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This is a reboot of the old 2015 thread.

I saw a lot of inputs in the old thread, I was wondering what we could come up with for 2017.

As before, no number inputs allowed (pi and e are allowed, though)

For an extra challenge, why not constrict it even further? Only one-line inputs are allowed, that can be run on the homescreen.

To start you off (29 bytes):


Code:
int(ππππππ+eeeeeee-πππ-ππ




[EDIT]
Let's try for all the years 2000-2017.
10 bytes.

Code:
int(π+π+e^(π^√(π
I'm gonna pretend that didn't happen and try for 2nd place.
18 bytes.

Code:
int(e)^int(πe+e)-int(πππ


Edit: 17 bytes, via "rule change" in IRC:


Code:
int(e)^int(πe+e)-int(π
2001= int(-ππ+e^(π^√(π
2007= int(-π+e^(π^√(π
2010= int(e^(π^√(π
2013= int(e+e^(π^√(π
2014= int(π+e^(π^√(π
2018= int(ee+e^(π^√(π
2019= int(eπ+e^(π^√(π

EDIT:

Code:
int(seq(A+e^(pi^sqrt(pi)),A,~pipi,ee,e
={2001 2003 2006 2009, 2011, 2014, 2017}

(22 bytes)/(7 years)=3.14 bytes/year
Now I'm going for trickshots, rather than a whole-in-one:
(20 bytes)

Code:
:sum(int(seq(X,X,e,π
:Ans+int(π

In one line:
(21 bytes)

Code:
:sum(int(seq(X,X,e,π)))+int(π


Note that π = (π)
JWinslow23 wrote:
10 bytes.

Code:
int(π+π+e^(π^√(π

damm
I got 15 bytes for 2017:

Code:
int(πe-e) nCr int(π)-int(e

and 12 bytes for 2016

Code:
int(-π+int(e)^int(sinh(π


Also, there used to be a topic specifically for the year 2015.
Here is an interesting answer from that topic:
If your calculator's time is correct,

Code:
getDate:Ans(π/π

Also, there was a codegolf topic for the year 2014

EDIT: Inspired from lirto's routine, 8 bytes for 2017:

Code:
int([i]+e^(cosh(e
You can get a list of all the years 2000-2017 with 26 bytes:

Code:
int(e^(pi[i])+seq(A+e^(pi^sqrt(pi)),A,~pipi,e


That works out to 1.44 bytes per year.
jonbush wrote:
You can get a list of all the years 2000-2017 with 31 bytes:

Code:
int(e^(pi[i])+seq(A+e^(pi^sqrt(pi)),A,~pipi,ee,~real(e^(pi[i]


That works out to 1.722 bytes per year.

You're over complicating this a lot.

Code:
seq(int(e)10^(int(π))+X,X,not(π),e^(sinh(sqrt(π

23 bytes (1.277 bytes per year)
EDIT:

Code:
int(e)10^(int(π))+cumSum(binomcdf(int(e-e),not(π

25 bytes (1.388 bytes per year)
The e token is actually two bytes, unfortunately.

I'm currently actually working on that computer search I talked about two years ago... Maybe it will find the optimal solution.
You people make this look easy.
I was able to shave a byte off of mr womp womp's solution, for 24 bytes:

Code:
seq(X,X,not(π),e-e)+int(e)10^(int(π
23 bytes:


Code:
int(seq(X,X,~π-e,e)+e^(π^sqrt(π


I think this can be improved further.
22 bytes:


Code:
int(seq(X,X,int(~π),e)+e^(π^√(π
Also 22:

Code:
int(seq(X,X,~πacosh(π),~π^-1)+e^(cosh(e
6 bytes for 2017:

Code:
int(√(10^(10^(tanh(π


7 bytes for 2015:

Code:
int(sinh(πe^(tan⁻(√(π


21 bytes for the list:

Code:
int(seq(X,X,π,π-π)+10^(cosh(sinh⁻(sinh⁻cosh(π


I believe the first two are the smallest possible, but 20 bytes may be possible for the list. There are certainly many 21-byte solutions.

EDIT: 20 bytes for the list. I believe this is optimal.

Code:
e^(sinh⁻(10^(int(π
int(seq(X,X,Ans,Ans+ππ√(π
  
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