|
|
A098295
|
|
((3/2)^n)/2^a(n) lies in the half-open interval [1,2).
|
|
1
|
|
|
0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43, 43
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
Stacking perfect fifths (the frequency ratio of a fifth is 3/2), a division by 2^a(n) leads the equivalent tone belonging to the first octave interval [1,2). For example, the third fifth, (3/2)^3, falls into the second octave. This means it lies in the interval [2^1,2^2)=[2,4). Hence ((3/2)^3)/2^1 belongs to the first octave, the interval [1,2).
This sequence coincides for the first 93 term with the floor of y(n)= 4*Pi*log(phi)*n/(Pi^2 + (2*log(phi)^2)), with phi:=(1+sqrt(5))/2. a(n) = floor(y(n)), for n=1..93. Note that y(n) is not the imaginary part of the zero of the Fibonacci function because of a different bracket setting. See A214656. - Wolfdieter Lang, Jul 24 2012
|
|
LINKS
|
|
|
FORMULA
|
a(n) = ceiling(tau*n)-1 with tau = log(3)/log(2)-1 = 0.58496250072..., n >= 1.
|
|
EXAMPLE
|
(3/2)^12 lies in the eighth octave [2^7,2^8) and
((3/2)^12)/2^a(12)= ((3/2)^12)/2^7 = 3^12/2^19 = 531441/524288 = 1.01363... belongs to the first octave [1,2). This ratio is called the Pythagorean comma.
|
|
PROG
|
|
|
CROSSREFS
|
This sequence differs from A074840 for the first time at entry a(41)=23: A074840(41)=24.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|