Sequence Without Sevens [duplicate]

Question

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Cantor's unspeakable numbers 25 answers

This challenge is based on a drinking game. I advise against alcohol consumption while programming.

In this game, the players count up in turns: the first player says 1, the second says 2 and so on. Here's the twist however: each number that is divisible by 7 or has a 7 in its digits (in base 10) is replaced by another counter. This means 7 is replaced by 1, 14 is replaced by 2, 17 is replaced by 3 and so on. For this "sub-sequence", the same rules apply! 35 would be replaced by 7, which in turn is replaced by 1. This process repeats until the number is "valid".

Here's a visualization of the first 100 numbers in the sequence:

n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
gets replaced by:
                1                  2        3           4                 5  6                    7     8              9             10    11                   12 13                14          15       16 17 18 19 20 21 22 23 24 25             26       27          28                29 30       
gets replaced by:
                                                                                                  1                                                                                   2                       3           4                                   5           6                            

final sequence:
    1 2 3 4 5 6 1 8 9 10 11 12 13  2 15 16  3 18 19 20  4 22 23 24 25 26  5  6 29 30 31 32 33 34  1 36  8 38 39 40 41  9 43 44 45 46 10 48 11 50 51 52 53 54 55 12 13 58 59 60 61 62  2 64 65 66 15 68 69 16  3 18 19 20  4 22 23 24 25 80 81 82 83 26 85 86  5 88 89 90  6 92 93 94 95 96 29 30 99 100

In the final sequence, no number has a seven in its digits or is divisible by seven.

Input description

An integer n (1 <= n <= 7777 or 0 <= n <= 7776)

You may choose whether you use a 0-indexed or 1-indexed input.

Output description

The nth number in the sequence

Examples

The examples are 1-indexed. For 0-indexed, subtract one from n.

n   ->   f(n)
1        1
5        5
7        1
35       1
77       23
119      1
278      86
2254     822
2255     2255
7777     3403

Scoring

This is code-golf, so shortest answer in bytes wins!

Hmm, is test case 278 correct? I get the correct result for each of them, but 278 becomes 86 instead of 5. :S — Kevin Cruijssen, Aug 23 '17 at 9:17
@KevinCruijssen You're right, 278 is supposed to be 86. 279 becomes 5. Fixed! — Arfie, Aug 23 '17 at 9:19
@Arnauld Yep that's an exact dupe...or not? That one asks you to print indefinitely, this one asks you for a specific item...it's somewhat like the "count up forever" question vs. my "count up folks!" question. Very different approaches. — Erik the Outgolfer, Aug 23 '17 at 11:56
I think it's similar enough to be a duplicate. We don't need both a "print at least n terms" and a "print the nth term" of every sequence challenge. — mbomb007, Aug 23 '17 at 13:33

Kevin Cruijssen · Accepted Answer · 2017-08-23 14:09:19Z

Java, 122 121 98 97 bytes

int c(int n){int a=0,b=0,c=n;for(;n-->0;c+=a=((--b+"").indexOf(55)&b%7)>>31);return a<0?-b:c(c);}

-23 bytes thanks to @Nevay.
-1 byte thanks to @Neil.

Explanation:

Try it here.

int c(int n){                            // Method with integer as parameter and return-type
  int a=0,                               //  Flag integer
      b=0,                               //  Regular counter
      c=n;                               //  7-counter
  for(;n-->0;                            //  Loop over the input
    c+=                                  //   Append `c` with:
       a=                                //    Set `a` to:
         ((--b+"").indexOf(55)&b&7)      //     If `b` is divisible by or contains 7:
                                         //     (and decrease `b` by 1 in the process)
                                   >>31  //      Bitwise right shift so `a` is either `-1` or `0`
  );                                     //  End of loop
  return a<0?                            //  If `a` is now -1:
    -b                                   //   Return the regular counter `b` (as positive)
   :                                     //  Else:
    c(c);                                //   Recursive call with 7-counter `c` as input
}                                        // End of method

102, I think: int c(int n){int q=0,r=1,a=1,b=0;for(;--n>0;b+=q=(++a+"").contains("7")|a%7<1?1:0);return q>0?c(b):a;} — Neil, Aug 23 '17 at 9:59
101 bytes: int c(int n){int r=1,a=1,b=1;for(;n-->0;r=(a+"").contains("7")|a++%7<1?b++:a-1);return--a==r?r:c(r);} — Leaky Nun, Aug 23 '17 at 10:15
98 bytes: int c(int n){int r=1,a=0,b=1;for(;n-->0;r=((--a+"").indexOf(55)&a%7)<0?a:b++);return a<r?c(r):-r;} — Nevay, Aug 23 '17 at 11:09
Actually my answer is only 98 bytes, once you remove the unused r=1,... — Neil, Aug 23 '17 at 12:06
97 bytes: int c(int n){int q=0,a=0,b=n;while(n-->0)b+=q=((--a+"").indexOf(55)&a%7)>>31;return q<0?-a:c(b);} — Neil, Aug 23 '17 at 12:23

Leaky Nun · Accepted Answer · 2017-08-23 09:34:38Z

6

Python 3, 77 bytes

g=lambda a:a%7<1or'7'in str(a)
f=lambda a:g(a)and f(sum(map(g,range(a))))or a

Try it online!

edited Aug 23 '17 at 9:34

answered Aug 23 '17 at 9:21

Leaky Nun

40.2k4 gold badges87 silver badges259 bronze badges

3

\$\begingroup\$ Does this not work in Python 2? Then you could do '7'in`a` instead of '7'in str(a) \$\endgroup\$ – Arfie Aug 23 '17 at 9:23
4

\$\begingroup\$ @Ruud Leaky doesn't like Python 2. \$\endgroup\$ – Mr. Xcoder Aug 23 '17 at 9:49
4

\$\begingroup\$ Shouldn't this be 23 bytes?! \$\endgroup\$ – Shaggy Aug 23 '17 at 11:16
2

\$\begingroup\$ @Shaggy Huh? Why? \$\endgroup\$ – Erik the Outgolfer Aug 23 '17 at 11:32
\$\begingroup\$ @EriktheOutgolfer (took me 2 hours to figure it out) because f(77) = 23 \$\endgroup\$ – Leaky Nun Aug 23 '17 at 13:55

add a comment |

Neil · Accepted Answer · 2017-08-23 12:01:08Z

5

JavaScript (ES6), 72 69 bytes

f=
n=>[...a=Array(n)].reduce(_=>a[++i]=i%7>7*/7/.test(i)?i:a[++j],i=j=0)

<input type=number min=1 oninput=o.textContent=f(+this.value)><pre id=o>

Edit: Saved 3 bytes thanks to @Shaggy.

edited Aug 23 '17 at 12:01

answered Aug 23 '17 at 10:09

Neil

90.2k8 gold badges46 silver badges190 bronze badges

\$\begingroup\$ 69 bytes: n=>[...a=Array(n)].reduce(_=>a[++i]=i%7>7*/7/.test(i)?i:a[++j],i=j=0) \$\endgroup\$ – Shaggy Aug 23 '17 at 10:33
\$\begingroup\$ @Shaggy I can't believe I forgot to use reduce... hangs head in shame \$\endgroup\$ – Neil Aug 23 '17 at 11:27

add a comment |

score 3 · Accepted Answer · 2017-08-23 11:51:21Z

3

JavaScript, 79 75 bytes

2 bytes saved thanks to Shaggy

f=a=>(z=a=>a%7<1|/7/.test(a))(a)?f([...Array(a)].reduce(a=>a+z(b++),b=0)):a

Try it online!

edited Aug 23 '17 at 11:51

answered Aug 23 '17 at 9:42

user72349

\$\begingroup\$ Just finished golfing mine down ... and then spotted you beat me to the identical solution. :\ \$\endgroup\$ – Shaggy Aug 23 '17 at 9:55
\$\begingroup\$ 75 bytes: f=a=>(z=a=>a%7<1|/7/.test(a))(a)?f([...Array(a)].reduce(a=>a+z(b++),b=0)):a \$\endgroup\$ – Shaggy Aug 23 '17 at 10:00
\$\begingroup\$ @Shaggy. I actually had tried using reduce, but before I rearranged variables, so it had turned out that it was longer. I forgot to recheck it again. Thanks. \$\endgroup\$ – user72349 Aug 23 '17 at 10:05

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fergusq · Accepted Answer · 2017-08-23 09:45:40Z

2

Röda, 65 bytes

f a{f#[seq(1,a)|[1]if g _]if g a else[a]}g a{[a%7<1 or"7"in`$a`]}

Try it online!

A port of Leaky Nun's Python answer.

answered Aug 23 '17 at 9:45

fergusq

4,6921 gold badge11 silver badges36 bronze badges

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JungHwan Min · Accepted Answer · 2017-08-23 12:48:31Z

2

Mathematica, 57 bytes

Last[i=1;If[DigitCount[#,10,7]>0||7∣#,i++,#]&~Array~#]&

answered Aug 23 '17 at 12:48

JungHwan Min

12.7k3 gold badges16 silver badges66 bronze badges

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Emigna · Accepted Answer · 2017-08-23 14:03:56Z

05AB1E, 26 23 bytes

LÐ7Ösε7å}~_*[¤0Ê#D_O£]¤

Try it online!

Explanation

L                         # push range [1 ... n]
 Ð                        # triplicate
  7Ö                      # check if number in one copy for equality to 0 after mod by 7
    sε7å}                 # check each number in one copy if they contain 7
         ~_               # OR these 2 lists and invert
           *              # multiply with the range, 
                          # giving a list with 0s where the numbers need to be replaced
            [¤0Ê#    ]    # loop until the last number in the list is not 0
                 D        # duplicate the list
                  _       # invert it, 
                          # giving a list with 1s for the numbers that need to be replaced
                   O      # sum
                    £     # take these many elements from the list
                      ¤   # RESULT: push the last element of the list

Previous 26 byte solutions

>GNN7ÖN7å~Dˆ[_#¯s£OÐ7Ös7å~ 
>G1UN[Ð7Ös7å~XiDˆ}_#¯s£O0U

Dave · Accepted Answer · 2017-08-23 15:06:37Z

2

Pyth, 36 bytes

L?!hbbhx.f|/`Z`7!%Z7b)b;WnyyQyQ=yQ;Q

Fixed completely, but is now an unholy mess of letters and can certainly be golfed quite a bit.

1-indexed

edited Aug 23 '17 at 15:06

answered Aug 23 '17 at 13:48

Dave

4322 silver badges10 bronze badges

\$\begingroup\$ Fixed, I put the h in front of the Q instead of G /)_- \$\endgroup\$ – Dave Aug 23 '17 at 14:18
\$\begingroup\$ It should work now... \$\endgroup\$ – Dave Aug 23 '17 at 14:53
\$\begingroup\$ I missed copy-pasting the leading L to define the lambda \$\endgroup\$ – Dave Aug 23 '17 at 15:06
\$\begingroup\$ Nice work! Looks good now! \$\endgroup\$ – Emigna Aug 23 '17 at 15:30
\$\begingroup\$ Thanks! Still needs some work though... \$\endgroup\$ – Dave Aug 23 '17 at 15:31

add a comment |

scottinet · Accepted Answer · 2017-08-24 07:39:06Z

C (gcc), 104 98 95 bytes

-6 bytes by getting rid of sprintf+index and looking manually for a 7 in an integer's digits
-3 bytes using the deprecated POSIX bzero instead of memset

A function writing the result in the global variable r. Usage: l(<int>)

Result is 1-indexed.

r,o[9];b(i,s){for(r=s=++o[i];s&&s%10^7;s/=10);r%7&&!s?:b(i+1);}l(f){for(bzero(o,36);f--;b(0));}

Try it online!

Explanation:

/*
  r = value of the nth element
  o = array of counters (o[0] == main counter)
 */
r,o[9];

// Recursive function, "i" is the counter index 
// "s" is used to determine if an int contain a 7 digit
b(i,s){
  /*
    1. Increases the value of the counter
    2. Stores the result into r and s
    3. Divide s by 10 until it is equal to 0 or its remainder is 7
   */ 
  for(r=s=++o[i];s&&s%10^7;s/=10);

  /*
    if:
      r%7 (r modulo 7) AND it contains no 7 digit
    then:
      nothing
    else:
      calls b with the next counter index
   */
  r%7&&!s?:b(i+1);
}

// Main function: initializes counters and plays
// the sequence until the nth member is calculated
l(f){
  for(bzero(o,36);f--;b(0));
}

Finally, this is not the longest solution anymore. Sorry, Java. — scottinet, Aug 24 '17 at 7:40

Xcali · Accepted Answer · 2017-08-23 14:37:49Z

Perl 5, 52 + 1 (-p) = 53 bytes

map{$s=0;do{$\=++$c[$s++]}until$\!~/7/&&$\%7}1..$_}{

Try it online!

How?

@c is an array of counters. $c[0] holds the main counter that determines which element we're working on. Each successive element holds another counter that is only accessed when the previous one contains a 7 or is divisible by 7.

map              # Loop to find the appropriate element
  $s=0;          # reset index of counters
  do{            # looping through the counters
    $\=++c[$s++] # increment the counter and the index; $\ is our output variable
  }until         # don't quit until
       $\!~/7/   # the result has no 7s
       &&$\%7    # and is indivisble by 7
}1..$_           # do this until the input is reached
}{  # close the implicit loop created by the -p flag and force output of $\

~~# Perl 5, 64 + 1 (-p) = 65 bytes~~

while($c[0]<$_){$s=0;do{$\=++$c[$s++]}while($\=~/7/||$\%7==0)}}{

Try it online!

Gregory Higley · Accepted Answer · 2017-08-24 03:35:25Z

1

J, 52 bytes

i=:('7'&e.@":+.0=7&|)"0
f=:]`(#@}.@I.@:i@i.@>:)@.i^:_

Result is 1-indexed. Using it is pretty straightforward:

f 7777

Try it online!

edited Aug 24 '17 at 3:35

answered Aug 23 '17 at 12:48

Gregory Higley

2951 silver badge11 bronze badges

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Sequence Without Sevens [duplicate]

Input description

Output description

Examples

Scoring

marked as duplicate by mbomb007, AdmBorkBork code-golf Aug 23 '17 at 13:34

11 Answers 11

Java, 122 121 98 97 bytes

Python 3, 77 bytes

JavaScript (ES6), 72 69 bytes

JavaScript, 79 75 bytes

Röda, 65 bytes

Mathematica, 57 bytes

05AB1E, 26 23 bytes

Pyth, 36 bytes

C (gcc), 104 98 95 bytes

Perl 5, 52 + 1 (-p) = 53 bytes

J, 52 bytes

Not the answer you're looking for? Browse other questions tagged code-golf number sequence counting or ask your own question.

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Sequence Without Sevens [duplicate]

Input description

Output description

Examples

Scoring

11 Answers 11

Java, 122 121 98 97 bytes

Python 3, 77 bytes

JavaScript (ES6), 72 69 bytes

JavaScript, 79 75 bytes

Röda, 65 bytes

Mathematica, 57 bytes

05AB1E, 26 23 bytes

Pyth, 36 bytes

C (gcc), 104 98 95 bytes

Perl 5, 52 + 1 (-p) = 53 bytes

J, 52 bytes

Not the answer you're looking for? Browse other questions tagged code-golf number sequence counting or ask your own question.

Linked

Related

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