A331112 Sum of the digits of the n-th prime number in balanced ternary.

0, 1, -1, 1, 1, 3, -1, 1, -1, 1, 3, 3, -3, -1, -1, -1, -1, 1, 3, -1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, -1, -3, -1, 1, -3, -1, 1, 1, -1, 1, -1, 1, 1, 3, 1, 3, 1, 1, 1, 3, -1, -1, 1, 1, -1, 1, 1, 3, 3, 3, 5, -1, 3, -1, 1, 1, 3, 5, 1, 3, 3, 3, -3, -1, -1, -3, -1, -1, -3, 1, -3, -1, -1, 1, 1, 1, -3, -1, -1, 1, -1, -1, 1, -1, 3, -1, -1, 1, 3, 1

Offset

1,6

Links

Table of n, a(n) for n=1..100.

Wikipedia, Balanced ternary

Formula

a(n) = A065363(A000040(n)). - Alois P. Heinz, Jan 09 2020

Example

Using T for -1 and _bt as suffix for balanced ternary: 2_10 = 1T_bt, sum of digits is zero; 3_10 = 10_bt, sum of digits is 1 and 5_10 = 1TT, sum of digits = -1.

Maple

b:= proc(n) `if`(n=0, 0, (d-> `if`(d=2,
      b(q+1)-1, d+b(q)))(irem(n, 3, 'q')))
    end:
a:= n-> b(ithprime(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 09 2020

Prog

(C)
#include <stdio.h>
#include <math.h>
#define N 1000 /* Largest prime considered - 1  */
char x[N];
int main()
{
  int i, n, v, s, r;
  for (i=4; i<N; i+=2)
    x[i] = 1;
  for (n=3; n<sqrt(N*1.0); n +=2)
    for (i=n+n; i<N; i+=n)
      x[i] = 1;
  for (n = 2; n < N; n++) {
    if (x[n] == 0) {
      v = n;
      s = 0;
      while (v != 0) {
        r = v % 3;
        if (r == 2)
          r = -1;
        s = s + r;
        v = (v - r) / 3;
      }
      printf("%d, ", s);
    }
  }
  printf("\n");
}

Crossrefs

See A007605 (sum of digits of primes in base 10); A239619 (sum of digits of primes in base 3).

Cf. A000040, A065363, A117966.

Sequence in context: A046555 A379484 A336467 * A029382 A073780 A124389

Adjacent sequences: A331109 A331110 A331111 * A331113 A331114 A331115

Keyword

sign,base,easy

Author

Thomas König, Jan 09 2020

Status

approved