Sylvester's dizisi OEIS A000058 , aşağıdaki gibi tanımlanmış bir tam sayı dizisidir:

Her üye bir önceki tüm üyelerin ürünüdür. Dizinin ilk üyesi 2'dir.

Görev

Bir n alır ve Sylvester Dizisinin birinci terimini hesaplayan mümkün olan en küçük programı oluşturun. Standart giriş, çıkış ve boşluklar uygulanır. Sonuç çok hızlı büyüdüğü için, sonucun seçtiğiniz dilde taşmasına neden olacak herhangi bir terim almanız beklenmez.

Test Kılıfları

Sıfır veya bir indeksleme kullanabilirsiniz. (Burada sıfır endeksleme kullanıyorum)

>>0
2
>>1
3
>>2
7
>>3
43
>>4
1807

Brain-Flak , 76 68 58 52 46 bayt

<>(()())<>{({}[()])<>({({}[()])({})}{}())<>}<>

Çevrimiçi deneyin!

Bunun yerine bu ilişkiyi kullanır:

sekansta sağlanandan modifiye edilmiş bu ilişkiden türetilir:

a(n+1) = a(n) * (a(n) - 1) + 1.

açıklama

Her komutun ne yaptığına dair bir dokümantasyon için lütfen GitHub sayfasını ziyaret edin .

Brain-Flak'ta sırasıyla Stack 1 ve Stack 2 olarak adlandırdığım iki yığın var.

Giriş, Yığın 1'de saklanır.

<>(()())<>             Store 2 in Stack 2.

{                      while(Stack_1 != 0){
  ({}[()])                 Stack_1 <- Stack_1 - 1;
  <>                       Switch stack.
  ({({}[()])({})}{}())     Generate the next number in Stack 2.
  <>                       Switch back to Stack 1.
}

<>                     Switch to Stack 2, implicitly print.

Üretim algoritması için:

({({}[()])({})}{}())      Top <- (Top + Top + (Top-1) + (Top-1) + ... + 0) + 1

(                  )      Push the sum of the numbers evaluated in the process:

 {            }               while(Top != 0){
  ({}[()])                        Pop Top, push Top-1    (added to sum)
          ({})                    Pop again, push again  (added to sum)
                              }

               {}             Top of stack is now zero, pop it.
                 ()           Evaluates to 1 (added to sum).

Alternatif 46 bayt sürüm

Bu sadece bir yığın kullanır.

({}<(()())>){({}<({({}[()])({})}{}())>[()])}{}

Çevrimiçi deneyin!

Jelly, 5 bytes

Ḷ߀P‘

This uses 0-based indexing and the definition from the challenge spec.

Try it online! or verify all test cases.

How it works

Ḷ߀P‘  Main link. Argument: n

Ḷ      Unlength; yield [0, ..., n - 1].
 ߀    Recursively apply the main link to each integer in that range.
   P   Take the product. This yields 1 for an empty range.
    ‘  Increment.

Hexagony, 27 bytes

1{?)=}&~".>")!@(</=+={"/>}*

Unfolded:

    1 { ? )
   = } & ~ "
  . > " ) ! @
 ( < / = + = {
  " / > } * .
   . . . . .
    . . . .

Try it online!

Explanation

Let's consider the sequence b(a) = a(n) - 1 and do a little rearranging:

b(a) = a(n) - 1
     = a(n-1)*(a(n-1)-1) + 1 - 1
     = (b(n-1) + 1)*(b(n-1) + 1 - 1)
     = (b(n-1) + 1)*b(n-1)
     = b(n-1)^2 + b(n-1)

This sequence is very similar but we can defer the increment to the very end, which happens to save a byte in this program.

So here is the annotated source code:

^{Created with Timwi's HexagonyColorer.}

And here is a memory diagram (the red triangle shows the memory pointer's initial position and orientation):

^{Created with Timwi's EsotericIDE.}

The code begins on the grey path which wraps the left corner, so the initial linear bit is the following:

1{?)(
1      Set edge b(1) to 1.
 {     Move MP to edge N.
  ?    Read input into edge N.
   )(  Increment, decrement (no-op).

Then the code hits the < which is a branch and indicates the start (and end) of the main loop. As long as the N edge has a positive value, the green path will be executed. That path wraps around the grid a few times, but it's actually entirely linear:

""~&}=.*}=+={....(

The . are no-ops, so the actual code is:

""~&}=*}=+={(
""             Move the MP to edge "copy".
  ~            Negate. This is to ensure that the value is negative so that &...
   &           ...copies the left-hand neighbour, i.e. b(i).
    }=         Move the MP to edge b(i)^2 and turn it around.
      *        Multiply the two copies of b(i) to compute b(i)^2.
       }=      Move the MP back to edge b(i) and turn it around.
         +     Add the values in edges "copy" and b(i)^2 to compute
               b(i) + b(i)^2 = b(i+1).
          ={   Turn the memory pointer around and move to edge N.
            (  Decrement.

Once this decrementing reduces N to 0, the red path is executed:

")!@
"     Move MP back to edge b(i) (which now holds b(N)).
 )    Increment to get a(N).
  !   Print as integer.
   @  Terminate the program.

J, 18 14 12 bytes

This version thanks to randomra. I'll try to write a detailed explanation later.

0&(]*:-<:)2:

J, 14 bytes

This version thanks to miles. Used the power adverb ^: instead of an agenda as below. More explanation to come.

2(]*:-<:)^:[~]

J, 18 bytes

2:`(1+*/@$:@i.)@.*

0-indexed.

Examples

   e =: 2:`(1+*/@$:@i.)@.*
   e 1
3
   e 2
7
   e 3
43
   e 4
1807
   x: e i. 10
2 3 7 43 1807 3263443 10650056950807 113423713055421862298779648 12864938683278674737956996400574416174101565840293888 1655066473245199944217466828172807675196959605278049661438916426914992848    91480678309535880456026315554816
   |: ,: x: e i. 10
                                                                                                        2
                                                                                                        3
                                                                                                        7
                                                                                                       43
                                                                                                     1807
                                                                                                  3263443
                                                                                           10650056950807
                                                                              113423713055421862298779648
                                                    12864938683278674737956996400574416174101565840293888
165506647324519994421746682817280767519695960527804966143891642691499284891480678309535880456026315554816

Explanation

This is an agenda that looks like this:

           ┌─ 2:
           │    ┌─ 1
       ┌───┤    ├─ +
       │   └────┤           ┌─ / ─── *
── @. ─┤        │     ┌─ @ ─┴─ $:
       │        └─ @ ─┴─ i.
       └─ *

(Generated using (9!:7)'┌┬┐├┼┤└┴┘│─' then 5!:4<'e')

Decomposing:

       ┌─ ...
       │
── @. ─┤
       │
       └─ *

Using the top branch as a the gerund G, and the bottom as the selector F, this is:

e n     <=>     ((F n) { G) n

This uses the constant function 2: when 0 = * n, that is, when the sign is zero (thus n is zero). Otherwise, we use this fork:

  ┌─ 1
  ├─ +
──┤           ┌─ / ─── *
  │     ┌─ @ ─┴─ $:
  └─ @ ─┴─ i.

Which is one plus the following atop series:

            ┌─ / ─── *
      ┌─ @ ─┴─ $:
── @ ─┴─ i.

Decomposing further, this is product (*/) over self-reference ($:) over range (i.).

Perl 6, 24 bytes

{(2,{1+[*] @_}...*)[$_]}

{(2,{1+.²-$_}...*)[$_]}

Explanation

# bare block with implicit parameter ｢$_｣
{
  (
    # You can replace 2 with 1 here
    # so that it uses 1 based indexing
    # rather than 0 based
    2,

    # bare block with implicit parameter ｢@_｣
    {
      1 +

      # reduce the input of this inner block with ｢&infix:<*>｣
      # ( the input is all of them generated when using a slurpy @ var )
      [*] @_

      # that is the same as:
      # ｢@_.reduce: &infix:<*>｣
    }

    # keep calling that to generate more values until:
    ...

    # forever
    *

  # get the value as indexed by the input
  )[ $_ ]
}

Usage:

my &code = {(2,{1+[*] @_}...*)[$_]}

say code 0; # 2
say code 1; # 3
say code 2; # 7
say code 3; # 43
say code 4; # 1807

# you can even give it a range
say code 4..7;
# (1807 3263443 10650056950807 113423713055421844361000443)

say code 8;
# 12864938683278671740537145998360961546653259485195807
say code 9;
# 165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185443
say code 10;
# 27392450308603031423410234291674686281194364367580914627947367941608692026226993634332118404582438634929548737283992369758487974306317730580753883429460344956410077034761330476016739454649828385541500213920807

my $start = now;
# how many digits are there in the 20th value
say chars code 20;
# 213441

my $finish = now;
# how long did it take to generate the values up to 20
say $finish - $start, ' seconds';
# 49.7069076 seconds

Haskell, 26 bytes

f n|n<1=2|m<-f$n-1=1+m*m-m

Usage example: f 4 -> 1807.

Java 7, 46 42 bytes

int f(int n){return--n<0?2:f(n)*~-f(n)+1;}

Uses 0-indexing with the usual formula. I swapped n*n-n for n*(n-1) though, since Java doesn't have a handy power operator, and the f() calls were getting long.

Haskell, 25 bytes

(iterate(\m->m*m-m+1)2!!)

S.I.L.O.S, 60 bytes

readIO 
p = 2
lblL
r = p
r + 1
p * r
i - 1
if i L
printInt r

Try it online!

Port of my answer in C.

Brain-Flak, 158 154 bytes

Leaky Nun has me beat here

({}<(()())>){({}[()]<<>(())<>([]){{}<>({}<<>(({}<>))><>)<>({<({}[()])><>({})<>}{})<>{}([])}{}<>({}())([]){{}({}<>)<>([])}{}<>>)}{}([][()]){{}{}([][()])}{} 

Try it Online!

Explanation

Put a two under the input a(0)

({}<(()())>) 

While the input is greater than zero subtract one from the input and...

{
({}[()]

Silently...

<

Put one on the other stack to act as a catalyst for multiplication <>(())<>

While the stack is non-empty

 ([])
 {
  {}

Move the top of the list over and copy

  <>({}<<>(({}<>))><>)

Multiply the catalyst by the copy

  <>({<({}[()])><>({})<>}{})<>{}
  ([])
 }
 {}

Add one

 <>({}())

Move the sequence back to the proper stack

 ([])
 {
 {}
 ({}<>)<>
 ([])
 }
 {}
 <>
>)
}{}

Remove all but the bottom item (i.e. the last number created)

([][()])
{
{}
{}
([][()])
}
{}

C, 32 bytes

f(n){return--n?f(n)*~-f(n)+1:2;}

Uses 1-based indexing. Test it on Ideone.

Actually, 9 bytes

2@`rΣτu`n

Try it online!

Uses this relationship instead:

which is derived from this relationship modified from that provided in the sequence:

a(n+1) = a(n) * (a(n) - 1) + 1.

R, 44 42 41 bytes

2 bytes save thanks to JDL

1 byte save thanks to user5957401

f=function(n)ifelse(n,(a=f(n-1))^2-a+1,2)

Oasis, 4 bytes (non-competing)

Probably my last language from the golfing family! Non-competing, since the language postdates the challenge.

Code:

²->2

Alternative solution thanks to Zwei:

<*>2

Expanded version:

a(n) = ²->
a(0) = 2

Explanation:

²    # Stack is empty, so calculate a(n - 1) ** 2.
 -   # Subtract, arity 2, so use a(n - 1).
  >  # Increment by 1.

Uses the CP-1252 encoding. Try it online!

Python, 38 36 bytes

2 bytes thanks to Dennis.

f=lambda n:0**n*2or~-f(n-1)*f(n-1)+1

Ideone it!

Uses this relationship modified from that provided in the sequence instead:

a(n+1) = a(n) * (a(n) - 1) + 1

Explanation

0**n*2 returns 2 when n=0 and 0 otherwise, because 0**0 is defined to be 1 in Python.

Cheddar, 26 bytes

n g->n?g(n-=1)**2-g(n)+1:2

Try it online!

Pretty idiomatic.

Explanation

n g ->    // Input n, g is this function
  n ?     // if n is > 1
    g(n-=1)**2-g(n)+1   // Do equation specified in OEIS
  : 2     // if n == 0 return 2

CJam, 10 bytes

2{_(*)}ri*

Uses 0-based indexing. Try it online!

05AB1E, 7 bytes

2sFD<*>

Explained

Uses zero-based indexing.

2         # push 2 (initialization for n=0)
 sF       # input nr of times do
   D<*    # x(x-1)
      >   # add 1

Try it online!

Prolog, 49 bytes

a(0,2).
a(N,R):-N>0,M is N-1,a(M,T),R is T*T-T+1.

SILOS 201 bytes

readIO 
def : lbl
set 128 2
set 129 3
j = i
if j z
print 2
GOTO e
:z
j - 1
if j Z
print 3
GOTO e
:Z
i - 1
:a
a = 127
b = 1
c = 1
:b
b * c
a + 1
c = get a
if c b
b + 1
set a b
i - 1
if i a
printInt b
:e

Feel free to try it online!

Jelly, 7 bytes

²_’
2Ç¡

Try it online!

Uses this relationship provided in the sequence instead: a(n+1) = a(n)^2 - a(n) + 1

Explanation

2Ç¡   Main chain, argument in input

2     Start with 2
  ¡   Repeat as many times as the input:
 Ç        the helper link.


²_’   Helper link, argument: z
²     z²
  ’   z - 1
 _    subtraction, yielding z² - (z-1) = z² - z + 1

C, 46 bytes

s(n,p,r){for(p=r=2;n-->0;p*=r)r=p+1;return r;}

Ideone it!

Uses p as the temporary storage of the product.

Basically, I defined two sequences p(n) and r(n), where r(n)=p(n-1)+1 and p(n)=p(n-1)*r(n).

r(n) is the required sequence.

R, 50 46 44 bytes

    n=scan();v=2;if(n)for(i in 1:n){v=v^2-v+1};v

Rather than tracking the whole sequence, we just keep track of the product, which follows the given quadratic update rule as long as n>1 n>0. (This sequence uses the "start at one zero" convention)

Using the start at zero convention saves a couple of bytes since we can use if(n) rather than if(n>1)

Jellyfish, 13 bytes

p
\Ai
&(*
><2

Try it online!

Explanation

Let's start from the bottom up:

(*
<

This is a hook, which defines a function f(x) = (x-1)*x.

&(*
><

This composes the previous hook with the increment function so it gives us a function g(x) = (x-1)*x+1.

\Ai
&(*
><

Finally, this generates a function h which is an iteration of the previous function g, as many times as given by the integer input.

\Ai
&(*
><2

And finally, we apply this iteration to the initial value 2. The p at the top just prints the result.

Alternative (also 13 bytes)

p
>
\Ai
(*
>1

This defers the increment until the very end.

C, 43, 34, 33 bytes

1-indexed:

F(n){return--n?n=F(n),n*n-n+1:2;}

Test main:

int main() {
  printf("%d\n", F(1));
  printf("%d\n", F(2));
  printf("%d\n", F(3));
  printf("%d\n", F(4));
  printf("%d\n", F(5));
}

Brachylog, 13 bytes

0,2|-:0&-y+*+

Try it online!

Uses this relationship instead:

which is derived from this relationship modified from that provided in the sequence:

a(n+1) = a(n) * (a(n) - 1) + 1.

Mathematica, 19 bytes

Nest[#^2-#+1&,2,#]&

Or 21 bytes:

Array[#0,#,0,1+1##&]&

Labyrinth, 18 bytes

Credits to Sp3000 who found the same solution independently.

?
#
)}"{!@
* (
(:{

Try it online!

Actually, 14 12 bytes

This used 0-indexing. Golfing suggestions welcome. Try it online!

2#,`;πu@o`nF

Ungolfing:

2#              Start with [2]
  ,`     `n     Take 0-indexed input and run function (input) times
    ;           Duplicate list
     πu         Take product of list and increment
       @o       Swap and append result to the beginning of the list
           F    Return the first item of the resulting list

GolfScript, 12 10 bytes

2 bytes thanks to Dennis.

~2\{.(*)}*

Try it online!

Uses a(n) = a(n-1) * (a(n-1)-1) + 1.