Padovan n-step number sequences

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Revision as of 03:36, 3 December 2021 by Alextretyak (talk | contribs) (Added 11l)
Task
Padovan n-step number sequences
You are encouraged to solve this task according to the task description, using any language you may know.

As the Fibonacci sequence expands to the Fibonacci n-step number sequences; We similarly expand the Padovan sequence to form these Padovan n-step number sequences.

The Fibonacci-like sequences can be defined like this: 

   For n == 2:
       start:      1, 1
       Recurrence: R(n, x) = R(n, x-1) + R(n, x-2); for n == 2
   For n == N:
       start:      First N terms of R(N-1, x)
       Recurrence: R(N, x) = sum(R(N, x-1) + R(N, x-2) + ... R(N, x-N))

For this task we similarly define terms of the first 2..n-step Padovan sequences as: 

   For n == 2:
       start:      1, 1, 1
       Recurrence: R(n, x) = R(n, x-2) + R(n, x-3); for n == 2
   For n == N:
       start:      First N + 1 terms of R(N-1, x)
       Recurrence: R(N, x) = sum(R(N, x-2) + R(N, x-3) + ... R(N, x-N-1))

The initial values of the sequences are:

Padovan -step sequences
Values OEIS Entry
2 1,1,1,2,2,3,4,5,7,9,12,16,21,28,37, ... A134816: 'Padovan's spiral numbers'
3 1,1,1,2,3,4,6,9,13,19,28,41,60,88,129, ... A000930: 'Narayana's cows sequence'
4 1,1,1,2,3,5,7,11,17,26,40,61,94,144,221, ... A072465: 'A Fibonacci-like model in which each pair of rabbits dies after the birth of their 4th litter'
5 1,1,1,2,3,5,8,12,19,30,47,74,116,182,286, ... A060961: 'Number of compositions (ordered partitions) of n into 1's, 3's and 5's'
6 1,1,1,2,3,5,8,13,20,32,51,81,129,205,326, ... <not found>
7 1,1,1,2,3,5,8,13,21,33,53,85,136,218,349, ... A117760: 'Expansion of 1/(1 - x - x^3 - x^5 - x^7)'
8 1,1,1,2,3,5,8,13,21,34,54,87,140,225,362, ... <not found>


Task
  1. Write a function to generate the first terms, of the first 2..max_n Padovan -step number sequences as defined above.
  2. Use this to print and show here at least the first t=15 values of the first 2..8 -step sequences.
    (The OEIS column in the table above should be omitted).



11l

Translation of: Nim

<lang 11l>F rn(n, k) -> [Int] 

  assert(k >= 2)
  V result = I n == 2 {[1, 1, 1]} E rn(n - 1, n + 1)
  L result.len != k
     result.append(sum(result[(len)-n-1 .< (len)-1]))
  R result

L(n) 2..8 

  print(n‘: ’rn(n, 15).map(it -> ‘#3’.format(it)).join(‘ ’))</lang>
Output:
2:   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3:   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4:   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5:   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6:   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7:   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8:   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

ALGOL 68

Translation of: ALGOL W

<lang algol68>BEGIN # show some valuies of the Padovan n-step number sequences # 

   # returns an array with the elements set to the elements of   #
   # the Padovan sequences from 2 to max s & elements 1 to max e #
   # max s must be >= 2                                          #
   PROC padovan sequences = ( INT max s, max e )[,]INT:
        BEGIN
           PRIO MIN = 1;
           OP   MIN = ( INT a, b )INT: IF a < b THEN a ELSE b FI;
           # sequence 2                                          #
           [ 2 : max s, 1 : max e ]INT r;
           FOR x TO max e MIN 3 DO r[ 2, x ] := 1 OD;
           FOR x FROM 4 TO max e DO r[ 2, x ] := r[ 2, x - 2 ] + r[ 2, x - 3 ] OD;
           # sequences 3 and above                               #
           FOR n FROM 3 TO max s DO
               FOR x TO max e MIN n + 1 DO r[ n, x ] := r[ n - 1, x ] OD;
               FOR x FROM n + 2 TO max e DO
                   r[ n, x ] := 0;
                   FOR p FROM x - n - 1 TO x - 2 DO r[ n, x ] +:= r[ n, p ] OD
               OD
           OD;
           r
        END # padovan sequences # ;
   # calculate and show the sequences                            #
   [,]INT r = padovan sequences( 8, 15 );
   print( ( "Padovan n-step sequences:", newline ) );
   FOR n FROM 1 LWB r TO 1 UPB r DO
       print( ( whole( n, 0 ), " |" ) );
       FOR x FROM 2 LWB r TO 2 UPB r DO print( ( " ", whole( r[ n, x ], -3 ) ) ) OD;
       print( ( newline ) )
   OD

END</lang>

Output:
Padovan n-step sequences:
2 |   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3 |   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4 |   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5 |   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6 |   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7 |   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8 |   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

ALGOL W

<lang algolw>begin % show some valuies of the Padovan n-step number sequences  % 

   % sets R(i,j) to the jth element of the ith padovan sequence  %
   % maxS is the number of sequences to generate and maxE is the %
   % maximum number of elements for each sequence                %
   % maxS must be >= 2                                           %
   procedure PadovanSequences ( integer array R ( *, * )
                              ; integer value maxS, maxE
                              ) ;
   begin
       integer procedure min( integer value a, b ) ; if a < b then a else b;
       % sequence 2                                              %
       for x := 1 until min( maxE, 3 ) do R( 2, x ) := 1;
       for x := 4 until maxE do R( 2, x ) := R( 2, x - 2 ) + R( 2, x - 3 );
       % sequences 3 and above                                   %
       for N := 3 until maxS do begin
           for x := 1 until min( maxE, N + 1 ) do R( N, x ) := R( N - 1, x );
           for x := N + 2 until maxE do begin
               R( N, x ) := 0;
               for p := x - N - 1 until x - 2 do R( N, x ) := R( N, x ) + R( N, p )
           end for_x
       end for_N
   end PadovanSequences ;
   integer MAX_SEQUENCES, MAX_ELEMENTS;
   MAX_SEQUENCES :=  8;
   MAX_ELEMENTS  := 15;
   begin % calculate and show the sequences                      %
       % array to hold the Padovan Sequences                     %
       integer array R ( 2 :: MAX_SEQUENCES, 1 :: MAX_ELEMENTS );
       % construct the sequences                                 %
       PadovanSequences( R, MAX_SEQUENCES, MAX_ELEMENTS );
       % show the sequences                                      %
       write( "Padovan n-step sequences:" );
       for n := 2 until MAX_SEQUENCES do begin
           write( i_w := 1, s_w := 0, n, " |" );
           for x := 1 until MAX_ELEMENTS do writeon( i_w := 3, s_w := 0, " ", R( n, x ) )
       end for_n
   end

end.</lang>

Output:
Padovan n-step sequences:
2 |   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3 |   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4 |   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5 |   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6 |   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7 |   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8 |   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

AppleScript

<lang applescript>use AppleScript version "2.4" use framework "Foundation" use scripting additions

PADOVAN N-STEP NUMBERS ----------------

-- padovans :: [Int] on padovans(n) 

   script recurrence
       on |λ|(xs)
           {item 1 of xs, ¬
               rest of xs & {sum(take(n, xs)) as integer}}
       end |λ|
   end script
   
   if 3 > n then
       set seed to |repeat|(1)
   else
       set seed to padovans(n - 1)
   end if
   
   if 0 > n then
       {}
   else
       unfoldr(recurrence, take(1 + n, seed))
   end if

end padovans


TEST -------------------------

on run 

   script nSample
       on |λ|(n)
           take(15, padovans(n))
       end |λ|
   end script
   
   script justified
       on |λ|(ns)
           concatMap(justifyRight(4, space), ns)
       end |λ|
   end script
   
   fTable("Padovan N-step Series:", str, justified, ¬
       nSample, enumFromTo(2, 8))

end run


FORMATTING ----------------------

-- fTable :: String -> (a -> String) -> (b -> String) -> -- (a -> b) -> [a] -> String on fTable(s, xShow, fxShow, f, xs) 

   set ys to map(xShow, xs)
   set w to maximum(map(my |length|, ys))
   script arrowed
       on |λ|(a, b)
           |λ|(a) of justifyRight(w, space) & " ->" & b
       end |λ|
   end script
   s & linefeed & unlines(zipWith(arrowed, ¬
       ys, map(compose(fxShow, f), xs)))

end fTable


GENERIC ------------------------

-- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c on compose(f, g) 

   script
       property mf : mReturn(f)
       property mg : mReturn(g)
       on |λ|(x)
           mf's |λ|(mg's |λ|(x))
       end |λ|
   end script

end compose


-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs) 

   set lng to length of xs
   set acc to {}
   tell mReturn(f)
       repeat with i from 1 to lng
           set acc to acc & (|λ|(item i of xs, i, xs))
       end repeat
   end tell
   if {text, string} contains class of xs then
       acc as text
   else
       acc
   end if

end concatMap


-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n) 

   if m ≤ n then
       set lst to {}
       repeat with i from m to n
           set end of lst to i
       end repeat
       lst
   else
       {}
   end if

end enumFromTo


-- intercalate :: String -> [String] -> String on intercalate(delim, xs) 

   set {dlm, my text item delimiters} to ¬
       {my text item delimiters, delim}
   set s to xs as text
   set my text item delimiters to dlm
   s

end intercalate


-- justifyRight :: Int -> Char -> String -> String on justifyRight(n, cFiller) 

   script
       on |λ|(v)
           set strText to v as text
           if n > length of strText then
               text -n thru -1 of ¬
                   ((replicate(n, cFiller) as text) & strText)
           else
               strText
           end if
       end |λ|
   end script

end justifyRight


-- length :: [a] -> Int on |length|(xs) 

   set c to class of xs
   if list is c or string is c then
       length of xs
   else
       (2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
   end if

end |length|


-- map :: (a -> b) -> [a] -> [b] on map(f, xs) 

   -- The list obtained by applying f
   -- to each element of xs.
   tell mReturn(f)
       set lng to length of xs
       set lst to {}
       repeat with i from 1 to lng
           set end of lst to |λ|(item i of xs, i, xs)
       end repeat
       return lst
   end tell

end map


-- maximum :: Ord a => [a] -> a on maximum(xs) 

   set ca to current application
   unwrap((ca's NSArray's arrayWithArray:xs)'s ¬
       valueForKeyPath:"@max.self")

end maximum


-- min :: Ord a => a -> a -> a on min(x, y) 

   if y < x then
       y
   else
       x
   end if

end min


-- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f) 

   -- 2nd class handler function lifted 
   -- into 1st class script wrapper. 
   if script is class of f then
       f
   else
       script
           property |λ| : f
       end script
   end if

end mReturn


-- repeat :: a -> Generator [a] on |repeat|(x) 

   script
       on |λ|()
           return x
       end |λ|
   end script

end |repeat|


-- Egyptian multiplication - progressively doubling a list, appending -- stages of doubling to an accumulator where needed for binary -- assembly of a target length -- replicate :: Int -> String -> String on replicate(n, s) 

   -- Egyptian multiplication - progressively doubling a list, 
   -- appending stages of doubling to an accumulator where needed 
   -- for binary assembly of a target length
   script p
       on |λ|({n})
           n ≤ 1
       end |λ|
   end script
   
   script f
       on |λ|({n, dbl, out})
           if (n mod 2) > 0 then
               set d to out & dbl
           else
               set d to out
           end if
           {n div 2, dbl & dbl, d}
       end |λ|
   end script
   
   set xs to |until|(p, f, {n, s, ""})
   item 2 of xs & item 3 of xs

end replicate


-- str :: a -> String on str(x) 

   x as string

end str


-- sum :: [Num] -> Num on sum(xs) 

   set ca to current application
   ((ca's NSArray's arrayWithArray:xs)'s ¬
       valueForKeyPath:"@sum.self") as real

end sum


-- take :: Int -> [a] -> [a] -- take :: Int -> String -> String on take(n, xs) 

   set c to class of xs
   if list is c then
       set lng to length of xs
       if 0 < n and 0 < lng then
           items 1 thru min(n, lng) of xs
       else
           {}
       end if
   else if string is c then
       if 0 < n then
           text 1 thru min(n, length of xs) of xs
       else
           ""
       end if
   else if script is c then
       set ys to {}
       repeat with i from 1 to n
           set v to |λ|() of xs
           if missing value is v then
               return ys
           else
               set end of ys to v
           end if
       end repeat
       return ys
   else
       missing value
   end if

end take


-- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v) 

   -- A lazy (generator) list unfolded from a seed value
   -- by repeated application of f to a value until no
   -- residue remains. Dual to fold/reduce.
   -- f returns either nothing (missing value),
   -- or just (value, residue).
   script
       property valueResidue : {v, v}
       property g : mReturn(f)
       on |λ|()
           set valueResidue to g's |λ|(item 2 of (valueResidue))
           if missing value ≠ valueResidue then
               item 1 of (valueResidue)
           else
               missing value
           end if
       end |λ|
   end script

end unfoldr


-- unlines :: [String] -> String on unlines(xs) 

   -- A single string formed by the intercalation
   -- of a list of strings with the newline character.
   set {dlm, my text item delimiters} to ¬
       {my text item delimiters, linefeed}
   set s to xs as text
   set my text item delimiters to dlm
   s

end unlines


-- until :: (a -> Bool) -> (a -> a) -> a -> a on |until|(p, f, x) 

   set v to x
   set mp to mReturn(p)
   set mf to mReturn(f)
   repeat until mp's |λ|(v)
       set v to mf's |λ|(v)
   end repeat
   v

end |until|


-- unwrap :: NSValue -> a on unwrap(nsValue) 

   if nsValue is missing value then
       missing value
   else
       set ca to current application
       item 1 of ((ca's NSArray's arrayWithObject:nsValue) as list)
   end if

end unwrap


-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] on zipWith(f, xs, ys) 

   set lng to min(length of xs, length of ys)
   set lst to {}
   if 1 > lng then
       return {}
   else
       tell mReturn(f)
           repeat with i from 1 to lng
               set end of lst to |λ|(item i of xs, item i of ys)
           end repeat
           return lst
       end tell
   end if

end zipWith</lang>

Output:
Padovan N-step Series:
2 ->   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3 ->   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4 ->   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5 ->   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6 ->   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7 ->   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8 ->   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

C

Translation of: Wren

<lang c>#include <stdio.h>

void padovanN(int n, size_t t, int *p) { 

   int i, j;
   if (n < 2 || t < 3) {
       for (i = 0; i < t; ++i) p[i] = 1;
       return;
   }
   padovanN(n-1, t, p);
   for (i = n + 1; i < t; ++i) {
       p[i] = 0;
       for (j = i - 2; j >= i - n - 1; --j) p[i] += p[j];
   }

}

int main() { 

   int n, i;
   const size_t t = 15;
   int p[t];
   printf("First %ld terms of the Padovan n-step number sequences:\n", t);
   for (n = 2; n <= 8; ++n) {
       for (i = 0; i < t; ++i) p[i] = 0;
       padovanN(n, t, p);
       printf("%d: ", n);
       for (i = 0; i < t; ++i) printf("%3d ", p[i]);
       printf("\n");
   }
   return 0;

}</lang>

Output:
First 15 terms of the Padovan n-step number sequences:
2:   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37 
3:   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129 
4:   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221 
5:   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286 
6:   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326 
7:   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349 
8:   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

F#

<lang fsharp> // Padovan n-step number sequences. Nigel Galloway: July 28th., 2021 let rec pad=function 2->Seq.unfold(fun(n:int[])->Some(n.[0],Array.append n.[1..2] [|Array.sum n.[0..1]|]))[|1;1;1|] 

                   |g->Seq.unfold(fun(n:int[])->Some(n.[0],Array.append n.[1..g] [|Array.sum n.[0..g-1]|]))(Array.ofSeq(pad(g-1)|>Seq.take(g+1)))

[2..8]|>List.iter(fun n->pad n|>Seq.take 15|>Seq.iter(printf "%d "); printfn "") </lang>

Output:
1 1 1 2 2 3 4 5 7 9 12 16 21 28 37
1 1 1 2 3 4 6 9 13 19 28 41 60 88 129
1 1 1 2 3 5 7 11 17 26 40 61 94 144 221
1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
1 1 1 2 3 5 8 13 21 34 54 87 140 225 362

Factor

Works with: Factor version 0.99 2021-02-05

<lang factor>USING: compiler.tree.propagation.call-effect io kernel math math.ranges prettyprint sequences ;

padn ( m n -- seq )
   V{ "|" 1 1 1 } over prefix clone over 2 -
   [ dup last2 + suffix! ] times rot pick 1 + -
   [ dup length 1 - pick [ - ] keepd pick <slice> sum suffix! ]
   times nip ;

"Padovan n-step sequences" print 2 8 [a..b] [ 15 swap padn ] map simple-table.</lang>

Output:
Padovan n-step sequences
2 | 1 1 1 2 2 3 4 5  7  9  12 16 21  28  37
3 | 1 1 1 2 3 4 6 9  13 19 28 41 60  88  129
4 | 1 1 1 2 3 5 7 11 17 26 40 61 94  144 221
5 | 1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
6 | 1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
7 | 1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
8 | 1 1 1 2 3 5 8 13 21 34 54 87 140 225 362

Go

Translation of: Wren

<lang go>package main

import "fmt"

func padovanN(n, t int) []int { 

   if n < 2 || t < 3 {
       ones := make([]int, t)
       for i := 0; i < t; i++ {
           ones[i] = 1
       }
       return ones
   }
   p := padovanN(n-1, t)
   for i := n + 1; i < t; i++ {
       p[i] = 0
       for j := i - 2; j >= i-n-1; j-- {
           p[i] += p[j]
       }
   }
   return p

}

func main() { 

   t := 15
   fmt.Println("First", t, "terms of the Padovan n-step number sequences:")
   for n := 2; n <= 8; n++ {
       fmt.Printf("%d: %3d\n", n, padovanN(n, t))
   }

}</lang>

Output:
First 15 terms of the Padovan n-step number sequences:
2: [  1   1   1   2   2   3   4   5   7   9  12  16  21  28  37]
3: [  1   1   1   2   3   4   6   9  13  19  28  41  60  88 129]
4: [  1   1   1   2   3   5   7  11  17  26  40  61  94 144 221]
5: [  1   1   1   2   3   5   8  12  19  30  47  74 116 182 286]
6: [  1   1   1   2   3   5   8  13  20  32  51  81 129 205 326]
7: [  1   1   1   2   3   5   8  13  21  33  53  85 136 218 349]
8: [  1   1   1   2   3   5   8  13  21  34  54  87 140 225 362]

Haskell

<lang haskell>import Data.Bifunctor (second) import Data.List (transpose, uncons, unfoldr)

PADOVAN N-STEP SERIES -----------------

padovans :: Int -> [Int] padovans n 

 | 0 > n = []
 | otherwise = unfoldr (recurrence n) $ take (succ n) xs
 where
   xs
     | 3 > n = repeat 1
     | otherwise = padovans $ pred n

recurrence :: Int -> [Int] -> Maybe (Int, [Int]) recurrence n = 

 ( fmap
     . second
     . flip (<>)
     . pure
     . sum
     . take n
 )
   <*> uncons

TEST -------------------------

main :: IO () main = 

 putStrLn $
   "Padovan N-step series:\n\n"
     <> spacedTable
       justifyRight
       ( fmap
           ( \n ->
               [show n <> " -> "]
                 <> fmap show (take 15 $ padovans n)
           )
           [2 .. 8]
       )

FORMATTING ----------------------

spacedTable :: 

 (Int -> Char -> String -> String) -> String -> String

spacedTable aligned rows = 

 unlines $
   fmap
     (unwords . zipWith (`aligned` ' ') columnWidths)
     rows
 where
   columnWidths =
     fmap
       (maximum . fmap length)
       (transpose rows)

justifyRight :: Int -> a -> [a] -> [a] justifyRight n c = drop . length <*> (replicate n c <>)</lang>

Output:
Padovan N-step series:

2 ->  1 1 1 2 2 3 4  5  7  9 12 16  21  28  37
3 ->  1 1 1 2 3 4 6  9 13 19 28 41  60  88 129
4 ->  1 1 1 2 3 5 7 11 17 26 40 61  94 144 221
5 ->  1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
6 ->  1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
7 ->  1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
8 ->  1 1 1 2 3 5 8 13 21 34 54 87 140 225 362

JavaScript

<lang javascript>(() => { 

   "use strict";

   // ---------- PADOVAN N-STEP NUMBER SERIES -----------

   // padovans :: Int -> [Int]
   const padovans = n => {
       // Padovan number series of step N
       const recurrence = ns => [
           ns[0],
           ns.slice(1).concat(
               sum(take(n)(ns))
           )
       ];



       return 0 > n ? (
           []
       ) : unfoldr(recurrence)(
           take(1 + n)(
               3 > n ? (
                   repeat(1)
               ) : padovans(n - 1)
           )
       );
   };

   // ---------------------- TEST -----------------------
   // main :: IO ()
   const main = () =>
       fTable("Padovan N-step series:")(str)(
           xs => xs.map(
               compose(justifyRight(4)(" "), str)
           )
           .join("")
       )(
           compose(take(15), padovans)
       )(
           enumFromTo(2)(8)
       );



   // --------------------- GENERIC ---------------------

   // compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
   const compose = (...fs) =>
       // A function defined by the right-to-left
       // composition of all the functions in fs.
       fs.reduce(
           (f, g) => x => f(g(x)),
           x => x
       );



   // enumFromTo :: Int -> Int -> [Int]
   const enumFromTo = m =>
       n => Array.from({
           length: 1 + n - m
       }, (_, i) => m + i);



   // repeat :: a -> Generator [a]
   const repeat = function* (x) {
       while (true) {
           yield x;
       }
   };



   // sum :: [Num] -> Num
   const sum = xs =>
       // The numeric sum of all values in xs.
       xs.reduce((a, x) => a + x, 0);



   // take :: Int -> [a] -> [a]
   // take :: Int -> String -> String
   const take = n =>
       // The first n elements of a list,
       // string of characters, or stream.
       xs => "GeneratorFunction" !== xs
       .constructor.constructor.name ? (
           xs.slice(0, n)
       ) : [].concat(...Array.from({
           length: n
       }, () => {
           const x = xs.next();

           return x.done ? [] : [x.value];
       }));



   // unfoldr :: (b -> Maybe (a, b)) -> b -> Gen [a]
   const unfoldr = f =>
       // A lazy (generator) list unfolded from a seed value
       // by repeated application of f to a value until no
       // residue remains. Dual to fold/reduce.
       // f returns either Nothing or Just (value, residue).
       // For a strict output list,
       // wrap with `list` or Array.from
       x => (
           function* () {
               let valueResidue = f(x);

               while (null !== valueResidue) {
                   yield valueResidue[0];
                   valueResidue = f(valueResidue[1]);
               }
           }()
       );

   // ------------------- FORMATTING --------------------

   // fTable :: String -> (a -> String) ->
   // (b -> String) -> (a -> b) -> [a] -> String
   const fTable = s =>
       // Heading -> x display function ->
       //           fx display function ->
       //    f -> values -> tabular string
       xShow => fxShow => f => xs => {
           const
               ys = xs.map(xShow),
               w = Math.max(...ys.map(y => [...y].length)),
               table = zipWith(
                   a => b => `${a.padStart(w, " ")} ->${b}`
               )(ys)(
                   xs.map(x => fxShow(f(x)))
               ).join("\n");

           return `${s}\n${table}`;
       };



   // justifyRight :: Int -> Char -> String -> String
   const justifyRight = n =>
       // The string s, preceded by enough padding (with
       // the character c) to reach the string length n.
       c => s => n > s.length ? (
           s.padStart(n, c)
       ) : s;



   // str :: a -> String
   const str = x => `${x}`;



   // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
   const zipWith = f =>
       // A list constructed by zipping with a
       // custom function, rather than with the
       // default tuple constructor.
       xs => ys => take(
           Math.min(xs.length, ys.length)
       )(
           xs.map((x, i) => f(x)(ys[i]))
       );

   // MAIN ---
   return main();

})();</lang>

Output:
Padovan N-step series:
2 ->   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3 ->   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4 ->   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5 ->   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6 ->   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7 ->   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8 ->   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

Julia

Translation of: Python

<lang julia> """ 

   First nterms terms of the first 2..max_nstep -step Padovan sequences.

""" function nstep_Padovan(max_nstep=8, nterms=15) 

   start = [[], [1, 1, 1]]     # for n=0 and n=1 (hidden).
   for n in 2:max_nstep
       this = start[n][1:n+1]     # Initialise from last
       while length(this) < nterms
           push!(this, sum(this[end - i] for i in 1:n))
       end
       push!(start, this)
   end
   return start[3:end]

end

function print_Padovan_seq(p) 

   println(strip("""
""")) for (n, seq) in enumerate(p) println("| $n || $(replace(string(seq[2:end]), r"[ a-zA-Z\[\]]+" => "")), ...\n|-") end println("|}") end print_Padovan_seq(nstep_Padovan()) </lang>
Output:
Padovan -step sequences
Values
Padovan -step sequences
Values
1 1,1,2,2,3,4,5,7,9,12,16,21,28,37, ...
2 1,1,2,3,4,6,9,13,19,28,41,60,88,129, ...
3 1,1,2,3,5,7,11,17,26,40,61,94,144,221, ...
4 1,1,2,3,5,8,12,19,30,47,74,116,182,286, ...
5 1,1,2,3,5,8,13,20,32,51,81,129,205,326, ...
6 1,1,2,3,5,8,13,21,33,53,85,136,218,349, ...
7 1,1,2,3,5,8,13,21,34,54,87,140,225,362, ...

Mathematica/Wolfram Language

<lang Mathematica>ClearAll[Padovan] Padovan[2,tmax_]:=Module[{start,a,m}, 

start={1,1,1};
start=MapIndexed[a[#21]==#1&,start];
RecurrenceTable[{a[m]==a[m-2]+a[m-3]}~Join~start,a,  {m,tmax}]

] Padovan[n_,tmax_]:=Module[{start,eq,a,m}, 

start=Padovan[n-1,n+1];
start=MapIndexed[a[#21]==#1&,start];
eq=Range[2,n+1];
eq=Append[start,a[m]==Total[a[m-#]&/@eq]];
RecurrenceTable[eq,a, {m,tmax}]

] Padovan[2,15] Padovan[3,15] Padovan[4,15] Padovan[5,15] Padovan[6,15] Padovan[7,15] Padovan[8,15]</lang>

Output:
{1,1,1,2,2,3,4,5,7,9,12,16,21,28,37}
{1,1,1,2,3,4,6,9,13,19,28,41,60,88,129}
{1,1,1,2,3,5,7,11,17,26,40,61,94,144,221}
{1,1,1,2,3,5,8,12,19,30,47,74,116,182,286}
{1,1,1,2,3,5,8,13,20,32,51,81,129,205,326}
{1,1,1,2,3,5,8,13,21,33,53,85,136,218,349}
{1,1,1,2,3,5,8,13,21,34,54,87,140,225,362}

Nim

<lang Nim>import math, sequtils, strutils

proc rn(n, k: Positive): seq[int] = 

 assert k >= 2
 result = if n == 2: @[1, 1, 1] else: rn(n - 1, n + 1)
 while result.len != k:
   result.add sum(result[^(n + 1)..^2])

for n in 2..8: 

 echo n, ": ", rn(n, 15).mapIt(($it).align(3)).join(" ")</lang>
Output:
2:   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3:   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4:   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5:   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6:   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7:   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8:   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362

Perl

<lang perl>use strict; use warnings; use feature <state say>; use List::Util 'sum'; use List::Lazy 'lazy_list';

say 'Padovan N-step sequences; first 25 terms:'; for our $N (2..8) { 

   my $pad_n = lazy_list {
       state $n  = 2;
       state @pn = (1, 1, 1);
       push @pn, sum @pn[ grep { $_ >= 0 } $n-$N .. $n++ - 1 ];
       $pn[-4]
   };

   print "N = $N |";
   print ' ' . $pad_n->next() for 1..25;
   print "\n"

}</lang>

Output:
N = 2 | 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 200 265 351 465 616
N = 3 | 1 1 1 2 3 4 6 9 13 19 28 41 60 88 129 189 277 406 595 872 1278 1873 2745 4023 5896
N = 4 | 1 1 1 2 3 5 7 11 17 26 40 61 94 144 221 339 520 798 1224 1878 2881 4420 6781 10403 15960
N = 5 | 1 1 1 2 3 5 8 12 19 30 47 74 116 182 286 449 705 1107 1738 2729 4285 6728 10564 16587 26044
N = 6 | 1 1 1 2 3 5 8 13 20 32 51 81 129 205 326 518 824 1310 2083 3312 5266 8373 13313 21168 33657
N = 7 | 1 1 1 2 3 5 8 13 21 33 53 85 136 218 349 559 895 1433 2295 3675 5885 9424 15091 24166 38698
N = 8 | 1 1 1 2 3 5 8 13 21 34 54 87 140 225 362 582 936 1505 2420 3891 6257 10061 16178 26014 41830

Phix

Translation of: Go
function padovann(integer n,t)
    if n<2 or t<3 then return repeat(1,t) end if
    sequence p = padovann(n-1, t)
    for i=n+2 to t do
        p[i] = sum(p[i-n-1..i-2])
    end for
    return p
end function
 
constant t = 15,
 fmt = "%d: %d %d %d %d %d %d %d %2d %2d %2d %2d %2d %3d %3d %3d\n"
printf(1,"First %d terms of the Padovan n-step number sequences:\n",t)
for n=2 to 8 do
    printf(1,fmt,n&padovann(n,t))
end for
Output:
First 15 terms of the Padovan n-step number sequences:
2: 1 1 1 2 2 3 4  5  7  9 12 16  21  28  37
3: 1 1 1 2 3 4 6  9 13 19 28 41  60  88 129
4: 1 1 1 2 3 5 7 11 17 26 40 61  94 144 221
5: 1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
6: 1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
7: 1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
8: 1 1 1 2 3 5 8 13 21 34 54 87 140 225 362

Python

Python: Procedural

Generates a wikitable formatted output <lang python>def pad_like(max_n=8, t=15): 

   """
   First t terms of the first 2..max_n-step Padovan sequences.
   """
   start = [[], [1, 1, 1]]     # for n=0 and n=1 (hidden).
   for n in range(2, max_n+1):
       this = start[n-1][:n+1]     # Initialise from last
       while len(this) < t:
           this.append(sum(this[i] for i in range(-2, -n - 2, -1)))
       start.append(this)
   return start[2:]

def pr(p): 

   print(
.strip()) for n, seq in enumerate(p, 2): print(f"| {n:2} || {str(seq)[1:-1].replace(' ', )+', ...'}\n|-") print('|}') if __name__ == '__main__': p = pad_like() pr(p)</lang>
Output:
Padovan -step sequences
Values
Padovan -step sequences
Values
2 1,1,1,2,2,3,4,5,7,9,12,16,21,28,37, ...
3 1,1,1,2,3,4,6,9,13,19,28,41,60,88,129, ...
4 1,1,1,2,3,5,7,11,17,26,40,61,94,144,221, ...
5 1,1,1,2,3,5,8,12,19,30,47,74,116,182,286, ...
6 1,1,1,2,3,5,8,13,20,32,51,81,129,205,326, ...
7 1,1,1,2,3,5,8,13,21,33,53,85,136,218,349, ...
8 1,1,1,2,3,5,8,13,21,34,54,87,140,225,362, ...


Python: Functional

Defined in terms of a generic anamorphism, using the unfold abstraction, which is dual to reduce.

Aims for a legible expression of the Padovan recurrence relation, and a good level of reliability and code reuse, as well as rapid drafting and refactoring.

Functional composition, while widely practiced and written about in the context of Python, is not the dominant Python tradition. The documentation of the Python itertools module does, however, acknowledge its debt to languages like ML and Haskell, both for an algebra of composition, and for a source of function-naming traditions.

This draft is thoroughly linted, for a high degree of compliance with Python language standards and layout.

Patterns of functional composition are constrained more by mathematical necessity than by arbitrary convention, but this still leaves room for alternative idioms of functional coding in Python. It is to be hoped that others will contribute divergent examples, enriching the opportunities for contrastive insight which Rosetta code aims to provide.

<lang python>Padovan n-step number sequences

from itertools import chain, islice, repeat


  1. nStepPadovan :: Int -> [Int]

def nStepPadovan(n): 

   Non-finite series of N-step Padovan numbers,
      defined by a recurrence relation.
   
   return unfoldr(recurrence(n))(
       take(1 + n)(
           repeat(1) if 3 > n else (
               nStepPadovan(n - 1)
           )
       )
   )


  1. recurrence :: Int -> [Int] -> Int

def recurrence(n): 

   Recurrence relation in Fibonacci,
      Padovan and Perrin sequences.
   
   def go(xs):
       h, *t = xs
       return h, t + [sum(take(n)(xs))]
   return go


  1. ------------------------- TEST -------------------------
  2. main :: IO ()

def main(): 

   First 15 terms each nStepPadovan(n) series
      where n is drawn from [2..8]
   
   xs = range(2, 1 + 8)
   print('Padovan n-step series:\n')
   print(
       spacedTable(list(map(
           lambda k, n: list(chain(
               [k + ' -> '],
               (
                   str(x) for x
                   in take(15)(nStepPadovan(n))
               )
           )),
           (str(x) for x in xs),
           xs
       )))
   )


  1. ----------------------- GENERIC ------------------------
  1. take :: Int -> [a] -> [a]
  2. take :: Int -> String -> String

def take(n): 

   The prefix of xs of length n,
      or xs itself if n > length xs.
   
   def go(xs):
       return (
           xs[0:n]
           if isinstance(xs, (list, tuple))
           else list(islice(xs, n))
       )
   return go


  1. unfoldr :: (b -> Maybe (a, b)) -> b -> [a]

def unfoldr(f): 

   Generic anamorphism.
      A lazy (generator) list unfolded from a seed value by
      repeated application of f until no residue remains.
      Dual to fold/reduce.
      f returns either None, or just (value, residue).
      For a strict output value, wrap in list().
   
   def go(x):
       valueResidue = f(x)
       while None is not valueResidue:
           yield valueResidue[0]
           valueResidue = f(valueResidue[1])
   return go


  1. ---------------------- FORMATTING ----------------------
  1. spacedTable :: String -> String

def spacedTable(rows): 

   A table with right-aligned columns.
   
   columnWidths = [
       max([len(x) for x in col])
       for col in zip(*rows)
   ]
   return '\n'.join(
       ' '.join(map(
           lambda x, w: x.rjust(w, ' '),
           row, columnWidths
       ))
       for row in rows
   )


  1. MAIN ---

if __name__ == '__main__': 

   main()

</lang>

Output:
Padovan n-step series:

2 ->  1 1 1 2 2 3 4  5  7  9 12 16  21  28  37
3 ->  1 1 1 2 3 4 6  9 13 19 28 41  60  88 129
4 ->  1 1 1 2 3 5 7 11 17 26 40 61  94 144 221
5 ->  1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
6 ->  1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
7 ->  1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
8 ->  1 1 1 2 3 5 8 13 21 34 54 87 140 225 362

Raku

<lang perl6>say 'Padovan N-step sequences; first 25 terms:';

for 2..8 -> \N { 

   my @n-step = 1, 1, 1, { state $n = 2; @n-step[ ($n - N .. $n++ - 1).grep: * >= 0 ].sum } … *;
   put "N = {N} |" ~ @n-step[^25]».fmt: "%5d";

}</lang>

Output:
Padovan N-step sequences; first 25 terms:
N = 2 |    1     1     1     2     2     3     4     5     7     9    12    16    21    28    37    49    65    86   114   151   200   265   351   465   616
N = 3 |    1     1     1     2     3     4     6     9    13    19    28    41    60    88   129   189   277   406   595   872  1278  1873  2745  4023  5896
N = 4 |    1     1     1     2     3     5     7    11    17    26    40    61    94   144   221   339   520   798  1224  1878  2881  4420  6781 10403 15960
N = 5 |    1     1     1     2     3     5     8    12    19    30    47    74   116   182   286   449   705  1107  1738  2729  4285  6728 10564 16587 26044
N = 6 |    1     1     1     2     3     5     8    13    20    32    51    81   129   205   326   518   824  1310  2083  3312  5266  8373 13313 21168 33657
N = 7 |    1     1     1     2     3     5     8    13    21    33    53    85   136   218   349   559   895  1433  2295  3675  5885  9424 15091 24166 38698
N = 8 |    1     1     1     2     3     5     8    13    21    34    54    87   140   225   362   582   936  1505  2420  3891  6257 10061 16178 26014 41830

REXX

Some additional code was added for this REXX version to minimize the width for any particular column. <lang rexx>/*REXX program computes and shows the Padovan sequences for M steps for N numbers. */ parse arg n m . /*obtain optional arguments from the CL*/ if n== | n=="," then n= 15 /*Not specified? Then use the default.*/ if m== | m=="," then m= 8 /* " " " " " " */ w.= 1 /*W.c: the maximum width of a column. */ 

       do #=2  for m-1
       @.= 0;    @.0= 1;    @.1= 1;    @.2= 1   /*initialize 3 terms of the Padovan seq*/
       $= @.0                                   /*initials the list with the zeroth #. */
              do k=2  for  n-1;      z= pd(k-1)
              w.k= max(w.k, length(z));  $= $ z /*find maximum width for a specific col*/
              end   /*k*/
       $.#= $                                   /*save each unaligned line for later.  */
       end          /*#*/

oW= 1 

       do col=1  for n;  oW= oW + w.col + 1     /*add up the width of each column.     */
       end   /*col*/
                   iW= length(m) + 2;       pad= left(, 20*(n<21))    /*maybe indent.*/

say pad center('M', iW, " ")"│"center('first ' n " Padovan sequence with step M", oW) say pad center(, iW, "─")"┼"center(, oW, "─") 

       do out=2  for m-1;   $=                  /*align columnar elements for outputs. */
            do j=1  for n;  $= $ right(word($.out, j),  w.j)      /*align the columns. */
            end     /*j*/
       say pad center(out,length(m)+2)'│'$      /*display a line of columnar elements. */
       end          /*out*/

say pad center(, length(m)+2, "─")"┴"center(, oW, "─") exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ pd: procedure expose @. #; parse arg x; if @.x\==0 then return @.x /*@.x defined?*/ 

                   do k=1  for  #;   _= x-1-k;    @.x= @.x + @._;    end;      return @.x</lang>
output   when using the default inputs:
                      M │ first  15  Padovan sequence with step  M
                     ───┼──────────────────────────────────────────
                      2 │ 1 1 1 2 2 3 4  5  7  9 12 16  21  28  37
                      3 │ 1 1 1 2 3 4 6  9 13 19 28 41  60  88 129
                      4 │ 1 1 1 2 3 5 7 11 17 26 40 61  94 144 221
                      5 │ 1 1 1 2 3 5 8 12 19 30 47 74 116 182 286
                      6 │ 1 1 1 2 3 5 8 13 20 32 51 81 129 205 326
                      7 │ 1 1 1 2 3 5 8 13 21 33 53 85 136 218 349
                      8 │ 1 1 1 2 3 5 8 13 21 34 54 87 140 225 362
                     ───┴──────────────────────────────────────────

Rust

<lang rust> fn padovan(n: u64, x: u64) -> u64 { 

   if n < 2 {
       return 0;
   }

   match n {
       2 if x <= n + 1 => 1,
       2 => padovan(n, x - 2) + padovan(n, x - 3),
       _ if x <= n + 1 => padovan(n - 1, x),
       _ => ((x - n - 1)..(x - 1)).fold(0, |acc, value| acc + padovan(n, value)),
   }

} fn main() { 

   (2..=8).for_each(|n| {
       print!("\nN={}: ", n);
       (1..=15).for_each(|x| print!("{},", padovan(n, x)))
   });

}

</lang>

Output:
N=2: 1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,      
N=3: 1,1,1,2,3,4,6,9,13,19,28,41,60,88,129,   
N=4: 1,1,1,2,3,5,7,11,17,26,40,61,94,144,221, 
N=5: 1,1,1,2,3,5,8,12,19,30,47,74,116,182,286,
N=6: 1,1,1,2,3,5,8,13,20,32,51,81,129,205,326,
N=7: 1,1,1,2,3,5,8,13,21,33,53,85,136,218,349,
N=8: 1,1,1,2,3,5,8,13,21,34,54,87,140,225,362,

Sidef

Translation of: Perl

<lang ruby>func padovan(N) { 

   Enumerator({|callback|
       var n = 2
       var pn = [1, 1, 1]
       loop {
           pn << sum(pn[n-N .. (n++-1) -> grep { _ >= 0 }])
           callback(pn[-4])
       }
   })

}

for n in (2..8) { 

   say "n = #{n} | #{padovan(n).first(25).join(' ')}"

}</lang>

Output:
n = 2 | 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 200 265 351 465 616
n = 3 | 1 1 1 2 3 4 6 9 13 19 28 41 60 88 129 189 277 406 595 872 1278 1873 2745 4023 5896
n = 4 | 1 1 1 2 3 5 7 11 17 26 40 61 94 144 221 339 520 798 1224 1878 2881 4420 6781 10403 15960
n = 5 | 1 1 1 2 3 5 8 12 19 30 47 74 116 182 286 449 705 1107 1738 2729 4285 6728 10564 16587 26044
n = 6 | 1 1 1 2 3 5 8 13 20 32 51 81 129 205 326 518 824 1310 2083 3312 5266 8373 13313 21168 33657
n = 7 | 1 1 1 2 3 5 8 13 21 33 53 85 136 218 349 559 895 1433 2295 3675 5885 9424 15091 24166 38698
n = 8 | 1 1 1 2 3 5 8 13 21 34 54 87 140 225 362 582 936 1505 2420 3891 6257 10061 16178 26014 41830

Wren

Library: Wren-fmt

<lang ecmascript>import "/fmt" for Fmt

var padovanN // recursive padovanN = Fn.new { |n, t| 

   if (n < 2 || t < 3) return [1] * t
   var p = padovanN.call(n-1, t)
   if (n + 1 >= t) return p
   for (i in n+1...t) {
       p[i] = 0
       for (j in i-2..i-n-1) p[i] = p[i] + p[j]
   }
   return p

}

var t = 15 System.print("First %(t) terms of the Padovan n-step number sequences:") for (n in 2..8) Fmt.print("$d: $3d" , n, padovanN.call(n, t))</lang>

Output:
First 15 terms of the Padovan n-step number sequences:
2:   1   1   1   2   2   3   4   5   7   9  12  16  21  28  37
3:   1   1   1   2   3   4   6   9  13  19  28  41  60  88 129
4:   1   1   1   2   3   5   7  11  17  26  40  61  94 144 221
5:   1   1   1   2   3   5   8  12  19  30  47  74 116 182 286
6:   1   1   1   2   3   5   8  13  20  32  51  81 129 205 326
7:   1   1   1   2   3   5   8  13  21  33  53  85 136 218 349
8:   1   1   1   2   3   5   8  13  21  34  54  87 140 225 362
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