A067180 -id:A067180 - cp's OEIS frontend

A187553 Smallest prime p such that the sum of the squares of the digits of p equals n (or 0 if no such prime exists).

0, 11, 0, 2, 101111, 211, 2111, 101111111, 3, 13, 113, 112121, 23, 21221, 1123, 11213, 41, 1223, 313, 10133, 241, 233, 112223, 21313, 5, 431, 151, 1151, 13313, 251, 2333, 11251, 2243, 53, 1433, 1153, 61, 523, 1523, 11161, 443, 541, 353, 33413, 2621, 163, 1163, 13523, 7, 17

Offset: 1

Michel Lagneau, Mar 26 2011

a(3) = 0 because the numbers of the form 10..010..01 are divisible by 3. Conjecture : except for the numbers 1 and 3, for every possible square digit sum there exists a prime.

			a(13) = 23 because 2^2 + 3^2 = 13, and 23 is the least such prime.

Cf. A003132, A020449, A055016, A067180.

with(numtheory):for k from 2 to 100 do: id:=0:for p from 1 to 100000 while(id=0)
  do:n:=ithprime(p):l:=length(n):n0:=n:s:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s:=s+u^2:od: if s=k then id:=1:printf(`%d, `,n):else fi:od:od:

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.

Original entry on oeis.org

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

Dimitris Valianatos, Feb 12 2017

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.

			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.

Cf. A067180.

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 *)

{
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", "));
        )
}

cp's OEIS Frontend

A187553 Smallest prime p such that the sum of the squares of the digits of p equals n (or 0 if no such prime exists).

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