp's OEIS frontend

A225035 Primes such that there is a nontrivial rearrangement of the digits which is a prime.

13, 17, 31, 37, 71, 73, 79, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 239, 241, 251, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 359, 367, 373, 379, 389, 397, 401, 419, 421

Offset: 1

Jayanta Basu, Apr 24 2013

The new prime is necessarily different from the original prime (so 11, for example) is not a term. - N. J. A. Sloane, Jan 22 2023
Permutations producing leading zeros are allowed: thus 101 is in the sequence because a nontrivial permutation of its digits is 011. - Robert Israel, Aug 13 2019
It seems reasonable to expect that the proportion of n-digit primes that are in this sequence approaches 1 as n increases. - Peter Munn, Sep 13 2022

			13 is a term since a nontrivial permutation of its digits yields 31, which is also a prime.

H.-E. Richert, On permutation prime numbers, Norsk. Mat. Tidsskr. 33 (1951), p. 50-53.
Joe Roberts, Lure of the Integers, Math. Assoc. of Amer., 1992, p. 293.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Second Edition, Cambridge University Press, p. 121.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A052902, A007933, A007935, A166681.

See A055387, A359136-A359139 for other versions.

dmax:=3: # for all terms of up to dmax digits
Res:= {}:
p:= 1:
do
  p:= nextprime(p);
  if p > 10^dmax then break fi;
  L:= sort(convert(p,base,10),`>`);
  m:= add(L[i]*10^(i-1),i=1..nops(L));
  if assigned(A[m]) then
    if ilog10(A[m])=ilog10(p) then
      Res:= Res union {A[m], p}
    else Res:= Res union {p}
    fi
  else A[m]:= p
  fi
od:
sort(convert(Res,list)); # Robert Israel, Aug 13 2019

t={}; Do[p = Prime[n]; list1 = Permutations[IntegerDigits[p]]; If[Length[ Select[Table[FromDigits[n], {n,list1}], PrimeQ]] > 1, AppendTo[t,p]], {n,84}]; t

is(p) = if(isprime(p), my(d=vecsort(digits(p))); d==vector(#d,x,1)&&return(1); forperm(d, e, my(c = fromdigits(Vec(e))); p!=c && isprime(c) && return(1))); \\ Ruud H.G. van Tol, Jan 22 2023

from sympy import isprime
from itertools import permutations
def ok(n):
    if not isprime(n): return False
    perms = (int("".join(p)) for p in permutations(str(n)))
    return any(isprime(t) for t in perms if t != n)
print([k for k in range(500) if ok(k)]) # Michael S. Branicky, Sep 14 2022

Edited by N. J. A. Sloane, Jan 22 2023

A358020 Least prime number > prime(n) (n >= 5) whose set of decimal digits coincides with the set of decimal digits of prime(n), or -1 if no such prime exists.

Original entry on oeis.org

1111111111111111111, 31, 71, 191, 223, 229, 113, 73, 4111, 433, 4447, 353, 599, 661, 677, 1117, 337, 97, 383, 8999, 797, 10111, 1013, 701, 1009, 131, 271, 311, 173, 193, 419, 1151, 571, 613, 617, 317, 197, 811, 199, 1193, 719, 911, 2111, 233, 277, 929, 2333, 293, 421, 521, 2557

Offset: 5

Jean-Marc Rebert, Oct 24 2022

			prime(6) = 13 and the prime number 31 have the same set of digits {1,3}, and 31 is the smallest such number, hence a(6) = 31.
prime(13) = 41 and the prime number 4111 have the same set of digits {1,4}, and 4111 is the smallest such number, hence a(13) = 4111.
prime(20) = 71 and the prime number 1117 have the same set of digits {1,7}, and 1117 is the smallest such number, hence a(20) = 1117.

Jean-Marc Rebert, Table of n, a(n) for n = 5..10000

Cf. A000040, A357096, A004022, A004023, A020451, A020455, A050288, A166681.

N:= 60: # for a(5)..a(N)
A:= Array(5..N):
R:= 1111111111111111111:
A[5]:= R: count:= 1:
for k from 6 while count < N-4 do
  p:= ithprime(k);
  S:= convert(convert(p,base,10),set);
  if assigned(V[S]) and V[S]<=N then A[V[S]]:= p; count:=count+1;  fi;
  V[S]:= k;
od:
convert(A,list); # Robert Israel, Oct 25 2022

a(n)=my(m=Set(digits(prime(n)))); if(n<5, return()); if(n==5,return(1111111111111111111));forprime(p=prime(n+1), , if(Set(digits(p))==m, return(p)))

from sympy import isprime, prime
from itertools import count, product
def a(n):
    pn = prime(n)
    s = str(pn)
    for d in count(len(s)):
        for p in product(set(s), repeat=d):
            if p[0] == "0": continue
            t = int("".join(p))
            if t > pn and isprime(t):
                return t
print([a(n) for n in range(5, 56)]) # Michael S. Branicky, Oct 25 2022

cp's OEIS Frontend

A225035 Primes such that there is a nontrivial rearrangement of the digits which is a prime.

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