A102703: Numbers k such that 100*k+99 is prime.

Noisy static sound with medium-high pitch. Minimal fluctuation in volume and pitch.

Residue counts:
2   [71967, 72048]
3   [0, 71977, 72038]
4   [35905, 36117, 36062, 35931]
5   [28947, 28799, 28721, 28821, 28727]
6   [0, 36042, 36032, 0, 35935, 36006]
7   [24019, 24085, 23948, 0, 23999, 24030, 23934]
8   [17937, 17986, 18033, 17997, 17968, 18131, 18029, 17934]
9   [0, 23980, 23970, 0, 23979, 24012, 0, 24018, 24056]
10   [14466, 14407, 14413, 14497, 14372, 14481, 14392, 14308, 14324, 14355]
11   [0, 14409, 14403, 14505, 14401, 14336, 14367, 14438, 14433, 14312, 14411]
12   [0, 18024, 18047, 0, 17920, 18093, 0, 18018, 17985, 0, 18015, 17913]

Density information:
A(x) at x=10,100,1000,...:
[3, 29, 230, 1968, 16657, 144015]

A(x)/x at x=10,100,1000,...:
0.3,0.29,0.23,0.1968,0.16657,0.144015,

First 100 first differences:
[3, 1, 8, 1, 2, 3, 1, 3, 3, 3, 3, 2, 6, 7, 2, 1, 3, 8, 1, 3, 3, 6, 2, 9, 1, 3, 
2, 9, 3, 1, 3, 5, 1, 3, 11, 1, 2, 3, 1, 3, 2, 7, 5, 1, 14, 4, 2, 3, 6, 7, 8, 7, 
2, 3, 6, 1, 2, 9, 4, 3, 2, 3, 6, 3, 10, 2, 1, 3, 3, 6, 5, 3, 4, 9, 2, 21, 1, 3, 
2, 7, 2, 4, 9, 6, 8, 1, 3, 3, 3, 3, 5, 1, 5, 4, 5, 3, 3, 3, 4, 2]

First 99 second differences:
[-2,7,-7,1,1,-2,2,0,0,0,-1,4,1,-5,-1,2,5,-7,2,0,3,-4,7,-8,2,-1,7,-6,-2,2,2,-4,2,8,
-10,1,1,-2,2,-1,5,-2,-4,13,-10,-2,1,3,1,1,-1,-5,1,3,-5,1,7,-5,-1,-1,1,3,-3,7,-8,-1,
2,0,3,-1,-2,1,5,-7,19,-20,2,-1,5,-5,2,5,-3,2,-7,2,0,0,0,2,-4,4,-1,1,-2,0,0,1,-2,]

There is a prominent spectral line at 14700=44100/3 hz. 

Since 99 is divisible by 3, 100k+99 is divisible by 3 if k is divisible by 3.
So no multiples of three appear in this sequence.

This sequence is infinite because of Dirichlet’s theorem on arithmetic progressions. 
The theorem states that if a and d are coprime integers (they have no common factors 
besides 1), then there are infinitely many primes of the form a + nd, where n is a 
positive integer. In this sequence, 100k + 99 can be rewritten as 99 + 100k, where 
a = 99 and d = 100. Since 99 and 100 are coprime, Dirichlet’s theorem guarantees 
that there are infinitely many primes of the form 99 + 100k. Therefore, there are 
infinitely many values of k in the sequence.

The values of A(x)/x gradually decrease as x increases, suggesting that the sequence 
becomes less dense among larger integers. In other words, valid choices for k become 
more spread out as the integers grow larger.

-- Lauren Yee