prepend,append-Sequence

Question

Task

The prepend,append-Sequence is defined recursively, like this

• a(1) = 1
• a(n) = a(n-1).n , if n is even
• a(n) = n.a(n-1) , if n is odd

where the . represents an integer concatenation.

So the first few terms are: 1,12,312,3124,53124,531246,7531246,... This is A053064.

Your task is, given an integer a > 0 to return n, such that the nth element in the prepend,append-Sequence is equal to a and if no such n exists return 0, a negative number or error out etc.

Rules

• Input can be taken as an integer, string, list of characters/digits etc.
• Output can be printed to STDOUT or returned (integer, string etc. is fine)
• On invalid input & in the case no such n exists your program may do anything but return a positive integer (eg. loop forever, return 0 etc.)
• You may choose to use 0-indexing, but then the output in case no n exists cannot be 0

Test cases

1 -> 1
12 -> 2
21 -> 0
123 -> 0
312 -> 3
213 -> 0
211917151311975312468101214161820 -> 21
2119171513119753102468101214161820 -> 0
333129272523211917151311975312468101214161820222426283031 -> 0
999795939189878583817977757371696765636159575553514947454341393735333129272523211917151311975312468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100 -> 100

Answer 1

JavaScript (ES6), 40 bytes

Takes input as a string. Throws a recursion error if no index is found.

f=(n,s=k='1')=>n==s?k:f(n,++k&1?k+s:s+k)

Demo

f=(n,s=k='1')=>n==s?k:f(n,++k&1?k+s:s+k)

console.log(f('1')) // 1
console.log(f('12')) // 2
console.log(f('312')) // 3
console.log(f('211917151311975312468101214161820')) // 21
console.log(f('999795939189878583817977757371696765636159575553514947454341393735333129272523211917151311975312468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100')) // 100

Answer 2

C# (.NET Core), 83, 80, 60 59 bytes

n=>{int i=0;for(var t="";t!=n;)t=++i%2<1?t+i:i+t;return i;}

Try it online!

Takes the input as a string into a lambda function. 1-indexed. Returns the index of the value for truthy, or infitnitely loops for a "falsey"

Answer 3

Python 2, 63 bytes

-1 byte thanks to @EriktheOutgolfer.

f=lambda x,i='1',j=2:i!=`x`and f(x,[i+`j`,`j`+i][j%2],j+1)or~-j

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Python 2, 64 bytes

-18 bytes thanks to @officialaimm, because I didn't notice erroring out was allowed!

x,i,j=input(),'1',1
while i!=x:j+=1;i=[i+`j`,`j`+i][j%2]
print j

Try it online!

Python 2, 82 bytes (does not loop forever)

This one returns 0 for invalid inputs.

def f(n,t="",i=1):
 while len(t)<len(n):t=[t+`i`,`i`+t][i%2];i+=1
 print(n==t)*~-i

Try it online!

Answer 4

Jelly, 12 bytes

Rs2ZU1¦ẎVµ€i

Try it online!

Explanation:

Rs2ZU1¦ẎVµ€i
         µ€  Eval this link for each (automatic [1..n] range)
R             Range
 s2           Split in pieces of: 2
   Z          Zip
    U1¦       Only keep index: 1 of: Vectorized reverse
       Ẏ      Flatten 1-deep
        V     Concatenate string versions and eval
           i Find index of y in x (y = implicit input)

Answer 5

05AB1E, 14 bytes

$vDNÌNFs}«})Ik

Try it online! or as a Test suite

Explanation

0-indexed.
Returns -1 if the input is not in the sequence.

$                 # push 1 and input
 v                # for each y,N (element, index) in input do:
  D               # duplicate top of stack
   NÌ             # push N+2
     NF }         # N times do:
       s          # swap the top 2 elements on the stack
         «        # concatenate the top 2 elements on the stack
          })      # end loop and wrap in a list
            Ik    # get the index of the input in this list

Answer 6

R, 73 bytes

p=paste0;n=scan(,'');l='';while(l!=n){F=F+1;l="if"(F%%2,p(F,l),p(l,F))};F

Reads from stdin and returns the value of the index (implicitly printed). Infinite loops when the value isn't in the sequence. F is by default FALSE which is cast to 0 when used in arithmetic.

Try it online!

Answer 7

Haskell, 75 71 57 bytes

f n=[i|i<-[1..],(show=<<reverse[1,3..i]++[2,4..i])==n]!!0

Takes n as a string.

Try it online!

Answer 8

Husk, ¹⁷ 15 bytes

£moiṁsṁṠehGJC2N

Try it online!

1-indexed. Returns 0 if not in the sequence.

-2 bytes from Leo, GJ!

Answer 9

Vyxal, 12 bytes

ɾɖ≬wJṘyRYvṅḟ

Try it Online! Uses 0-indexing, outputs -1 for invalid.

           ḟ # Find index of input in...
 ɖ           # Scan
ɾ            # range(1, input)
  ≬---       # by
   wJ        # Concatenate
     Ṙ       # Reverse
      yRY    # Reverse every other element
         vṅ  # Concatenate each list into a number

Answer 10

Mathematica, 135 bytes

s=t={};x=1;While[x<5!,{s~AppendTo~#&,s~PrependTo~#&}[[x~Mod~2+1]]@x;AppendTo[t,FromDigits@Flatten[IntegerDigits/@s]];x++];t~Position~#&

Answer 11

Jelly,  19 18  15 bytes

+ḂḶṚm2;RḤ$ṁµ€Vi

A monadic link taking and returning integers.

Try it online! (very slow - takes ~50s on TIO just to confirm that 3124 is at index 4)

For a much faster version use the previous 18 byter (only checks up to the length of the input, which is sufficient).

How?

+ḂḶṚm2;RḤ$ṁµ€Vi - Link: number, v
           µ€   - perform the monadic link to the left for €ach k in [1,2,3,...v]
                -                 (v can be big, lots of k values makes it slow!)
 Ḃ              -   modulo k by 2  = 1 if odd 0 if even
+               -   add to k = k+isOdd(k)
  Ḷ             -   lowered range = [0,1,2,...,k+isOdd(k)]
   Ṛ            -   reverse = [k+isOdd(k),...,2,1,0])
    m2          -   modulo slice by 2 = [k+isOdd(k),k+isOdd(k)-2,...,3,1]
         $      - last two links as a monad:
       R        -   range(k) = [1,2,3,...,k]
        Ḥ       -   double = [2,4,6,...,2k]
     ;          - concatenate = [k+isOdd(k),k+isOdd(k)-2,...,3,1,2,4,6,...,2k]
         ṁ      - mould like range(k) = [k+isOdd(k),k+isOdd(k)-2,...,3,1,2,4,6,...,k-isOdd(k)]
                -   (this is a list of the integers to be concatenated for index k)
             V  - evaluate as Jelly code (yields a list of the concatenated integers)
              i - first index of v in that (or 0 if not found)

Answer 12

Swift 4, 92 bytes

This loops infinitely for invalid cases, so I didn't include them in the testing link.

func f(x:String){var i="1",j=1;while i != x{j+=1;i=[i+String(j),String(j)+i][j%2]};print(j)}

Test Suite.

Amusingly, it is longer with a closure:

var f:(String)->Int={var i="1",j=1;while i != $0{j+=1;i=[i+String(j),String(j)+i][j%2]};return j}

Test Suite.

Answer 13

Haskell, 115 85 bytes

s=read.(show=<<)
f 1=1
f x|odd x=s[x,f$x-1]
f x=s[f$x-1,x]
g x=[n|n<-[1..],x==f n]!!0

Try it online!

Answer 14

Perl 5, 54 + 1 (-n) = 55 bytes

$a=++$,%2?$,.$a:$a.$,while length$a<length;say/$a/&&$,

Try it online!

Returns nothing if not found.

Answer 15

Japt, 17 bytes

@P=PiXv n X;¥P}a1

Try it online! or check (valid) test cases

Takes input as a string or integer. On invalid input, continues "forever" looking for a solution (thus why I didn't include them in the test cases).

Explanation:

@P=PiXv n X;¥P}a1    
                     # Implicitly start with P = ""
@             }a1    # Repeat the function for each integer X > 0 until one returns true
 P=Pi     X          #  Insert X into P at index:
     Xv              #   X is divisible by 2?
        n            #   Times -1
           ;¥P       #  Return whether P now equals the input
                     # Implicitly output the last value of X

The "insert at index" portion might be a bit confusing, so I'll add more detail here. In Japt, NvD is a function which returns 1 if N is divisible by D, and 0 otherwise. If D is not provided, it defaults to 2. Thus Xv here is equal to 1 if X is even, and 0 if X is odd. NnD is a function that returns D - N. If D is not provided it defaults to 0, effectively returning -N. In this program, that results in -1 if X is even, and still 0 if X is odd. Finally, in Japt indexing negative numbers count from the end of a string or array. Thus the segment PiXv n X evaluates to Pi0 X if X is odd, prepending X, but it evaluates to Pi-1 X if X is even, appending it instead.