A277577 a(n) is the number of primes of the form (6n+3) +/- 3^k +/- 1, 1<=k<(t-2) and t is the number of digits of (6n+3)'s balanced ternary representation.

4, 7, 7, 6, 6, 6, 9, 8, 8, 8, 9, 8, 8, 7, 6, 9, 8, 8, 5, 8, 9, 8, 10, 9, 10, 9, 9, 9, 8, 10, 11, 8, 10, 7, 6, 8, 8, 11, 9, 7, 8, 9, 8, 10, 9, 10, 8, 7, 5, 8, 10, 6, 10, 5, 5, 10, 7, 9, 9, 9, 8, 12, 10, 8, 11, 10, 9, 10, 10, 9, 11, 11, 12, 8, 8, 11, 10, 9, 9, 10, 10, 10, 8, 7, 12, 10, 8, 8, 8, 7, 9, 6, 9, 11, 10, 11, 8, 11, 10, 11, 9, 10, 10, 9, 8, 10, 11, 11, 8, 9, 9

Offset: 1

Author

Lei Zhou, Oct 20 2016

Keywords

• nonn
• base

Comments

The first a few of a(n)=0 terms of this sequence appear on n=72030, 228439, 282059, 383424, 384204, 593477, 710787, 736541, 1057465, 1628226, 1831745, 1892512, 1922647, 2128995, 2244660, 2260650, 2276272, 2289706, 2374644, 2507484, 2633477, 2681747, 2695747, 2875824, 2889472, 2913659 up to n<=3000000.

Examples

			n=1: 6n+3=9. 9=(100)_bt. 1<=k<=(3-2)=1. When k=1, (6n+3)+/-3^1+/-1 = 5, 7, 11, 13, all primes. So a(1) = 4;
n=2: 6n+3=15. 15=(1TT0)_bt. 1<=k<=(4-2)=2. When k=1, (6n+3)+/-3^1+/-1 = 11, 13, 17, 19, all primes; when k=2, (6n+3)+/-3^2+/-1 = 5, 7, 23, 25, three primes.  Totals 7. So a(2) = 7.
		

Links

• Lei Zhou, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A196698.

Programs

Mathematica

BTDigits[m_Integer,g_] :=
  Module[{n = m, d, sign, t = g},
   If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n];
    d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];
    While[Length[t] < d, PrependTo[t, 0]];
    t[[Length[t] + 1 - d]] = sign;
    t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
Table[trib = 6*n + 3; t = BTDigits[trib,{}]; l = Length[t]; ct = 0;
Do[If[PrimeQ[trib - 3^j - 1], ct++];
  If[PrimeQ[trib - 3^j + 1], ct++]; If[PrimeQ[trib + 3^j + 1], ct++];
  If[PrimeQ[trib + 3^j - 1], ct++], {j, 1, l - 2}]; ct, {n, 1, 111}]