Denote by a[0], a[1], ... the continued fraction terms of pi (= 3,7,15,1,... = A001203). If there are rationals A,B,C,D such that pi = (Ae+B)/(Ce+D), then there is a quasiperiod q and a preamble p such that the second differences a[n+2q]-2a[n+q]+a[n] = 0 for all n>p. For e itself, q=3 and p=1. (Or any positive multiple of 3 and any positive integer). E.g., define a second-differencer: In[13]:= dif2[L_List, q_Integer:1] := Drop[L, 2*q] - 2*Drop[Drop[L, -q], q] + Drop[L, -2*q] In[16]:= dif2[{a, b, c, d, e}, 2] Out[16]= {a - 2 c + e} Then In[163]:= dif2[ContinuedFraction[E, 22], 3] Out[163]= {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=1,q=3). In[166]:= dif2[ContinuedFraction[E^(2/3), 22], 5]
Out[166]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=0,q=5). In[171]:= dif2[ContinuedFraction[(E - 1)/(1 - E/6), 333], 130]
Out[171]= {2, 1, 11, -1, 0, 62, 1, 3, 0, -1, 0, 1, -5, 5, 5, 1, -2, \ -153, -2, 0, 1, -58, 5, 2, 1, 152, 2, 57, -6, 3, -2, -21, 6, -10, 3, \ 22, -18, 3, 18, 4, 0, 3, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=44, q=130)
By second-differencing Neil's 458585379 terms one can contrive at most seven(!) terminal 0s, with, e.g., p=458585379-2q, q=16400301. But not even these seven can signal an incipient quasiperiod, because annihilating all future second differences requires that some yet-to-be-computed terms be nonpositive! A program has confirmed that *all* terminal bursts, even of length one, are similarly doomed, thereby precluding any e-like (Hurwitz) behavior with p+2q < 458585379. Thus at least one of A,B,C,D has a big numerator or denominator. (How big?) --rwg Maybe not huge. For 3.141595... = 4-6e/19, p=447 and q=676.
Was this checked for all N & Q where N+2Q < 458M? The nominal work is ~10^17. I can see some algorithm possibilities to improve this, but I'm curious about what's actually been done. Rich ---------- Quoting Bill Gosper <billgosper@gmail.com>:
Denote by a[0], a[1], ... the continued fraction terms of pi (= 3,7,15,1,... = A001203). If there are rationals A,B,C,D such that pi = (Ae+B)/(Ce+D), then there is a quasiperiod q and a preamble p such that the second differences
a[n+2q]-2a[n+q]+a[n] = 0
for all n>p. For e itself, q=3 and p=1. (Or any positive multiple of 3 and any positive integer). E.g., define a second-differencer: In[13]:= dif2[L_List, q_Integer:1] := Drop[L, 2*q] - 2*Drop[Drop[L, -q], q] + Drop[L, -2*q]
In[16]:= dif2[{a, b, c, d, e}, 2]
Out[16]= {a - 2 c + e} Then In[163]:= dif2[ContinuedFraction[E, 22], 3]
Out[163]= {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=1,q=3). In[166]:= dif2[ContinuedFraction[E^(2/3), 22], 5]
Out[166]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=0,q=5). In[171]:= dif2[ContinuedFraction[(E - 1)/(1 - E/6), 333], 130]
Out[171]= {2, 1, 11, -1, 0, 62, 1, 3, 0, -1, 0, 1, -5, 5, 5, 1, -2, \ -153, -2, 0, 1, -58, 5, 2, 1, 152, 2, 57, -6, 3, -2, -21, 6, -10, 3, \ 22, -18, 3, 18, 4, 0, 3, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} (p=44, q=130)
By second-differencing Neil's 458585379 terms one can contrive at most seven(!) terminal 0s, with, e.g., p=458585379-2q, q=16400301. But not even these seven can signal an incipient quasiperiod, because annihilating all future second differences requires that some yet-to-be-computed terms be nonpositive! A program has confirmed that *all* terminal bursts, even of length one, are similarly doomed, thereby precluding any e-like (Hurwitz) behavior with p+2q < 458585379. Thus at least one of A,B,C,D has a big numerator or denominator. (How big?) --rwg Maybe not huge. For 3.141595... = 4-6e/19, p=447 and q=676.
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