Padovan n-step number sequences

From Rosetta Code
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[edit]

Translation of: Nim
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(‘ ’))
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[edit]

Translation of: ALGOL W
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
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[edit]

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.
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[edit]

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
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[edit]

Translation of: Wren
#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;
}
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#[edit]

// 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 "")
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[edit]

Works with: Factor version 0.99 2021-02-05
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.
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[edit]

Translation of: Wren
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))
    }
}
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[edit]

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 <>)
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[edit]

(() => {
    "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();
})();
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[edit]

Translation of: Python
"""
    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("""
:::: {| style="text-align: left;" border="4" cellpadding="2" cellspacing="2"
|+ Padovan <math>n</math>-step sequences
|- style="background-color: rgb(255, 204, 255);"
! <math>n</math> !! Values
|-
          """))
    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())
Output:
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[edit]

ClearAll[Padovan]
Padovan[2,tmax_]:=Module[{start,a,m},
 start={1,1,1};
 start=MapIndexed[a[#2[[1]]]==#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[#2[[1]]]==#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]
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[edit]

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(" ")
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[edit]

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"
}
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[edit]

Translation of: Go
with javascript_semantics
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[edit]

Python: Procedural[edit]

Generates a wikitable formatted output

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('''
:::: {| style="text-align: left;" border="4" cellpadding="2" cellspacing="2"
|+ Padovan <math>n</math>-step sequences
|- style="background-color: rgb(255, 204, 255);"
! <math>n</math> !! Values
|-
          '''.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)
Output:
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[edit]

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.

'''Padovan n-step number sequences'''

from itertools import chain, islice, repeat


# 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)
            )
        )
    )


# 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


# ------------------------- TEST -------------------------
# 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
        )))
    )


# ----------------------- GENERIC ------------------------

# take :: Int -> [a] -> [a]
# 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


# 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


# ---------------------- FORMATTING ----------------------

# 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
    )


# MAIN ---
if __name__ == '__main__':
    main()
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[edit]

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";
}
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[edit]

Some additional code was added for this REXX version to minimize the width for any particular column.

/*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
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[edit]

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)))
    });
}
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[edit]

Translation of: Perl
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(' ')}"
}
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[edit]

Library: Wren-fmt
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))
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