Sum of (at most) 5 primes

Question

Terence Tao recently proved a weak form of Goldbach's conjecture! Let's exploit it!

Given an odd integer n > 1, write n as a sum of up to 5 primes. Take the input however you like, and give output however you like. For example,

def g(o):
    for l in prime_range(o+1):
        if l == o:
            return l,
        for d in prime_range(l+1):
            for b in prime_range(d+1):
                if l+d+b == o:
                    return l,d,b
                for c in prime_range(b+1):
                    for h in prime_range(c+1):
                        if l+d+b+c+h == o:
                            return l,d,b,c,h

is Sage code that takes an integer as input, and returns a list of integers as output whose sum is n. By Tao's theorem, this will always terminate!

Input

An odd integer n. You decide how to take the input, but if it's weird, explain it.

Output

Rather open-ended. Return a list. Print a string. Gimme one, a few, or all. Leave crap lying around on the stack (GS, Piet, etc) or in a consecutive (reachable) memory block (BF, etc) in a predictable manner. For these later cases, explain the output. In all cases, what you return / print / whathaveyou should be a straightforward representation of a partition of n into primes with fewer than 6 parts.

Scoring

This is code golf, smallest byte count wins.

Bonus! if the word 'goldbach' appears as a subsequence (not necessarily consecutive; just in order. Case doesn't matter) of your program subtract 8 points. The code above is an example of this.

Answer 1

Mathematica, 38

IntegerPartitions[n,5,Prime~Array~n,1]

Answer 2

C, 192-8 = 184 chars

Contains "Goldbach" consecutively (excluding punctuation), and "Tao" as well.
When the sum is less than 5 primes (i.e. always), prints zeros (16 = 0+0+0+3+13)
Read the number from standard input: echo 30 | ./prog.

#define T(x)for(x=0;x<=s;b=&x,c(++x))
G,o,l,d,*b,a;c(h)
{(*b-1?h<3:++*b)||c(*b%--h?h:++*b);}
main(s){
    scanf("%d",&s);
    T(G)T(o)T(l)T(d)T(a)o+G+l+d+a-s?0:exit(printf("%d+%d+%d+%d+%d\n",G,o,l,d,a));
}

Old version (179 chars), which can find only sums of exactly 5 primes (and therefore fails for x<10):

#define T(x)for(x=2;x<s;b=&x,c(++x))
G,o,l,d,*b,a;c(h)
{h<3||c(*b%--h?h:++*b);}
main(s){
    scanf("%d",&s);
    T(G)T(o)T(l)T(d)T(a)o+G+l+d+a-s?0:exit(printf("%d+%d+%d+%d+%d\n",G,o,l,d,a));
}

Explanation:
c sets *b to the next prime (including *b itself if it's prime).
T builds a for loop, which advances one of the variables G,o,l,d,a to the next prime.
Within all for loops, we check if the sum matches, and print&exit if it does.

Answer 3

Brachylog, 9 bytes

~+.ṗᵐl≤5∧

Try it online!

~+.          Output (.) should sum to the input,
   ṗᵐ        consist of all primes,
     l≤5     and have length ≤ 5.
        ∧    (Don't unify that 5 with the implicit output variable.)

Answer 4

J, 29

(#~y=+/@>),{5$<0,p:i._1 p:>:y

Assumes input is in y. Value of expression is list of boxes of list of 5 primes or 0 that sum to y.

   y =. 16
   (#~y=+/@>),{5$<0,p:i._1 p:>:y
+----------+----------+----------+----------+----------+---------+----------+----------+----------+----------+----------+----------+----------+---------+---------+----------+----------+----------+----------+----------+----------+----------+---------+------...
|0 0 0 3 13|0 0 0 5 11|0 0 0 11 5|0 0 0 13 3|0 0 2 3 11|0 0 2 7 7|0 0 2 11 3|0 0 3 0 13|0 0 3 2 11|0 0 3 11 2|0 0 3 13 0|0 0 5 0 11|0 0 5 11 0|0 0 7 2 7|0 0 7 7 2|0 0 11 0 5|0 0 11 2 3|0 0 11 3 2|0 0 11 5 0|0 0 13 0 3|0 0 13 3 0|0 2 0 3 11|0 2 0 7 7|0 2 0 ...
+----------+----------+----------+----------+----------+---------+----------+----------+----------+----------+----------+----------+----------+---------+---------+----------+----------+----------+----------+----------+----------+----------+---------+------...

Not enough letters to earn any bonus points.

Answer 5

Ruby 138 124 117 - 8 = 109

require'mathn'
def g(o,l=[])
p l if l.inject(:+)==o#db
(l.last||1..o).each{|d|d.prime?and g(o,l+[d])if l.count<5}
end

Call with g(<number>). Sample output:

[2, 2, 2, 2, 19]
[2, 2, 3, 3, 17]
[2, 2, 3, 7, 13]
...

Test: http://ideone.com/rua7A

Answer 6

PHP 143 122 - 8 = 114

EDIT: Saved a few bytes on output, removed the explicit function call.

<?function g($o,$l,$d,$b){for(;$o>=$b=gmp_intval(gmp_nextprime(+$b));)echo$b^$o?$l<4&&g($o-$b,$l+1,"$d$b,",$b-1):"$d$b
";}

Unrolled:

<?
function g($o,$l,$d,$b){
  for(;$o>=$b=gmp_intval(gmp_nextprime(+$b));)
    echo$b^$o?$l<4&&g($o-$b,$l+1,"$d$b,",$b-1):"$d$b
";}

Call with @g(<number>); Sample output for n=27:

2,2,2,2,19
2,2,3,3,17
2,2,3,7,13
2,2,5,5,13
2,2,5,7,11
2,2,23
2,3,3,19
2,3,5,17
2,3,11,11
2,5,7,13
2,7,7,11
3,3,3,5,13
3,3,3,7,11
3,3,5,5,11
3,3,7,7,7
3,5,5,7,7
3,5,19
3,7,17
3,11,13
5,5,5,5,7
5,5,17
5,11,11
7,7,13

Answer 7

Ruby 2 -rmathn, 66 bytes - 8 = 58

g=->o,*l{o==l.reduce(:+)?p(l):l[5]||b=Prime.each(o){|x|g[o,*l,x]}}

Heavily based on GolfWolf's answer, but since it's 6 years old I'm going to post my own instead of nitpicking. Advances in technology include the stabby lambda, using reduce instead of inject for the free d, a terse way of stopping at partitions of 5, and Prime.each(o), which iterates over all primes less than or equal to o (and furnishes an ach). Maybe in another 6 years there'll be a better way of using the b.

Answer 8

Scala 137-8=129

def g(o:Int)={val l=0+:(2 to o).filterNot(d=>(2 to d-1).exists(d%_==0))
for(b<-l;a<-l;c<-l;h<-l;e<-l;if(b+a+c+h+e==o))yield{(b,a,c,h,e)}}

After boothby's hint: eliminated one function call, allow to interpret 3 as the sum of 3 and nothing, remove input from output - saves another 20 chars.

Bonus emphasizing:

def g(o:Int)={val l=0+:(2 to o).filterNot(d=>(2 to d-1).exists(d%_==0)) for(b<-l;a<-l;c<-l;h<-l;e<-l;if(b+a+c+h+e==o))yield{(b,a,c,h,e)}}

Invocation and result:

println (l(17)) 
Vector((17,0,0,2,2,13), (17,0,0,2,13,2), (17,0,0,3,3,11), ...

The output repeats x for every list to sum up to x, and then shows the 5 summands. 0 for missing summand, i.e. 2+2+13.

Ungolfed:

// see if there is some x, such that o%x is 0.
def dividable (o:Int) = (2 to o-1).exists (x=> o % x == 0)

// +: is a kind of cons-operator for Vectors
def primelist (d: Int) = {
  val s = 0 +: (2 to d).filterNot (b => dividable (b))
  for (a <- s;
    b <- s;
    c <- s;
    h <- s;
    e <- s;
    if (a+b+c+h+e == d)) yield {(a,b,c,h,e)}
}

Answer 9

MuPAD 113 - 8 = 105

g:=[0,ithprime(i)$i=1..n]:f:=_for_in:f(l,g,f(d,g,f(b,g,f(a,g,f(c,g,if l+d+b+a+c=n then print(l,d,b,a,c)end)))))

This version will also print all permutations of every solution:

0, 0, 0, 0, 7
0, 0, 0, 2, 5
0, 0, 0, 5, 2
0, 0, 0, 7, 0
0, 0, 2, 0, 5
...

And yes, it creates a way too long list g. Who cares? :-)

Ungolfed version:

g:=[0].select([$1..n],isprime):
for l in g do
  for d in g do
    for b in g do
      for a in g do
        for c in g do
          if l+d+b+a+c=n then print(l,d,b,a,c); end;
        end
      end
    end
  end
end

Answer 10

Jelly, 19 bytes (but very slow -- advice wanted)

ÆR;0x5Œ!ḣ€5¹©S€i³ị®

Try it online!

ÆR;0x5Œ!ḣ€5¹©€i³ị®     main link, takes one argument N
ÆR                     get all the primes less than N
  ;0x5                 add zero, and then repeat the entire list 5 times
      Œ!               get all the permutations of this huge list (takes a long time!)
        ḣ€5            for each permutation, just take the first 5 numbers
                       (this gives us all possible length-5 combinations of the primes plus zero, with some repeats)
           ¹©          save that list to register
              S€       take the sum of every permutation in the list...
                i³     and find the index of our first argument N in that list of sums
                  ị®   then recall our list of permutations, and get the correct permutation at that index!

If you have any ideas to make it faster and shorter, please let me know!

Answer 11

Vyxal, 9 bytes

Ṅ'L5≤næA∧

Try it Online!

Outputs all possible sums.

Explained

Ṅ'L5≤næA∧   # Takes a single integer n and returns a list of lists
Ṅ           # Push all the integer partitions (ways to sum to) of n
 '          # and keep only those where:
  L5≤       #   the length of the partition is less than or equal to 5 (can also be 6<)
        ∧   #   and
     næA    #   all integers in the partition are prime

Answer 12

Japt, 19 bytes

Hideous, and slow as hell! Will crap out on inputs higher than 5.

ÆõÃc fj à l§5 æ@¶Xx

Try it

ÆõÃc fj à l§5 æ@¶Xx     :Implicit input of integer U
Æ                       :Map the range [0,U)
 õ                      :  Range [1,U]
  Ã                     :End map
   c                    :Flatten
     f                  :Filter
      j                 :  Prime
        à               :Combinations
          l             :Filter by length
           §5           :  <=5
              æ         :Get the first to return true
               @        :When passed through the following function as X
                ¶       :  U is equal to
                 Xx     :  Sum of X