I need to get this list
{1,0,2,1,3,2,4,3,5,4,6,5,7,6,8,7,9,8}
with a very small code (< 30 characters) using Table and List functions.
On each step, either it
I have tried
Range[1,18] + Table[(-i)*2^n, {i, -1, 1}, {n, 0, 1}
because I want $i = -1$ while $n = 0$ and $i = 1$ while $n = 1$, but I don't know how to set it up in one line.
Well, without Table and such:
Riffle[Range[9], Range[0, 8]]
Or, we can use SequenceFunction:
FindSequenceFunction[Riffle[Range[9], Range[0, 8]]]
(* 1/4 (-1)^#1 (-3 + (-1)^(1 + #1) + 2 (-1)^#1 #1) & *)
So:
1/4 (-1)^#1 (-3 + (-1)^(1 + #1) + 2 (-1)^#1 #1) & /@ Range[18]
If you have to use Table you can get the desired result with 23 characters:
## &[i, i - 1]~Table~{i, 9}
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
"##&[i,i-1]~Table~{i,9}" // StringLength
22
If not you can save a few key strokes using
## &[#, # - 1] &~Array~9
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
"##&[#,#-1]&~Array~9" // StringLength
19
or using
## &[#, # - 1] & /@ Range @ 9
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
"##&[#,#-1]&/@Range@9" // StringLength
20
26 characters is the shortest I can think of:
{#, # - 1} & /@ Range@9 // Flatten
Table as requested, Table[{n, n - 1}, {n, 9}] // Flatten is still below 30.
$\endgroup$
Commented
Feb 18, 2020 at 20:13
f[n_]:= Sequence[n, n - 1]
Array[f, 9]
(*{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}*)
or
Flatten@Array[{#, #-1}&, 9]
(*{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}*)
Update
From this by @azerbajdzan, which introduced me to SubstitutionSystem
SubstitutionSystem[{n_ :> n+1}, {1,0}, 8]//Flatten
(* {1,0,2,1,3,2,4,3,5,4,6,5,7,6,8,7,9,8} *)
Original Answer
Flatten@NestList[#+1&,{1,0},8]
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
In addition
Array[{#,#-1}&,9,1,Join]
and
Array[{#+1,#}&, 9,0,Join]==Array[{#,#-1}&, 9,1,Join]
Original Post
NestList[#+{1,1}&, {1,0},8]//Flatten
Sow, Reap and (recursion with) #0: (i) If[#1[[1]]>9, Nothing, #0[Sow[#1]+1]]&[{1,0}]//Reap//Flatten and (ii) Nest[Sow[#]+1&, {1,0},9]//Reap//Rest//Flatten
$\endgroup$
with a very small code (< 30 characters) using Table and List functions.
If requirement to use Table only, then one way is
Flatten@Table[Table[i, {i, j, j - 1, -1}], {j, 1, 9}]
Gives
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
Or this (inspired by the FindSequenceFunction reported by @march):
ClearAll[aList, sf,testList];
aList = {1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8};
sf = FindSequenceFunction[aList];
testList = sf /@ Table[i, {i, 1, Length[aList]}]
aList == testList
(*{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}*)
(*True*)
Table[i, {i, 1, Length[aList]}] is equivalent (but potentially slower) than Range[Length[aList]]
$\endgroup$
Using Splice (new in 12.1) - 24 characters
Splice @ {#, # - 1} & /@ Range @ 9
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8}
f[n_] := Riffle[Range[n], Range[0, n - 1]]
f[10] will yield 20 terms.
{1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8, 10, 9}
Can crop as desired
a[n_] := 1/2 + (-1)^n 3/2
FoldList[#1 + a[#2] &, 1, Range[20]]
-> {1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8, 10, 9, 11}
or in one line:
FoldList[#1 + (1 + 3(-1)^#2)/2 &, 1, Range[20]]
Table[(n+3Mod[n,2])/2-1,{n,18}]$\endgroup$Table[Floor[n/2]-(-1)^n,{n,18}]$\endgroup$