With everyone clamoring for an explanation of the sequence I mentioned in a recent post,
I can't continue to ignore all the minds thirsting for knowledge.
Starting from x in (0,1), set
N(1) = floor(-ln(x)) and
x(1) = frac(-ln(x))
(where frac(c) := c - floor(c)).
Continue in the same way:
N(k+1) = floor(-ln(x(k)) and
x(k+1) = frac(-ln(x(k)).
Then the continued logarithm expansion of x is by definition N(1), N(2), N(3), . . . .
According to Mathematica, the continued log expansion of pi/4 is this:
0,1,0,0,1,0,2,0,1,1,2,2,1,0,0,1,0,1,0,2,0,0,1,0,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,2,1,0,1,1,0,0,0,0,0,3,1,0,1,1,4,4,1,0,1,1,0,1,1,0,0,2,1,0,1,0,1,0,4,4,0,0,0,1,0,1,0,3,1,0,0,1,0,1,2,2,0,1,3
(first 100 terms, anyhow).
I am not totally sure of their accuracy; I got Mathematica to create the exact expression for N(k), 1 <= k <= 100,
and then asked Mathematica to evaluate it:
If the quoted sequence is accurate, this idea does not appear in OEIS. (Otherwise, it may well.)
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In[1]:= x[0] = Pi/4
In[2]:= For[k=1,k<=100,k=k+1,{tmp=-Log[x[k-1]],n[k]=Floor[tmp],x[k]=tmp-n[k]}]
In[3]:= For[k=1,k<=100,k=k+1,Print[k,n[k] ]]
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Note: This would be slightly more analogous to a continued fraction expansion if the
function replacing 1/x -- that is, -ln(x) -- instead of mapping (0,1) to (0,oo), mapped
(0,1) to (1,oo). This is easily achieved by using 1 - ln(x) in lieu of -ln(x). The only difference
would be that all the terms of the sequence of integers would be greater than they are by 1:
1,2,1,1,2,1,3,1,2,2,3,3,2,1,1,2,1,2,1,3,1,1,2,1,3,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,3,2,1,2,2,1,1,1,1,1,4,2,1,2,2,5,5,2,1,2,2,1,2,2,1,1,3,2,1,2,1,2,1,5,5,1,1,1,2,1,2,1,4,2,1,1,2,1,2,3,3,1,2,4
(This is also not in OEIS.)
--Dan
