A282342 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

0, 2, 3, 13, 23, 0, 43, 53, 0, 73, 83, 0, 139, 149, 0, 277, 359, 0, 379, 389, 0, 499, 599, 0, 997, 1889, 0, 1999, 2999, 0, 4999, 6899, 0, 17989, 18899, 0, 29989, 39989, 0, 49999, 59999, 0, 79999, 98999, 0, 199999, 389999, 0, 598999, 599999, 0, 799999, 989999, 0, 2998999

Offset

1,2

Comments

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

Other terms:

a(97) = 79999999999;

a(98) = 98999999999;

a(100) = 298999999999;

a(1000) = 299989999999999999999999999999999999999999999999999999999999999999

9999999999999999999999999999999999999999999999.

Links

Table of n, a(n) for n=1..55.

Example

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

Mathematica

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 *)

Prog

(PARI) {
print1(0", "2", ");
n=3; p=3; sp=3;
while(p<1000000,
        while(sp<>n,
                  p=nextprime(p+1);
                  sp=sumdigits(p);
                );
                 print1(p", ");
                 n++; if(n%3==0, n++; print1(0", "));
        )
}

Crossrefs

Cf. A067180.

Sequence in context: A138699 A341713 A077248 * A137248 A355438 A358427

Adjacent sequences: A282339 A282340 A282341 * A282343 A282344 A282345

Keyword

nonn,base

Author

Dimitris Valianatos, Feb 12 2017

Status

approved