Concatenating n with n + 1

Question

Introduction

OEIS sequence A127421 is the sequence of numbers whose decimal expansion is a concatenation of 2 consecutive increasing non-negative numbers. Put simply, every number in the sequence is formed by putting together n with n+1 for some non-negative, integer value of n. The first several terms are:

1, 12, 23, 34, 45, 56, 67, 78, 89, 910, 1011, 1112, 1213, 1314, 1415, 1516, 1617, 1718, 1819, 1920, 2021, 2122, 2223, 2324, 2425, 2526, 2627, 2728, 2829, 2930, 3031, 3132, 3233, 3334, 3435, 3536, 3637, 3738, 3839, 3940, 4041, 4142, 4243, 4344, 4445, 4546, …

Challenge

Given a single positive integer n, print the first n entries of OEIS sequence A127421 in increasing order.

• Input and output can be in any acceptable format. Strings or numbers are fine for output.
• Leading zeroes are not permitted.
• Either a full program or function is permitted.
• For the purposes of this challenge, n will be positive and under 100.
• Standard loopholes are disallowed by default.
• This question is code golf, so lowest byte-count wins.

• Here is some sample input and output:

  1 => 1
  2 => 1, 12
  3 => 1, 12, 23
  10 => 1, 12, 23, 34, 45, 56, 67, 78, 89, 910
  

If you have any questions, don't hesitate to ask. Good luck.

P.S this is my first challenge, so hopefully this all makes sense.

EDIT: Removed output restriction to allow numbers or strings.

Answer 1

Shakespeare Programming Language, 310 bytes

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy.Ajax:You cat.Open heart.[Exit Ajax][Enter Page]Scene V:.Ford:Am I worse Ajax?If notlet usScene X.You be twice the sum ofa cat a big big cat.Speak thy.Page:Open heart.You be the sum ofyou a cat.Open heart.Let usScene V.Scene X:.[Exeunt]

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Answer 2

Runic Enchantments, 33 bytes

/01iR1-:0)?;{1+:
\$1<\}$q}:+1{$ '

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Uncompressing the entry sequence and moving the reflectors to the other side:

>1$01iR1-:0)?;{1+:\
      \}$q}:+1{$ '/

Entry sequence (>1$01i) is fairly straight forward. Push and print 1, push 0, push 1, read input and push it to the top of the stack.

At R we enter the program's main loop (unrolled with directional control characters removed):

1-:0)?;{1+:' ${1+:}q$}

At this point the stack is [0,1,i] where i is the input value.

The loop subtracts 1 from the input value (1-), compares it to greater than 0 (if true, skip terminator, else terminate; :0)?;).

Then a series of stack manipulations ({1+:{1+:}) and increments to result in [2,(i-1),1,1,2] as well as printing a space (' $). q then concats the top two items on the stack, which is then printed (giving 12 in the output stream).

Finally the stack is rotated once more, leaving [1,2,(i-1)] as the input to the next loop iteration.

Bonus challenge: using two IPs? 40 bytes

>1$0iR1-:0)?;1{+:' \
> F1iU }$~?=am$?=9m/

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As there's no way to clone the input to a second instruction pointer (I have thought about stack cloning, so this may be possible in the future, but the spec for it would be difficult to implement), we have to read it from the input stream twice.

Flow results in the second pointer being a step behind the first (avoiding merging) and the Fizzle lets us distinguish the two IPs, letting one print a space and the other discards it. I can't figure out a shorter way of performing this check.

However if it allowable to print two spaces as a separator it can be reduced to this (30 bytes):

>1$0iR1-:0)?;1\
> F1iU}$:+{$ '/

input: 4 4
output: 1  12  23  34

But this is an admittedly dubious answer due to how it has to take input, but 3 bytes shorter than the single IP answer, which is interesting.

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Update: Stack transfer

Getting the two pointers to enter the T command in the right execution order is a huge pain. The remaining two spaces in this program can't be removed, as it messes with the timing, but it avoids having to supply the input value twice. Prints 2 spaces between each entry in the sequence (35 bytes).

>1$0y TR1-:0)?;1\
 >1i:1/U}$:+{$ '/

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Answer 3

MathGolf, 8 6 5 4 bytes

{└§p

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Outputs a newline separated series.

Explanation:

{└§p
{       Start a loop over the range(0, input)
 └      Push the top of stack (implicitly the index of the loop) + 1
  §     Concatenate the two (this removes leading 0s)
   p    And print the value

Answer 4

K (ngn/k), 13 bytes

{.,/$x+!2}'!:

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! generate the list 0 1 ... n-1

{ } is a function with argument x

' applied to each

!2 is 0 1

x+!2 is x, x+1

$ format as strings

,/ concatenate

. evaluate (convert to number)

Answer 5

MBASIC, 102 94 bytes

1 INPUT N:FOR I=0 TO N-1:IF I>0 THEN PRINT MID$(STR$(I),2);
2 PRINT MID$(STR$(I+1),2);" ";:NEXT

Output:

? 3
1 12 23

? 10
1 12 23 34 45 56 67 78 89 910

? 16
1 12 23 34 45 56 67 78 89 910 1011 1112 1213 1314 1415 1516

Could have been much cleaner, but the STR$(n) number-to-string conversion function returns with a leading space that had to be dealt with.

Turns out the variable for NEXT and the trailing PRINT are not needed, saving 8 bytes.

Answer 6

><>, 15 13 bytes

ln1-:?!;aoln:

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We take in the input as a command line argument.

Thanks to @jo king for 2 byte loss (noticing the input loop value could be used as stack length.

Explanation (simple):

ln1-:?!;aoln:
ln              : Add the length of the stack to the stack and print.
   1-:?!;       : Take 1 off the input loop and check if zero, if 0, end.
         ao     : Print new line
           ln:  : Add the length of the stack to the stack, print and duplicate the input loop.

Answer 7

Nim, 70 bytes

import sugar,sequtils,math
s(x=>toSeq(1..x).map y=>(y-1)*10^len($y)+y)

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Answer 8

jq, 30 27 characters

1,(range(1;.)|"\(.)\(.+1)")

Sample run:

bash-5.1$ jq -r '1,(range(1;.)|"\(.)\(.+1)")' <<< 5
1
12
23
34
45

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Answer 9

J-uby, 26 25 bytes

-1 byte thanks to Razetime

:*|:*&(-[I,:+&1]|:join|Z)

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Explanation

:* |              # Range 0..n-1, then
:* & (            # map with ...
  -[I, :+ & 1] |  #   [m, m+1] (I is identity), then
  :join |         #   join, then
  Z               #   convert to int (for special case "01")
)

Answer 10

Vyxal, 5 bytes

ʀ¨pJ⌊

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Explained

ʀ¨pJ⌊
ʀ     # Inclusive zero range
 ¨p   # each overlapping pair
   J  # merge
    ⌊ # floor

Answer 11

Nim, 46 bytes

proc(x:any)=
 echo 1;for i in(2..x):echo i-1,i

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Answer 12

Rockstar, 76 bytes

listen to N
X's 0
while N-X
let S be X
let X be+1
let S be+""+X
cast S
say S

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listen to N       :Read input string into variable N
X's 0             :Initialise X as 0
while N-X         :While X is less than N
let S be X        :  Assign X to S
let X be+1        :  Increment X
let S be+""+X     :  Append X to S
cast S            :  Cast S to an integer
say S             :  Output S

Answer 13

Julia, 28 bytes

!n=["1";2:n.|>i->"$(i-1)$i"]

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based on the answer by Sundar R

Answer 14

Tcl, 59 bytes

proc C n {puts [incr i]
time {puts $i[incr i]} [expr $n-1]}

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PS: I will golf it more later!

Answer 15

Haskell, 57 55 bytes

f n=read<$>zipWith((.show).(++).show)[0..][1..n]::[Int]

A function which takes a number and returns a list.

EDIT: -2 Bytes thanks to Cat Wizard

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Answer 16

CJam, 11 bytes

{{_)s+si}%}

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Answer 17

Python 2, 43 bytes

lambda i:[1]+[`x`+`x+1`for x in range(1,i)]

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First element is an integer, the rest are strings.

Answer 18

Perl 5, 29 bytes

say"@{[1,map$_-1 .$_,2..<>]}"

Answer 19

RAD, 28 27 bytes

(⊢(⊢+⊣×10*1+∘⌊10⍟⊢)1+⊢)¨0,⍳

Repository

Must be called like: ((⊢(⊢+⊣×10*1+∘⌊10⍟⊢)1+⊢)¨0,⍳) <value>, as I broke assignment of functions when I implemented shy results.

Note: this should also work in Dyalog APL, and due to that: -1 byte thanks to @Adám!

The ⍳ can either be the APL symbol or the greek one if run in RAD, but must be the APL version ifran in Dyalog APL

Answer 20

C# (Visual C# Compiler), 54 bytes

Uses space as seperator character.

n=>{var r="";for(;n-->1;)r=" "+n+(n+1)+r;return"1"+r;}

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Ungolfed full code:

class P
{
    static void Main()
    {
        System.Func<int, string> f =
            n =>
            {
                var r = "";        //Declare variable for the result
                for (; n--> 1;)    //Loop until n is 1

                    r =            //Set the result to:
                        " " +      //Seperator +
                        n +        //Value of n as string
                        (n + 1) +  //Value of n + 1 as string
                        r;         //Previous content of the result

                return "1" + r;    //Return 1 (first item) + the result.
            }
            ;

        System.Console.WriteLine(f(1));
        System.Console.WriteLine(f(2));
        System.Console.WriteLine(f(3));
        System.Console.WriteLine(f(10));
        System.Console.WriteLine(f(100));
    }
}

Answer 21

C# (Visual C# Compiler), 46 bytes

If you are allowed to have some using static directives in the top whose bytes do not count:

n=>1+Concat(Range(1,n-1).Select(x=>","+x+++x))

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Will explode if input n is 0.

It is interesting that, while the C# rules say ","+x+++x means ( "," + (x++) ) + x, if it had meant ( "," + x ) + (++x) instead, the program would still work!

Answer 22

Clojure, 54 bytes

#(reduce(fn[a n](conj a(str n(inc n))))[1](range 1 %))

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Just a reduction over a range between 1 (inclusive) and n (exclusive).

(defn concat-n-n+1 [n]
  (reduce (fn [acc m]
            (conj acc (str m (inc m))))
          [1]
          (range 1 n)))

The first element of the list is a number, the rest are strings. This seems to be allowed by the spec.

Answer 23

K (oK), 12 bytes

Solution:

,/'$1,2':1+!

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Examples:

,/'$1,2':1+!1
,,"1"
,/'$1,2':1+!2
(,"1"
 "12")
,/'$1,2':1+!3
(,"1"
 "12"
 "23")
,/'$1,2':1+!10 
(,"1"
 "12"
 "23"
 "34"
 "45"
 "56"
 "67"
 "78"
 "89"
 "910")

Explanation:

,/'$1,2':1+! / the solution
           ! / range 0..n
         1+  / add 1
      2':    / sliding window of 2
    1,       / prepend 1
   $         / convert to strings
,/'          / flatten each

Answer 24

C Sharp 51 bytes

n=>new int[n].Select((_,o)=>int.Parse($"{o}{o+1}"))

Answer 25

Attache, 15 bytes

{N!_'-~_}=>Iota

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Explanation

{N!_'-~_}=>Iota
{       }=>        map the inside function:
           Iota    (over each number k from 0 to n-1)
    '              array concatenate:
   _               k
     -~_           k+1
 N!                convert [k, k + 1] to an integer (concatenates and returns a number)

Answer 26

Rutger, 108 bytes

x=$Input;
f=For[$x];
f=f[@i];
f=f[{p=Subtract[$i];r=Concat[Str[p[1]]];Print[Integer[r[Str[$i]]]];}];
Do[$f];

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Ungolfed

input = $Input;
for = For[$input];
for = for[@index];

for = for[{
    dec = Subtract[$index];
    ret = Concat[Str[dec[1]]];
    Print[Integer[ret[Str[$index]]]];
}];

Do[$for];

The basic concept of Rutger is the lack of multi-adic commands: every command takes a single argument, and if the command needs multiple argument, each new call returns a curried function.

First, we take evaluated input and store it in \$x\$. We then create a For object, a tri-adic command. The first two calls create a variable \$f\$, that will loop \$x\$ times, using the iteration variable \$i\$, starting at \$i := 1\$. The third call indicates the code to be run:

{p=Subtract[$i];r=Concat[Str[p[1]]];Print[Integer[r[Str[$i]]]];}

This creates a block of code containing 3 statements. Our first two are variable assignments. First, we create a curried function Subtract[$i], which prepares to subtract a value from \$i\$. This curried function is saved in the \$p\$ variable. Next, we create a second curried variable, \$r\$, which starts by calling p[1], subtracting \$1\$ from \$i\$. We then convert \$i-1\$ to a string and pass it as the first argument to the Concat function. Finally, we convert \$i\$ to a string, concatenate it to \$str(i - 1)\$ and convert it to an integer to reomve the leading \$0\$ when \$i = 1\$. This is then printed.

The last line, Do[$f];, runs the for loop.

Answer 27

RAD, 72 bytes

0,(⊢+10×⊣)⍁¨∊¨(⌽∘⌊(⊢|⍨10*(⍳(1+∘⌊10⍟⊢)))÷10*1-⍨(⍳(1+∘⌊10⍟⊢)))¨¨(⊢,1+⊢)¨⍳⎕

Link to repository

I wanted to try to see if I could do it in a "non-mathematical" way by splitting a number up into its digits.

Answer 28

C, 39 bytes

x(n){n&&x(n-1);printf("%d%d, ",n,n+1);}

eliminated the ternary conditional and shaved off 3 bytes.

Answer 29

ARBLE, 23 20 bytes

Saved a handful of bytes by using better named variables.

range(0,n)//(x..y|0)

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Answer 30

Python 2.7, 52 bytes

I tried my best. Not yet familiar with this golfing thing.

def a(n):
    for e in range(n):
        print int(`e`+`e+1`)

The int() removes the leading zero in the first output, as specified in the rules.