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

1,1

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.

Links

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

Example

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.

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}]

Crossrefs

Cf. A196698.

Sequence in context: A201940 A075113 A197739 * A011222 A157298 A070326

Adjacent sequences: A277574 A277575 A277576 * A277578 A277579 A277580

Keyword

nonn,base

Author

Lei Zhou, Oct 20 2016

Status

approved