A240623 Prime numbers n such that replacing each digit d in the decimal expansion of n with d^d produces a prime. Zeros are allowed with the convention 0^0 = 1.

11, 13, 17, 31, 53, 61, 71, 79, 151, 167, 229, 233, 251, 263, 311, 313, 331, 337, 349, 367, 389, 409, 419, 443, 673, 751, 947, 971, 991, 1433, 1531, 1699, 1733, 1993, 2011, 2027, 2053, 2063, 2081, 2111, 2141, 2153, 2221, 2333, 2393, 2503, 2521, 2833, 2963

Offset: 1

Author

Michel Lagneau, Apr 09 2014

Keywords

• nonn
• base

Examples

			263 is in the sequence because 263 becomes 44665627 which is also prime, where 44665627 is the concatenation of the numbers (2^2, 6^6, 3^3) = (4, 46656, 27).
2503 is in the sequence because 2503 becomes 43125127 which is also prime, where 43125127 is the concatenation of the numbers (2^2, 5^5, 0^0, 3^3) = (4, 3125, 1, 27).
		

Crossrefs

Cf. A068492, A240624.

Programs

Maple

with(numtheory):T:=array(1..10):L:=array(1..10):
   for n from 1 to 1000 do:
     p:=ithprime(n):k:=0:s:=0:j:=0:
     x:=convert(p,base,10):n1:=nops(x):
       for m from n1 by -1 to 1 do:
       k:=k+1:T[k]:=x[k]^x[k]:L[k]:=length(T[k]):
       od:
       n2:=sum('L[j]', 'j'=1..n1):s2:=0:
         for u from n1 by -1 to 1 do:
         s2:=s2+T[u]*10^(n2-L[u]):n2:=n2-L[u]:
         od:
            if type(s2,prime)=true
            then
            printf(`%d, `,p):
            else
            fi:
     od: