Sum pyramid of primes

Question

Given a number N, the challenge is to get the sum of the pyramid of N primes. To clarify things, here is an example:

Input: 4

We will list the first 4 primes, and then compute the sum of them. After that, we will compute the sums of the sums, and so on:

  2
    > 5
  3     > 13
    > 8      > 33
  5     > 20
    > 12
  7

You can see that the final result is 33. Here is another example, with N = 8:

   2
     >   5
   3       >  13
     >   8       >  33
   5       >  20       >  83
     >  12       >  50       > 205
   7       >  30       > 122       > 495
     >  18       >  72       > 290       > 1169
  11       >  42       > 168       > 674
     >  24       >  96       > 384
  13       >  54       > 216
     >  30       > 120
  17       >  66
     >  36
  19

You can see that the final result is 1169.

Here is another example with an odd N, N = 3:

 2
   > 5
 3     > 13
   > 8
 5

This gives us 13 as the result

Your task is to write a program or a function, that takes an integer greater than 0, and output the final result.

Here are some test results:

1:  2
2:  5
3:  13
4:  33
5:  83
6:  205
7:  495
8:  1169
9:  2707
10: 6169
11: 13889
12: 30993
13: 68701
14: 151469
15: 332349
16: 725837
17: 1577751
18: 3413221
19: 7349029
20: 15751187
21: 33616925
22: 71475193
23: 151466705
24: 320072415
25: 674721797
26: 1419327223
27: 2979993519
28: 6245693407
29: 13068049163
30: 27297614797
31: 56929779663
32: 118543624847
33: 246475746269
34: 511766428817
35: 1061264813321
36: 2198298700845
37: 4548996804811
38: 9405003164065
39: 19429190057417
40: 40107799133677
41: 82736199371081
42: 170553108953473
43: 351333736092089
44: 723224546040181
45: 1487710742395387
46: 3058157261678325
47: 6282142186547177
48: 12896743408107403
49: 26460652594917673
50: 54262186256186881
51: 111224391050741687
52: 227896496141836195
53: 466805185374509003
54: 955904519939662217
55: 1956988697590280537
56: 4005572366722212927
57: 8196803221276230093
58: 16769645303734608963
59: 34300013739423719561
60: 70136585692535099353
61: 143371352962891226373
62: 292978031452308375001
63: 598482012866917021541
64: 1222083126601616763473
65: 2494459637841415902073
66: 5089478703050176444803
67: 10379794709536133386939
68: 21160351440305258275579
69: 43119914481530819445497
70: 87833066190052490228187
71: 178841897161848754603319
72: 364014682565128163812791
73: 740654046243174781813209
74: 1506496270380756958474835
75: 3063280375436290387756263
76: 6227039507615221644290617
77: 12655020557561801933128885
78: 25712267089927372837530869
79: 52230425385198423845305957
80: 106076955379202815098486497
81: 215397386589448754140867649
82: 437308717912632286770415395
83: 887706233370396897803709611
84: 1801721089699452657985592689
85: 3656329898231436156162865559
86: 7418972676822310377574227797
87: 15051599987013574096449515927
88: 30532404546282900804722616529
89: 61926565462373271494414919017
90: 125582269494835615524470915169
91: 254631689768733901573206365479
92: 516210444730946464864091626473
93: 1046330617753410129672316234861
94: 2120493010460433691014704829565
95: 4296639990460140795780826898943
96: 8704509990931940668688755806845
97: 17631229933967301681217551193565
98: 35706243541395815998303171050377
99: 72298621492552303967009812018997

This is code-golf, so the shortest amount of bytes wins!

Answer 1

J, 15 bytes

p:@i.+/ .*i.!<:

Explanation:

Basically the same as my Mathematica answer.

p:@i.+/ .*i.!<:
          i.!<:    binomial coefficients
p:@i.              first n primes
     +/ .*         dot product

Answer 2

Mathematica, 38 36 35 bytes

Prime[r=Range@#].Binomial[#-1,r-1]&

Answer 3

Minkolang 0.14, 17 bytes

n[i3M$i1-i6M*+]N.

Try it here and check all test cases here.

Explanation

n                    Take number from input (N)
 [                   Open for loop that repeats N times
  i                  Loop counter (n)
   3M                Pop n and push nth prime (where 2 is the 0th prime)
     $i1-            Max iterations - 1 (which is N-1)
         i           Loop counter (n)
          6M         Pop n,k and push kCn (binomial)
            *+       Multiply and add
              ]      Close for loop
               N.    Output as number and stop.

I use basically the same algorithm as several of the earlier answers that use binomial coefficients. Whenever you see such a pyramid of numbers being added, Pascal's triangle should be the first thing to come to mind. I don't see that any of the other answers have explained why this works, so I'll do that.

MORE explanation

2
  > [2,3]
3         > [2,3,3,5]
  > [3,5]             > [2,3,3,3,5,5,5,7]
5         > [3,5,5,7]
  > [5,7]
7

As you can see, the primes 2,3,5,7 appear 1,3,3,1 times in the final result. Lemme change the layout a bit.

_ _ _ 7
_ _ 5
_ 3
2

The number of times that the 3 will contribute to the final result is the same as the number of paths from the 3 to the upper-left corner, moving only up and left. Here, there are three such paths for the 3:

_    _    _ _
_    _ _    _
_ 3    3    3

Note that I can reverse the direction without loss of generality. So I want to know how many paths there are from the upper-left corner to each position along the jagged edge. I can count them like so...

1 1 1 1 1 . . .
1 2 3 4
1 3 6
1 4   .
1       .
.         .
.
.

For every number in this triangle, if it is X units from the left and Y units from the top, then the number at that position is

The way I use it, though, X+Y = N is constant and X ranges from 0 to N, which goes along one diagonal. I multiply each coefficient with the corresponding prime number and then add it all up.

See the Wikipedia article on Pascal's triangle for more on this.

Answer 4

JavaScript ES7 107

Abusing the fixed limit at 27 - how boring is actually finding the primes.

n=>eval("t=2;for(p=[for(v of'012242424626424662642646842')t-=-v];--n;)p=p.slice(0,n).map((v,i)=>v+p[i+1])")

Test snippet (using array comprehension it will work only in Firefox)

F=n=>eval("t=2;for(p=[for(v of'012242424626424662642646842')t-=-v];--n;)p=p.slice(0,n).map((v,i)=>v+p[i+1])")

// Less golfed

Q=n=>{
  t=2;
  // Note: the golfed version will return the last computed value, that is p if the loop is entered, else t=2
  p=[for(v of '012242424626424662642646842') t-=-v] // build the array of first 27 primes in p
  while(--n) p = p.slice(0,n).map((v,i)=>v+p[i+1])  
  return p
}  

//TEST
console.log=x=>O.innerHTML+=x+'\n'

for(i=1;i<28;i++)console.log(i+' : '+F(i))

<pre id=O></pre>

Answer 5

Pyth, 18 bytes

husM.:G2tQ.f}ZPZQ0

Try it online: Demonstration

Explanation:

husM.:G2tQ.f}ZPZQ0   implicit: Q = input number
          .f    Q0   find the first Q numbers Z >= 0, which satisfy
            }ZPZ        Z appears in the prime factorization of Z
                     this gives the first Q prime numbers
 u      tQ           assign this list to G and repeat the following Q-1 times:
    .:G2               create all subarrays of length 2
  sM                   sum them up and update G
h                    take the first element (=result) and print

Answer 6

Pyth - 16 15 bytes

Uses reduce and first_n filter.

u+VGtGtQ.f}ZPZQ

Test Suite.

Answer 7

Pyth, 16 bytes

s*V.cLtQQ.f}ZPZQ

Very simple actually:

s*V          ; Dot product of
  .cLtQQ     ; the binomial coefficients for n
  .f}ZPZQ    ; and the first n prime numbers.

Answer 8

Haskell, 74 bytes

import Data.Numbers.Primes
f n=([]:iterate(zipWith(+)=<<tail)primes)!!n!!0

Usage example:

*Main> map f [1..12]
[2,5,13,33,83,205,495,1169,2707,6169,13889,30993]

How it works: repeatedly calculate the neighbor sums of all primes. Take the head of the nth iteration.

[2,3,5,7,11,13,17,19,23,29,...]             -- plain primes (and 1st iteration)
[5,8,12,18,24,30,36,42,52,60,...]           -- 2nd iteration of neighbor sums
[13,20,30,42,54,66,78,94,112,128,...]       -- 3rd iteration
[33,50,72,96,120,144,172,206,240,274,...]
...

As the index operator !! is zero based, I'm prepending an empty list to avoid having to use !!(n-1).

Answer 9

Matlab, 76 bytes

Thanks to David for saving a lot of bytes!

n=input('');x=primes(103);
for s=2:n,x=conv(x,[1 1]);end
disp(num2str(x(n)))

Old version, 98 bytes

n=input('');m=1;x=[];while nnz(x)<n
m=m+1;x=primes(m);end
for s=2:n,x=conv(x,[1 1]);end
disp(x(n))

Answer 10

JavaScript (ES6), 121 bytes

n=>eval(`for(p=[],c=0,x=1;c<n;s?p[c++]=x:0)for(s=i=++x;--i>1;)x%i?0:s=0;for(;--c;p=s)for(i=c,s=[];i;)s[c-i]=p[i]+p[--i]`)

Explanation

Most of the size comes from finding the prime numbers.

n=>
  eval(`                   // eval used to enable for loops without {} or return

    // Get primes up to n
    for(                   // loop from range 2 to n
      p=[],                // p = primes
      c=0,                 // c = count of primes
      x=1;                 // x = current number to check for primality
      c<n;
      s?p[c++]=x:0         // add the number to the primes if it has no divisors
    )
      for(                 // loop from range 2 to x to check for divisors
        s=                 // s = true if x is a prime
          i=++x;
        --i>1;
      )
        x%i?0:s=0;         // check if x has a divisor

    // Sum primes
    for(;--c;p=s)          // while the new pyramid has pairs to sum
      for(i=c,s=[];i;)     // loop through each pair of the pyramid
        s[c-i]=p[i]+p[--i] // push the sum of the pair to the new pyramid s
  `)                       // implicit: return the final sum

Test

<input type="number" id="input" value="0" oninput="result.textContent=(

n=>eval(`for(p=[],c=0,x=1;c<n;s?p[c++]=x:0)for(s=i=++x;--i>1;)x%i?0:s=0;for(;--c;p=s)for(i=c,s=[];i;)s[c-i]=p[i]+p[--i]`)

)(+this.value)" />
<pre id="result"></pre>

Answer 11

Shell + GNU and BSD utilities, 92

echo `primes 1|sed $1q`|sed -r ':
s/(\w+) (\w+)/$((\1+\2)) \2/
t
s/ \w+$//
s/^/echo /e
/ /b'

Answer 12

Seriously, 23 bytes

,r`P`M;lD`;pX@dXZ'Σ£M`n

Outputs the result as a list of length 1: 8 -> [1169]

Try it online

Explanation:

,r`P`M    push the first n primes as a list
;lD       push 1 minus the length of the list (we'll call this k) ([2,3,5,7],3)
`...`n    call the following function k times:
    ;pX      duplicate the list, pop and discard the first element
    @dX      swap, pop and discard the last element
    Z        zip the two lists
    'Σ£      push the string "Σ" and convert to function
    M        map the function over the list

Answer 13

Mathematica 73 bytes

NestWhile[Plus@@@Partition[#,2,1]&,Prime@n~Table~{n,#},Length@#>1&][[1]]&

---

How it works

Prime@n~Table~{n,#}& gives a list of the first # primes.

Partition[#,2,1]& rearranges a list of numbers, {a, b, c, d ...} as {{a,b}, {b,c}, {c,d}...}}.

Plus@@@ then returns {a+b, b+c, c+d...}.

NestWhile begins with the list of # primes and repeatedly applies Plus@@@Partition... as long as there is more than one number in the list.

---

NestWhile[Plus@@@Partition[#,2,1]&,Prime@n~Table~{n,#},Length@#>1&][[1]]&[4]

33

---

NestWhile[Plus @@@ Partition[#, 2, 1] &, Prime@n~Table~{n, #}, Length@# > 1 &][[1]] &[5]

83

---

It takes about 1/5 sec to solve for the first 1000 primes.

NestWhile[Plus @@@ Partition[#, 2, 1] &, Prime@n~Table~{n, #}, 
 Length@# > 1 &][[1]] &[10^3] // AbsoluteTiming

{0.185611, 1917231113909474354152581359443368948301825453723617274940459548079399 7849439430405641625002631859205971635284844253657654843025188471660669 0868945436580032828177831204066809442374364181056590286849530757875874 9185665854180901580438781223737728559484382552514103542932932981340942 3918431043908415228663677}

Answer 14

Jelly, 7 bytes

ÆN€+ƝƬṀ

Try it online!

Originally I wrote a Brachylog answer, 1<|~lṗᵐ≠≜{s₂ᶠ+ᵐ}ⁱ~g, but when it came out to 19 bytes I decided that I should probably try a different language.

      Ṁ    The largest value from
     Ƭ     every stage of repeatedly
   +       adding
    Ɲ      adjacent values, starting with
ÆN         nth prime
  €        mapped over the input.

Apparently, mapping over a number treats it as a range from 1 to itself inclusive, and integers sort as greater than lists or whatever '' is.

Answer 15

APL(NARS), 41 chars, 82 bytes

{1=≢⍵:↑⍵⋄∇+/¨¯1↓⍵,¨1⌽⍵}∘{⍵↑v/⍨0πv←⍳1+⍵×⍵}

In input if one want to use big number has to enter the type number_x as 47x. There could be something not ok: i here write that n primes are in the set 1..n^2 Test:

  h←{1=≢⍵:↑⍵⋄∇+/¨¯1↓⍵,¨1⌽⍵}∘{⍵↑v/⍨0πv←⍳1+⍵×⍵}
  h 1
2
  h 2
5
  h 9
2707
  h 24
320072415
  h 47x
6282142186547177 
  h 99x
72298621492552303967009812018997 
  h 200x
433205808657246411262213593770934980590715995899633306941417373

Answer 16

Perl 6, 52 bytes

{grep(&is-prime,1..*)[^$_],{[|$_]Z+.skip}...1& &say}

Try it online!

Anonymous code block that takes an argument and prints out a one element list containing the result.

Answer 17

Python 2, 159 bytes

m=int(input())
q=[]
x=2
while len(q)<m:
 if not any([x%g<1 for g in q]):q+=[x]
 x+=1
for i in range(m-1):
 for p in q:q+=[q[1]+q[0]];q.pop(0)
 print(q.pop())
print q

Answer 18

Ceylon, 169 bytes

alias I=>Integer;I s(I*l)=>l.size<2then(l[0]else 0)else s(*l.paired.map((I[2]i)=>i[0]+i[1]));I p(I n)=>s(*loop(2)(1.plus).filter((c)=>!(2:c-2).any((d)=>c%d<1)).take(n));

This defines two functions – s calculates the pyramid sum of a sequence of integers, while p calls this on the sequence of the first n primes.

Looks like about half the size is finding the first n primes, other half is calculating the pyramid sum.

Here is a formatted/commented version:

// Sum pyramid of primes
//
// Question:  http://codegolf.stackexchange.com/q/65822/2338
// My answer: http://codegolf.stackexchange.com/a/65879/2338

alias I => Integer;

// Calculate the pyramid sum of some sequence.
I s(I* l) =>
        // If less than two elements ...
        l.size < 2
        // then use the first (only element), or 0 if no such.
        then (l[0] else 0)
        // otherwise,
        else s(*
               // take the iterable of pairs of consecutive elements,
               l.paired
               // and add each of them together.
                .map((I[2] i) => i[0] + i[1])
               // then apply s (recursively) on the result.
               );

// Calculate the pyramid sum of the first n primes.
I p(I n) => s(*
              // the infinite sequence of integers, starting with 2.
              loop(2)(1.plus)
              // filter by primality (using trial division)
              .filter((c) => !(2 : c-2)
                              .any((d) => c%d < 1))
              // then take the first n elements
              .take(n)
              // then apply s on the result.
             );

Answer 19

Milky Way 1.4.8, 26 25 bytes

This answer is not competing. Some of the operations were created after this question was posted (but not necessarily for this challenge).

'E&{~F§{G}:y1ba?{_^_}};!

I was able to remove a byte after reading through the comments. Output is a single-element list.

---

Explanation

'                        #  read input from the command line
 E                       #  push a list of the first N primes
  &{~                }   #  while loop
     F                   #  push the sum of TOS elements i.e. [A, B, C] => [[A,B], [B,C]]
      §{ }               #  mapping
        G                #  sum i.e. [1, 2, 3] => 6
          :              #  duplicate the TOS
           y             #  push the length of the TOS to the stack
            1            #  push 1 to the stack
             b           #  evaluate equality of the TOS and STOS
              a          #  logical not
               ?{_ _}    #  if-else statement
                  ^      #  pop the TOS
                     ;   #  swap the TOS and STOS
                         #  dump the TOS to the stack
                      !  #  output the TOS

---

Usage

python3 milkyway.py <path-to-code> -i <input-integer>

Answer 20

Stax, 13 bytes

ä/($◘é∟o¶Vƒm^

Run and debug it