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%I #18 Jul 10 2018 02:20:18
%S 13,31,11313331,11333131,13111333,13131133,13131331,13133311,13311313,
%T 31133131,33113131,1113131333,1131131333,1131311333,1131331133,
%U 1133111333,1133113133,1133133311,1133311313,1133313113,1133313131
%N Primes consisting only of digits 1 and 3 occurring with equal frequency.
%C There are 19 digit pairs which can produce such primes. (1, 0), (7, 0), (1, 3), (1, 4), (1, 6), (1, 7), (1, 9), (2, 3), (2, 9), (3, 4), (3, 5), (3, 7), (3, 8), (4, 7), (4, 9), (5, 9), (6, 7), (7, 9), (8, 9). - corrected by _Robert Israel_, Jul 10 2018
%C The number of digits is even and not divisible by 3. - _Robert Israel_, Jul 09 2018
%H Robert Israel, <a href="/A087511/b087511.txt">Table of n, a(n) for n = 1..10000</a>
%p sort(select(isprime,[seq(seq((10^(2*d)-1)/9+2*add(10^i, i=s), s=combinat:-choose([$0..(2*d-1)], d)), d=[1,2,4,5,7,8,10])])); # _Robert Israel_, Jul 09 2018
%t Union[FromDigits/@Select[Flatten[Table[Tuples[{1,3},k],{k,10}],1], PrimeQ[FromDigits[#]] && Count[#,1]==Count[#,3] &]] (* _Jayanta Basu_, May 19 2013 *)
%o (PARI) d1=1; d2=3; k=0; a=vector(100); for(n=1, 3000, B=binary(n); L=length(B); s=sum(j=1, length(B), B[j]); if(L%2==0 & s==L/2, C=vector(L, n, (d2-d1)*B[n]+d1); p=subst(Pol(C), x, 10); if(isprime(p), if(k<100, k++; a[k]=p)); D=vector(L, n, d2-(d2-d1)*B[n]); q=subst(Pol(D), x, 10); if(isprime(q ), if(k<100, k++; a[k]=q))); ); a=vector(k, n, a[n]); vecsort(a)
%Y Cf. A087510, A087512, A087513.
%K base,nonn
%O 1,1
%A _Paul D. Hanna_ and _Amarnath Murthy_, Sep 11 2003
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