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.) ---------------------------------------------------------------------------------------------------------- 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] ]] ---------------------------------------------------------------------------------------------------------- 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