Jumping Numbers

Question

A jumping number is defined as a positive number n which all pairs of consecutive decimal digits differ by 1. Also, all single digit numbers are considered jumping numbers. eg. 3, 45676, 212 are jumping numbers but 414 and 13 are not. The difference between 9 and 0 is not considered as 1

The challenge Create a program that output one of the following results:

• Given an input n output the first n jumping numbers.
• Given an input n output the n^th term of the sequence.

Note

• Any valid I/O format is allowed
• 1-index or 0-index is allowed (please specify)

Here are some jumping numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 101, 121, 123, 210, 212, 232, 234, 321, 323, 343, 345, 432, 434, 454, 456, 543, 545, 565, 567, 654, 656, 676, 678, 765, 767, 787, 789, 876, ...

This is also A033075

Answer 1

Haskell, 57 bytes

(l!!)
l=[1..9]++[x*10+t|x<-l,t<-[0..9],(mod x 10-t)^2==1]

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Answer 2

Jelly, 8 bytes

1DI*`ƑƊ#

A full program accepting an integer, n, from STDIN which prints a list of the first n positive jumping numbers.

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How?

Acceptable incremental differences between digits are 1 and -1 while others from [-9,-2]+[2,9] are not. This lines up with integers which are invariant when raised to themselves. i.e. \$x^x=x\$ since:

$$0^0=1$$ $$1^1=1$$ $$2^2=4$$ $$\cdots$$ $$-1^{-1}=-1$$ $$-2^{-2}=-\frac{1}{4}$$ $$\cdots$$

1DI*`ƑƊ# - Main Link: no arguments (accepts a line of input from STDIN)
       # - count up keeping the first (input) n matches...
1        - ...start with n equal to: 1
      Ɗ  - ...match function: last three links as a monad:  e.g. 245       777      7656
 D       -   convert to a list of decimal digits                 [2,4,5]   [7,7,7]  [7,6,5,6]
  I      -   incremental differences                             [2,1]     [0,0]    [-1,-1,1]
     Ƒ   -   invariant under?:
    `    -     using left argument as both inputs of:
   *     -       exponentiation (vectorises)                     [4,1]     [1,1]    [-1,-1,1]
         -                                            --so we:   discard   discard  keep
         - implicitly print the list of collected values of n

Answer 3

05AB1E (legacy), 5 bytes

The input is 1-indexed.

Code:

µN¥ÄP

Uses the 05AB1E encoding. Try it online!

Explanation

µ          # Get the nth number, starting from 0, such that...
   Ä       #   The absolute values
 N¥        #   Of the delta's of N
    P      #   Are all 1 (product function, basically acts as a reduce by AND)

Answer 4

Python 2, 79 75 bytes

-4 bytes by xnor

f=lambda n,i=1:n and-~f(n-g(i),i+1)
g=lambda i:i<10or i%100%11%9==g(i/10)>0

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Derived from Chas Brown's answer. The helper function g(i) returns whether i is a jumping number. Iff the last two digits of a number n have absolute difference 1, then n%100%11 will be either 1 or 10, so n%100%11%9 will be 1.

Answer 5

APL (Dyalog Unicode), 36 bytes^SBCS

1-indexed. Thanks to dzaima for their help with golfing this.

Edit: -15 bytes from ngn.

1+⍣{∧/1=|2-/⍎¨⍕⍺}⍣⎕⊢0

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Explanation

We have f⍣g⍣h, where, as ⍣ is an operator, APL translates this to (f⍣g)⍣h. (In contrast with functions where 2×3+1 is translated 2×(3+1))

1+⍣{...}⍣⎕⊢0  This is equivalent to 
               "do {check} we find the n-th integer that fulfils {check}"

1+⍣{...}   ⊢0  Start with 0 and keep adding 1s until the dfn 
               (our jumping number check in {}) returns true.
        ⍣⎕    We take input n (⎕) and repeat (⍣) the above n times 
               to get the n-th jumping number.

{∧/1=|2-/⍎¨⍕⍺}  The dfn that checks for jumping numbers.

         ⍎¨⍕⍺   We take the base-10 digits of our left argument
                 by evaluating each character of the string representation of ⍺.
     |2-/        Then we take the absolute value of the pairwise differences of the digits
 ∧/1=            and check if all of the differences are equal to 1.

Answer 6

C (gcc), 90 bytes

f(n,K,b,k){for(K=0;n;b&&printf("%d,",K,n--))for(b=k=++K;k/10;)b*=abs(k%10-(k/=10)%10)==1;}

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Answer 7

Japt, 14 bytes

Outputs the first nth term, 1-indexed.

_ì äa dÉ ªU´}f

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(I know, I know, I'm supposed to be taking a break but I'm in golf withdrawal!)

Answer 8

Python 2, 88 87 bytes

f=lambda n,i=2:n and f(n-g(i),i+1)or~-i
g=lambda i:i<10or abs(i/10%10-i%10)==1==g(i/10)

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Returns the 0-indexed jumping number (i.e., f(0) => 1, etc).

Answer 9

Haskell, 69 bytes

• Thanks to Joseph Sible for enforcing the challenge rules and saving three bytes.
• Saved two bytes thanks to nimi.

(filter(all((==1).abs).(zipWith(-)<*>tail).map fromEnum.show)[1..]!!)

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Answer 10

JavaScript (ES7), 60 bytes

Returns the \$n\$^th term of the sequence (1-indexed).

f=(n,k)=>[...k+''].some(p=x=>(p-(p=x))**2-1)||n--?f(n,-~k):k

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Answer 11

Jelly, 9 bytes

-1 by Jonathan Allan

1DạƝ=1ẠƲ#

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1-indexed.

Answer 12

Swift, 228 bytes

func j(n:Int){
var r:[Int]=[]
for x in 0...n{
if x<=10{r.append(x)}else{
let t=String(x).compactMap{Int(String($0))}
var b=true
for i in 1...t.count-1{if abs(t[i-1]-t[i]) != 1{b=false}}
if b{r.append(x)}
}
}
print(r)
}
j(n:1000)

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Answer 13

Python 3, 122 121 bytes

g=lambda s:len(s)==1or 1==abs(ord(s[0])-ord(s[1]))and g(s[1:])
def f(n,i=1):
	while n:
		if g(str(i)):n-=1;yield i
		i+=1

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-1 byte by changing f from printing, to a generator function.

g is a recursive helper function which determines if a string s is a "jumping string" (this works since the character codes for 0 to 9 are in order and contiguous).

f is a generator function that takes in n and yields the first n jumping numbers.

Answer 14

R, 85 bytes

i=scan();j=0;while(i)if((j=j+1)<10|all(abs(diff(j%/%10^(0:log10(j))%%10))==1))i=i-1;j

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Suspect this can be golfed more. Reads the number using scan() and outputs the appropriate jumping number.

Answer 15

Perl 5, 56 bytes

map{1while++$n=~s|.(?=(.))|abs$&-$1!=1|ger>9}1..<>;say$n

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1-indexed, outputs the n^th jumping number

Answer 16

Wolfram Language (Mathematica), 85 bytes

If[#<10,#,t=n=1;While[t<=#,{-1,1}~SubsetQ~Differences@IntegerDigits@n++~If~t++];n-1]&

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returns the n-th number

Answer 17

Factor, 129 bytes

: f ( x -- ) 1 [ [ dup 10 >base >array differences [ abs 1 = ] all? ] [ 1 + ] until
dup . 1 + [ 1 - ] dip over 0 > ] loop 2drop ;

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Outputs the first n jumping numbers

Answer 18

Catholicon, 5 bytes

ρHṘḃǰ

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