The On-Line Encyclopedia of Integer Sequences (OEIS)

A282342 revision #36

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

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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) = 2999899999999999999999999999999999999999999999999999999999999999999999999

999999999999999999999999999999999999999.

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", "));

)