A282342 - OEIS

%I #39 Mar 07 2017 05:08:40

%S 0,2,3,13,23,0,43,53,0,73,83,0,139,149,0,277,359,0,379,389,0,499,599,

%T 0,997,1889,0,1999,2999,0,4999,6899,0,17989,18899,0,29989,39989,0,

%U 49999,59999,0,79999,98999,0,199999,389999,0,598999,599999,0,799999,989999,0,2998999

%N a(n) is the smallest prime number, with sum of digits equals n and a(n) is greater than previous nonzero terms, except if this is not possible in which case a(n)=0

%C I conjecture that there are prime numbers for every n, if n is not divisible by 3.

%C Other terms:

%C a(97) = 79999999999;

%C a(98) = 98999999999;

%C a(100) = 298999999999;

%C a(1000) = 299989999999999999999999999999999999999999999999999999999999999999

%C           9999999999999999999999999999999999999999999999.

%e a(23) = 599 because 599 is a prime number greater than a(22) = 499 and the sum of its digits is 5 + 9 + 9 = 23.

%e a(24) = 0 because 24 (mod 3) = 0.

%t a = {1}; Do[If[n != 3 && Divisible[n, 3], AppendTo[a, 0], p = NextPrime@ Max@ a; While[Total@ IntegerDigits@ p != n, p = NextPrime@ p]; AppendTo[a, p]], {n, 2, 57}]; a (* _Michael De Vlieger_, Feb 12 2017 *)

%o (PARI) {

%o print1(0", "2", ");

%o n=3;p=3;sp=3;

%o while(p<1000000,

%o         while(sp<>n,

%o                   p=nextprime(p+1);

%o                   sp=sumdigits(p);

%o                 );

%o                  print1(p", ");

%o                  n++;if(n%3==0,n++;print1(0", "));

%o         )

%o }

%Y Cf. A067180.

%K nonn,base

%O 1,2

%A _Dimitris Valianatos_, Feb 12 2017