

A277577


a(n) is the number of primes of the form (6n+3) +/ 3^k +/ 1, 1<=k<(t2) and t is the number of digits of (6n+3)'s balanced ternary representation.


1



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
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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<=(32)=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<=(42)=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



