%I #29 Jun 29 2022 01:50:13
%S 69,105,263,311,341,425,491,515,561,575,725,993,1091,1133,1193,1521,
%T 1583,1991,2085,2153,2289,2351,2379,2415,2433,2441,2543,2601,3203,
%U 3225,3285,4245,4263,4541,4713,4955,5039,5073,5463,5499,5523,5651,5675,6779,7295
%N Numbers k such that k^3 + 2 is an emirp.
%H K. D. Bajpai, <a href="/A227271/b227271.txt">Table of n, a(n) for n = 1..1101</a>
%e a(1)=69: 69^3 + 2 = 328511; reversing its digits gives 115823, which is distinct from 328511, and both are primes, so 328511 is an emirp.
%p with(StringTools):K := proc(n) local a,b; a :=(n^3+2): b:=parse(Reverse(convert((a),string))):if (isprime(a)and isprime(b) and a<>b) then RETURN (n) fi: end: seq(K(n), n=1..100000); # _K. D. Bajpai_, Jul 04 2013
%p with(StringTools):K:=proc()local n,a,b,c; c:=1; for n from 1 to 500000 do; a:= (n^3+2); b:=parse(Reverse(convert((a),string))); if isprime(a)and isprime(b)and a<>b then lprint(c,n); c:=c+1;fi; od; end: K(); # _K. D. Bajpai_, Jul 04 2013
%t emirpQ[n_]:=Module[{idn=IntegerDigits[n^3+2],rev},rev=Reverse[idn];idn != rev && And@@PrimeQ[FromDigits/@{idn,rev}]]; Select[Range[7500],emirpQ] (* _Harvey P. Dale_, Jul 04 2013 *)
%Y Cf. A006567 (Emirps, primes whose reversal is a different prime).
%K nonn,base
%O 1,1
%A _K. D. Bajpai_, Jul 04 2013