19
\$\begingroup\$

Definition of Additive Primes:

  • Numbers which have exactly 2 divisors are called Prime numbers.

  • Numbers which are prime and their sum of digits is also a prime number are called Additive Primes

Task:

Given an integer x, compute all the additive primes amongst the first x prime numbers, with 2 being considered both the first prime and additive prime number. The numbers are represented in base 10.

Rules:

  • The output consists of all the additive primes amongst the first x primes
  • 0 < x < 151, for this challenge, for functionality purposes
  • Since the additive primes are all integers, decimals are not allowed (e.g.: you should output 2, not 2.0) and they must not be displayed as a fraction.

Examples:

10 -> 2 3 5 7 11 23 29

Explanation:

The first 10 primes are 2 3 5 7 11 13 17 19 23 29, and only 2 3 5 7 11 23 29 have their sum of digits prime numbers, those being, respectively 2,3,5,7,2,5,11, so they are additive primes

Following the explanation from example 1, other test cases may be:

2 -> 2 3 25 -> 2 3 5 7 11 23 29 41 43 47 61 67 83 89 7 -> 2 3 5 7 11

Leaderboard:

var QUESTION_ID=112088,OVERRIDE_USER=59487;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left;font-family:"Helvetica Neue"}table thead{font-weight:700;font-family:"Helvetica Neue"}table td{padding:5px;font-family:"Helvetica Neue"}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

NOTE: Please read the newly-edited rule 1, it brings changes to the output format slightly

Your code should be as short as possible, since this is , so the shortest answer in bytes wins. Good luck!

Improve this question
\$\endgroup\$
1
  • \$\begingroup\$ That's fine. I'd recommend waiting around 24 hours though, because every time you accept the answer they get 15 rep, but they lose it when you un-accept. It's somewhat frustrating sometimes to ride the rollercoaster and continuously lose and gain rep. \$\endgroup\$ Commented Mar 5, 2017 at 15:19

23 Answers 23

Reset to default
9
\$\begingroup\$

Röda, 136 135 bytes

f n{P=[2]S=[2]seq 3,863|{|i|{P|{P+=i;s=0;((""..i)/"")|parseInteger _|s+=_;S+=i if[s in P and not(i in S)]}if{|p|[i%p>0]}_}if[#P<n]}_;S}

Try it online!

It's a function that returns the requested additive primes.

Usage: main { f(25) | print ap for ap } The code uses version 0.12, which is in branch roda-0.12.

Ungolfed:

function f(n) { primes := [2] ultraprimes := [2] seq(3, 863) | for i do break if [ #primes = n ] if [ i%p != 0 ] for p in primes do primes += i sum := 0 ((""..i)/"") | parseInteger _ | sum += digit for digit ultraprimes += i if [ sum in primes and not (i in ultraprimes) ] done done ultraprimes }
Improve this answer
\$\endgroup\$
10
  • 1
    \$\begingroup\$ Nice language! You made this yourself a long time ago it looks like? 10/10, looks pretty cool. \$\endgroup\$ Commented Mar 5, 2017 at 16:14
  • \$\begingroup\$ Neat language! How do you run the program? \$\endgroup\$ Commented Mar 5, 2017 at 16:26
  • \$\begingroup\$ Was just about to ask the same thing. Although I looked over the documentation, I cannot run or compile the source at all. What is your approach? \$\endgroup\$ Commented Mar 5, 2017 at 16:40
  • \$\begingroup\$ @KritixiLithos @Xcoder123 It requires Java 8 and Gradle. The version I use in this answer is 0.12 (in its own branch). The repository must be cloned recursively. To make a runnable jar, invoke gradle fatJar. Do you get any errors when compiling? \$\endgroup\$ Commented Mar 5, 2017 at 16:43
  • \$\begingroup\$ @fergusq Running gradle fatJar doesn't seem to create a jar for me \$\endgroup\$ Commented Mar 5, 2017 at 16:45
8
\$\begingroup\$

Pyke, 9 7 bytes

~p>#Yss

Try it online!

The single byte is_prime was only pushed 3 hours ago. Github commit.

~p - All the prime numbers > - first input of them #Yss - filter(^) Y - digits(^) s - sum(^) s - is_prime(^)
Improve this answer
\$\endgroup\$
5
  • 3
    \$\begingroup\$ Did you just edit your language to suit this challenge? :D \$\endgroup\$ Commented Mar 5, 2017 at 17:43
  • \$\begingroup\$ so... s means is_prime on numbers, and sum on lists? \$\endgroup\$ Commented Mar 5, 2017 at 20:15
  • \$\begingroup\$ @ConorO'Brien yes, I overload it for lists and integers \$\endgroup\$ Commented Mar 5, 2017 at 21:36
  • \$\begingroup\$ @Džuris no, I've been meaning to for a while because I haven't had a single node for doing prime checking, only factorising into primes and divisors. Before I would have had to do _P which is 1 byte longer in this case \$\endgroup\$ Commented Mar 5, 2017 at 21:38
  • 1
    \$\begingroup\$ a new contender for "youngest language feature to win a challenge"? under the wire by ~2 hours? \$\endgroup\$ Commented Mar 5, 2017 at 22:06
8
\$\begingroup\$

Python 2, 124 118 bytes

With help from Riker:

n,f,P=input(),filter,lambda n:all(n%i for i in range(2,n)) f(lambda x:P(sum(map(int,`x`)))&P(x),f(P,range(2,n*n))[:n])

Original:

n,o,c,P=input(),0,2,lambda n:all(n%i for i in range(2,n)) while o<n: o+=P(c) if P(sum(map(int,`c`)))and P(c):print c c+=1

Checking primality in Python ain't fun.

Improve this answer
\$\endgroup\$
2
  • \$\begingroup\$ I (read: got conor to write cool J code for me) have tested this with 9n, doesn't work. :/ n**2 does work, but at a cost of 1 byte. \$\endgroup\$ Commented Mar 5, 2017 at 17:58
  • \$\begingroup\$ Try n*n for n**2 \$\endgroup\$ Commented Mar 5, 2017 at 17:59
5
\$\begingroup\$

Perl 6, 53 bytes

{grep *.comb.sum.is-prime,grep(*.is-prime,0..*)[^$_]}

Try it

Expanded:

{ grep *.comb.sum.is-prime, # find the ultra primes from: grep( *.is-prime, # find the primes 0..* # from all integers )[ ^$_ ] # grab only the first x primes }

If this challenge were changed so that you took the first x ultraprimes this could be shortened to just

{grep({($_&.comb.sum).is-prime},0..*)[^$_]}
Improve this answer
\$\endgroup\$
5
\$\begingroup\$

Python 2, 96 87 bytes

p=-input(),0;m=k=1 while sum(p): m*=k*k;k+=1;p+=m%k, if m%k*p[int(`k`,36)%35]:print k

Thanks to @xnor for golfing off 9 bytes!

Try it online!

Improve this answer
\$\endgroup\$
4
  • \$\begingroup\$ Looks like using a list of indicator variables is shorter. \$\endgroup\$ Commented Mar 6, 2017 at 1:52
  • \$\begingroup\$ The digit sum can be done shorter as is int(`k`,36)%35. All the inputs will be small enough that this suffices. \$\endgroup\$ Commented Mar 6, 2017 at 2:12
  • \$\begingroup\$ More golfing \$\endgroup\$ Commented Mar 6, 2017 at 2:23
  • \$\begingroup\$ Wow! I'm not sure how I thought of a Boolean dict but not a Boolean tuple (hindsight is 20/20), but sum(p) and int(`k`,36)%35 are something else... Thanks! \$\endgroup\$ Commented Mar 6, 2017 at 2:36
5
\$\begingroup\$

Mathematica, 61 47 bytes

Prime@Range@#~Select~PrimeQ@*Tr@*IntegerDigits&
Improve this answer
\$\endgroup\$
2
  • \$\begingroup\$ Not entirely familiar with Mathematica's shorthand syntaxes - what's that @*? The * doesn't look like it's in the right place to be multiplication? \$\endgroup\$ Commented Mar 6, 2017 at 3:23
  • 3
    \$\begingroup\$ @numbermaniac it's function composition. f@*g is essentially f@g@##&. \$\endgroup\$ Commented Mar 6, 2017 at 7:08
4
\$\begingroup\$

Jelly, 10 bytes

ÆNDS$€ĖÆPM

Try it online!

How?

A slightly different approach...

ÆNDS$€ĖÆPM - Main link: n (>0) e.g. 10 ÆN - nth prime number 29 € - for each in range(1,nth prime) [1, 2, 3, ..., 27, 28, 29] $ - last two links as a monad D - decimal digit list [[1], [2], [3], ...,[2,7], [2,8], [2,9]] S - sum [1, 2, 3, ..., 9, 10, 11] Ė - enumerate [[1,1],[2,2],[3,3],...,[9,27],[10,28],[11,29]] ÆP - is prime? (vectorises) [[0,0],[1,1],[1,1],...,[0,1], [0,0], [1,1]] M - indices of maximal elements [ 2, 3, ..., 29]
Improve this answer
\$\endgroup\$
1
  • 3
    \$\begingroup\$ Nice use of Ė. \$\endgroup\$ Commented Mar 5, 2017 at 16:12
3
\$\begingroup\$

05AB1E, 9 bytes

ÝبvySOp—

Uses the CP-1252 encoding. Try it online!

Improve this answer
\$\endgroup\$
0
3
\$\begingroup\$

Jelly, 11 bytes

ÆN€DS$ÆP$Ðf

Try it online!

Explanation:

ÆN€DS$ÆP$Ðf Main link (args: z) ÆN€ Generate first z primes. DS$ Take the digital sum. ÆP Check if it's prime. $ Join last two links and make a monad. Ðf Only keep elements which conform to the criterion above.

I got outgolfed.

Improve this answer
\$\endgroup\$
0
2
\$\begingroup\$

MATL, 15 13 bytes

2 bytes saved thanks to @Luis

:Yq"@V!UsZp?@

Try it at MATL Online

Explanation

 % Implicitly grab input as a number (N) : % Create an array [1...N] Yq % Get the k-th prime for each element k in that array " % For each element in this list @ % Get the current element V!U % Break it into digits s % Sum up the digits Zp % Determine if this is a prime number ?@ % If it is, push the value to the stack % Implicit end of for loop and implicit display of the stack
Improve this answer
\$\endgroup\$
1
  • \$\begingroup\$ @LuisMendo Ah! I knew there was some way to shorten that first part. Thanks \$\endgroup\$ Commented Mar 11, 2017 at 0:30
2
\$\begingroup\$

Fig, \$12\log_{256}(96)\approx\$ 9.877 bytes

FtxFmC@p'pSf

Try it online!

Explanation:

 t # take the first x # x items from mC # all positive integers F # filtered by @p # is prime, F ' # then filter those by S # the sum of f # the digits p # is prime
Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Bash + coreutils, 97 bytes

p()(factor $1|wc -w) for((;++n,c<$1;)){((`p $n`-2||(c++,`p $[n%10+n/10%10+n/100]`-2)))||echo $n;}

Try it online!

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Ohm, 10 bytes (CP437)

@▓_π;░_}Σp

This would be much shorter if I had vectorization or a component for the first N primes, but alas, I did not before this challenge (but I do now!).

Explanation:

@▓_π;░_}Σp Main wire, arguments: a @▓ ; Map...over the range (1..n) _π nth prime ░ Select from ToS where... _}Σ The sum of all digits p is prime
Improve this answer
\$\endgroup\$
1
\$\begingroup\$

PowerShell, 120 bytes

for($n=$args[0];$n){for(;'1'*++$i-notmatch($s='^(?!(..+)\1+$)..')){}if('1'*([char[]]"$i"-join'+'|iex)-match$s){$i};$n--}

Try it online!

Prime checking in PowerShell sucks.

The outer for loop goes from input $n down to 0. In the inner loop, we use a prime generator on $i, then check if the digit-sum (-join'+'|iex) is also a prime. If so, we put $i on the pipeline. In either case, we decrement $n-- and the outer for loop continues. The resulting $is are gathered from the pipeline and an implicit Write-Output happens at program completion.

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Bash + GNU utilities + bsd-games package, 69

primes 2|sed -rn 'h;s/./ + &/g;s/.*/expr &|factor/e;/\w\s/!{x;p};'$1q

Try it online.

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Pari/GP, 59 bytes

f(x)=[p|p<-vector(x,i,prime(i)),isprime(vecsum(digits(p)))]

Try it online!

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

Vyxal, 5 bytes

ʁǎ'∑æ

Try it online, or verify more test cases.

Explanation:

ʁ # Range from 0 to n-1 ǎ # Nth prime (vectorising) ' # Filtered by: ∑ # The digit sum æ # Is prime
Improve this answer
\$\endgroup\$
1
\$\begingroup\$

MathGolf, 9 bytes

♪╒g¶<gÅΣ¶

Try it online.

Explanation:

Since the input is guaranteed to be within the range \$[1,150]\$, where the \$150^{th}\$ prime is \$877\$, we'll use this to our advantage. (Although using ó (\$2^n\$ builtin using the implicit input) instead of ♪ would have worked as well, but would be way too slow for larger inputs.)

♪╒ # Push a list in the range [1,1000] g # Filter it by: ¶ # Check if the number is a prime < # Then only keep the first (implicit) input amount of prime numbers g # Filter it again, Å # using 2 characters as inner code-block: Σ # Get the digit-sum of the current integer ¶ # Check if it's also a prime # (after which the entire stack is output implicitly as result)
Improve this answer
\$\endgroup\$
0
\$\begingroup\$

Husk, 8 bytes

fȯṗΣd↑İp

Try it online!

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

Japt, 12 bytes

Èj}jU fÈìx j

Try it

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

Factor + math.primes math.unicode math.text.utils, 47 bytes

[ nprimes [ 1 digit-groups Σ prime? ] filter ]

Try it online!

Explanation:

It's a quotation (anonymous function) that takes an integer from the data stack as input and leaves a sequence of integers on the data stack as output.

  • nprimes Obtain a list of the first n primes.
  • [ ... ] filter Take elements from the list that match a predicate.
  • 1 digit-groups Obtain a list of digits from an integer.
  • Σ Sum a sequence.
  • prime? Is it prime?
Improve this answer
\$\endgroup\$
0
\$\begingroup\$

05AB1E, 6 bytes

ÅpʒSOp

Try it online or verify all test cases.

Explanation:

Åp # Get a list of the first (implicit) input amount of primes ʒ # Filter this list by: S # Convert the prime to a list of digits O # Sum them together p # Check if this sum is a prime # (after which the filtered list is output implicitly as result)
Improve this answer
\$\endgroup\$
0
\$\begingroup\$

MATL, 13 bytes

:YqtFYA!XsZp)

Try it at MATL Online!

Explanation

: % Range [1 2 ... n], where n is implicit input Yq % Array of first n prime numbers t % Duplicate FYA % Convert to decimal digits. Gives a matrix, where each original % number corresponds to a row. Left-pads with zeros if needed !Xs % Sum of rows Zp % Is prime? (element-wise) ) % Use as logical index into the array of the first n prime numbers % Implicitly display
Improve this answer
\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.