[comment, no answer]
I've some time ago been interested in the same thing and after finding an article of Gosper (see "Hakmem") I noted the following, which might also be interesting: perhaps a path to a proof for your "special case" $e^1$ can be derived.
> (...)
> http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item101b
>
>Keith Ramsay
The Gosper-article inspired me to check some powers of e.
If you allow negative coefficients, you seem to get a pretty general scheme
for
1/k
e for k= ... -3,-2,-1,{0},1,2,3,...
------------------------------------------------------------------------------------------
- 1 3 5 7 9 11
------------------------------------------------------------------------------------------
cf(e^(1/-2)): [1,-3, 1, 1, -7, 1, 1,-11, 1, 1,-15, 1, 1,-19, 1, 1,-23, 1 ]
cf(e^(1/-1)): [1,-2, 1, 1, -4, 1, 1, -6, 1, 1, -8, 1, 1,-10, 1, 1,-12, 1 ]
cf(e^(1/ 0)): [1,-1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1 ] divergent (oscillates on 0 and 1)
cf(e^(1/ 1)): [1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1 ]
cf(e^(1/ 2)): [1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1 ]
cf(e^(1/ 3)): [1, 2, 1, 1, 8, 1, 1, 14, 1, 1, 20, 1, 1, 26, 1, 1, 32, 1 ]
cf(e^(1/ 4)): [1, 3, 1, 1, 11, 1, 1, 19, 1, 1, 27, 1, 1, 35, 1, 1, 43, 1 ]
cf(e^(1/ 5)): [1, 4, 1, 1, 14, 1, 1, 24, 1, 1, 34, 1, 1, 44, 1, 1, 54, 1 ]
cf(e^(1/ 6)): [1, 5, 1, 1, 17, 1, 1, 29, 1, 1, 41, 1, 1, 53, 1, 1, 65, 1 ]
------------------------------------------------------------------------------------------
+ 1 3 5 7 9 11
------------------------------------------------------------------------------------------
for
2/k
e for k= ... -3,-2,-1,{0},1,2,3,...
-----------------------------------------------------------------------------------------------------------------------
delta: - 1 12 5 7 36 11 13 60 17 19 84 23 25
-----------------------------------------------------------------------------------------------------------------------
cf(e^(2/-1)); [1,-1, -6, -3, 1, 1, -4, -18, -6, 1, 1, -7, -30, -9, 1, 1, -10, -42, -12, 1, 1,-23 ]
cf(e^(2/1)); [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12 ]
cf(e^(2/3)); [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90, 25, 1, 1, 28, 126, 34, 1, 1, 37 ]
cf(e^(2/5)); [1, 2, 30, 12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57, 1, 1, 62 ]
cf(e^(2/7)); [1, 3, 42, 17, 1, 1, 24, 126, 38, 1, 1, 45, 210, 59, 1, 1, 66, 294, 80, 1, 1, 87 ]
cf(e^(2/9)); [1, 4, 54, 22, 1, 1, 31, 162, 49, 1, 1, 58, 270, 76, 1, 1, 85, 378, 103, 1, 1, 112 ]
cf(e^(2/11)); [1, 5, 66, 27, 1, 1, 38, 198, 60, 1, 1, 71, 330, 93, 1, 1, 104, 462, 126, 1, 1, 95 ]
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delta: + 1 12 5 7 36 11 13 60 17 19 84 23 25
-----------------------------------------------------------------------------------------------------------------------
Also, allowing fractions for coefficients, the primary expansion of e = e^(1/1) = e^(2/2) can be inserted in the previous table:
-----------------------------------------------------------------------------------------------------------------------
...
cf(e^(2/1)): [1, 0, 6, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1 ...
cf(e^(2/2)): [1, 0.5, 12, 4.5, 1, 1, 6.5, 36, 10.5, 1, 1, 12.5, 60, 16.5, 1, 1, 18.5, 84, 22.5, 1 ...
cf(e^(2/3)): [1, 1, 18, 7, 1, 1, 10, 54, 16, 1, 1, 19, 90, 25, 1, 1, 28, 126, 34, 1 ...
...
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delta + 0.5 6 2.5 3.5 18 5.5 6.5 60 8.5 9.5 42 11.5
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Perhaps this allowing of negative and/or fractional coefficients enables also to find
more simple regularities for e^k with abs(k)>2
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