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 Padowan 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).

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

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 

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]

Julia

Translation of: Python

<lang julia> """

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

""" function nstep_Padowan(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_Padowan_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_Padowan_seq(nstep_Padowan()) </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, ...

Phix

Translation of: Go

<lang Phix>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</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

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 Padowan 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.

<lang python>Padovan N-step series

from itertools import islice, repeat


  1. padovans :: Int -> [Int]

def padovans(n):

   Non-finite series of N-step Padovan numbers,
      defined by a recurrence relation.
   
   def recurrence(ns):
       return ns[0], ns[1:] + [sum(take(n)(ns))]
   return unfoldr(recurrence)(
       take(1 + n)(
           repeat(1) if 3 > n else padovans(n - 1)
       )
   )


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

def main():

   First 15 terms each n-step Padovan(n) series
      where n is drawn from [2..8]
   
   def sample(n):
       return take(15)(padovans(n))
   def columnWidth(n):
       def go(xs):
           return .join([
               str(x).rjust(n, ' ') for x in xs
           ])
       return go
   print(
       fTable('Padovan n-step series:')(repr)(
           columnWidth(4)
       )(sample)(range(2, 1 + 8))
   )


  1. ----------------------- GENERIC ------------------------
  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 list().
   
   def go(x):
       valueResidue = f(x)
       while None is not valueResidue:
           yield valueResidue[0]
           valueResidue = f(valueResidue[1])
   return go


  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. ---------------------- FORMATTING ----------------------
  1. fTable :: String -> (a -> String) ->
  2. (b -> String) -> (a -> b) -> [a] -> String

def fTable(s):

   Heading -> x display function ->
      fx display function -> f -> xs -> tabular string.
   
   def gox(xShow):
       def gofx(fxShow):
           def gof(f):
               def goxs(xs):
                   ys = [xShow(x) for x in xs]
                   w = max(map(len, ys))
                   def arrowed(x, y):
                       return y.rjust(w, ' ') + ' ->' + (
                           fxShow(f(x))
                       )
                   return s + '\n' + '\n'.join(
                       map(arrowed, xs, ys)
                   )
               return goxs
           return gof
       return gofx
   return gox


  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

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

say center('M', iw, " ")"│"center('first ' n " Padovan sequence with step M", ow) say center(, iw, "─")"┼"center( , ow, "─")

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

say 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
───┴────────────────────────────────────────── 

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