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Prime numbers with every digit a perfect cube, i.e., consisting of only digits 0, 1 and 8.
11

%I #29 Sep 08 2022 08:45:03

%S 11,101,181,811,881,1181,1801,1811,8011,8081,8101,8111,10111,10181,

%T 11801,18181,80111,81001,81101,81181,88001,88801,88811,100801,100811,

%U 101081,101111,108011,108881,110881,118081,118801,180001,180181,180811

%N Prime numbers with every digit a perfect cube, i.e., consisting of only digits 0, 1 and 8.

%C The intersection with A007500 is listed in A199328. - _M. F. Hasler_, Nov 05 2011

%H Robert Israel, <a href="/A061247/b061247.txt">Table of n, a(n) for n = 1..17482</a>

%e a(6) = 1801, 1801 is a prime and consists of only 1, 8 and 0.

%p N:= 1000: # to get the first N entries

%p count:= 0:

%p allowed:= {0,1,8}:

%p nallowed:= nops(allowed):

%p subst:= seq(i=allowed[i+1],i=0..nallowed-1);

%p for d from 1 while count < N do

%p for x1 from 1 to nallowed-1 while count < N do

%p for t from 0 to nallowed^d-1 while count < N do

%p L:= subs(subst,convert(x1*nallowed^d+t,base,nallowed));

%p X:= add(L[i]*10^(i-1),i=1..d+1);

%p if isprime(X) then

%p count:= count+1;

%p A[count]:= X;

%p fi

%p od od od:

%p seq(A[n],n=1..N); # _Robert Israel_, Apr 20 2014

%t Select[Prime[Range[50000]],Length[Union[{0,1,8},IntegerDigits[ # ]]] == 3&] (* _Stefan Steinerberger_, Jun 10 2007 *)

%t Select[FromDigits/@Tuples[{0,1,8},6],PrimeQ] (* _Harvey P. Dale_, Jan 12 2016 *)

%o (PARI) a(n=50, L=[0, 1, 8], show=0)={my(t); for(d=1, 1e9, u=vector(d, i, 10^(d-i))~; forvec(v=vector(d, i, [1+(i==1 && !L[1]), #L]), ispseudoprime(t=vector(d, i, L[v[i]])*u) || next; show && print1(t", "); n-- || return(t)))} \\ _M. F. Hasler_, Nov 05 2011

%o (Magma) [NthPrime(n): n in [1..2*10^4] | forall{d: d in Intseq(NthPrime(n)) | d in [0, 1, 8]}]; // _Vincenzo Librandi_, May 15 2019

%Y Cf. A061246, A020449-A020472, A199325-A199329.

%K nonn,base,less

%O 1,1

%A _Amarnath Murthy_, Apr 23 2001

%E Corrected and extended by _Stefan Steinerberger_, Jun 10 2007