A005187
a(n) = a(floor(n/2)) + n,a(0)=0; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.
Has a very screaming sound to it. you can also hear some rhythmic clicks in the audio.


Residue counts:
2   [499999, 500001]
3   [332863, 332985, 334152]
4   [249947, 249353, 250052, 250648]
5   [200125, 199880, 200070, 200009, 199916]
6   [166431, 166493, 167076, 166432, 166492, 167076]
7   [138067, 144156, 147077, 139676, 140040, 147280, 143704]
8   [139951, 147388, 142166, 126853, 109996, 101965, 107886, 123795]
9   [110947, 110990, 111382, 110952, 110993, 111386, 110964, 111002, 111384]
10   [100062, 99940, 100035, 100005, 99958, 100063, 99940, 100035, 100004, 99958]
11   [90898, 90918, 90902, 90916, 90901, 90911, 90912, 90908, 90910, 90906, 90918]
12   [84121, 82440, 83655, 83850, 82405, 84331, 82310, 84053, 83421, 82582, 84087, 82745]

Density information:
A(x) at x=10,100,1000,...:
[6, 51, 503, 5003, 50003, 500004,]

A(x)/x at x=10,100,1000,...:
0.6,0.51,0.503,0.5003,0.50003,0.500004

First 100 first differences:
[2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1]

First 99 second differences:
-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,4,-4,1,-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,5,-5,1,-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,4,-4,1,-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,6,-6,1,-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,4,-4,1,-1,2,-2,1,-1,3,-3,1,-1,2,-2,1,-1,5,-5,1,-1,2,-2,


The spectrogram has a very blocky look and a very uniform look all around. Some interesting frequencies are at 5500,2800,1400 hz and there is a big distinctive absence of the pitch
~7300 as seen by the black line. 

We can prove this sequence is strictly increasing through induction: 
    base case: a(0)=0, a(1)=1, a(2)=3 so a(n) is strictly increasing for all n<3
    Inductive hypo: suppose a(n) is strictly increaing for all n<m that is for all n<m, a(n)<a(n+1). We want to show that a(m)<a(m+1).
    if m is even, m=2k so a(m)=a(k)+m, a(m+1)=a(k)+m+1 do a(m)<a(m+1)
    if m is odd, m=2k+1 so a(m)=a(k)+m, a(m+1)=a(k+1)+m+1. By induction hypo, we have a(k)<a(k+1) so a(k)+m<a(k+1)+m+1 or a(m)<a(m+1).
    Thus by the inductive hypothesis a(n) is strictly increasing for all non negative n.

Another interesting fact is that a(n)+1 = a(n+1) iff n is even. 

--Jayme Kim