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Leonardo numbers

Leonardo numbers
You are encouraged to solve this task according to the task description, using any language you may know.

Leonardo numbers   are also known as the   Leonardo series.

The   Leonardo numbers   are a sequence of numbers defined by:

```       L(0) = 1                                          [1st equation]
L(1) = 1                                          [2nd equation]
L(n) = L(n-1)  +    L(n-2)   +  1                 [3rd equation]
─── also ───
L(n) =      2  *  Fib(n+1)   -  1                 [4th equation]
```
where the   + 1   will herein be known as the   add   number.
where the   FIB   is the   Fibonacci numbers.

This task will be using the 3rd equation (above) to calculate the Leonardo numbers.

Edsger W. Dijkstra   used   Leonardo numbers   as an integral part of his   smoothsort   algorithm.

The first few Leonardo numbers are:

```    1   1   3   5   9   15   25   41   67   109   177   287   465   753   1219   1973   3193   5167   8361  ···
```

•   show the 1st   25   Leonardo numbers, starting at L(0).
•   allow the first two Leonardo numbers to be specified   [for L(0) and L(1)].
•   allow the   add   number to be specified   (1 is the default).
•   show the 1st   25   Leonardo numbers, specifying 0 and 1 for L(0) and L(1), and 0 for the add number.

(The last task requirement will produce the Fibonacci numbers.)

11l

Translation of: C++
`F leo_numbers(cnt, =l0 = 1, =l1 = 1, add = 1)   L 1..cnt      print(l0, end' ‘ ’)      (l0, l1) = (l1, l0 + l1 + add)   print() print(‘Leonardo Numbers: ’, end' ‘’)leo_numbers(25)print(‘Fibonacci Numbers: ’, end' ‘’)leo_numbers(25, 0, 1, 0)`
Output:
```Leonardo Numbers: 1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Fibonacci Numbers: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

`with Ada.Text_IO; use Ada.Text_IO; procedure Leonardo is    function Leo     (N      : Natural;      Step   : Natural := 1;      First  : Natural := 1;      Second : Natural := 1) return Natural   is       L : array (0..1) of Natural := (First, Second);	begin		for i in 1 .. N loop			L := (L(1), L(0)+L(1)+Step);		end loop;		return L (0);	end Leo; begin   Put_Line ("First 25 Leonardo numbers:");   for I in 0 .. 24 loop      Put (Integer'Image (Leo (I)));   end loop;   New_Line;   Put_Line ("First 25 Leonardo numbers with L(0) = 0, L(1) = 1, " &             "step = 0 (fibonacci numbers):");   for I in 0 .. 24 loop      Put (Integer'Image (Leo (I, 0, 0, 1)));   end loop;   New_Line;end Leonardo;`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers with L(0) = 0, L(1) = 1, step = 0 (fibonacci numbers):
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

ALGOL 68

`BEGIN    # leonardo number parameters #    MODE LEONARDO = STRUCT( INT l0, l1, add number );    # default leonardo number parameters #    LEONARDO leonardo numbers = LEONARDO( 1, 1, 1 );    # operators to allow us to specify non-default parameters #    PRIO WITHLZERO = 9, WITHLONE = 9, WITHADDNUMBER = 9;    OP   WITHLZERO     = ( LEONARDO parameters, INT l0         )LEONARDO:         LEONARDO( l0, l1 OF parameters, add number OF parameters );    OP   WITHLONE      = ( LEONARDO parameters, INT l1         )LEONARDO:         LEONARDO( l0 OF parameters, l1, add number OF parameters );    OP   WITHADDNUMBER = ( LEONARDO parameters, INT add number )LEONARDO:         LEONARDO( l0 OF parameters, l1 OF parameters, add number );    # show the first n Leonardo numbers with the specified parameters #    PROC show = ( INT n, LEONARDO parameters )VOID:         IF n > 0 THEN            INT l0         = l0         OF parameters;            INT l1         = l1         OF parameters;            INT add number = add number OF parameters;            print( ( whole( l0, 0 ), " " ) );            IF n > 1 THEN                print( ( whole( l1, 0 ), " " ) );                INT lp := l0;                INT ln := l1;                FROM 2 TO n - 1 DO                    INT next = ln + lp + add number;                    lp := ln;                    ln := next;                    print( ( whole( ln, 0 ), " " ) )                OD            FI         FI # show # ;     # first series #    print( ( "First 25 Leonardo numbers", newline ) );    show( 25, leonardo numbers );    print( ( newline ) );    # second series #    print( ( "First 25 Leonardo numbers from 0, 1 with add number = 0", newline ) );    show( 25, leonardo numbers WITHLZERO 0 WITHADDNUMBER 0 );    print( ( newline ) )END`
Output:
```First 25 Leonardo numbers
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers from 0, 1 with add number = 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

AppleScript

Translation of: Python
(Generator version)

Drawing N items from a non-finite generator:

`------------------------ GENERATOR ------------------------- -- leo :: Int -> Int -> Int -> Generator [Int]on leo(L0, L1, delta)    script        property x : L0        property y : L1        on |λ|()            set n to x            set {x, y} to {y, x + y + delta}            return n        end |λ|    end scriptend leo  --------------------------- TEST ---------------------------on run    set leonardo to leo(1, 1, 1)    set fibonacci to leo(0, 1, 0)     unlines({"First 25 Leonardo numbers:", ¬        twoLines(take(25, leonardo)), "", ¬        "First 25 Fibonacci numbers:", ¬        twoLines(take(25, fibonacci))})end run  ------------------------ FORMATTING ------------------------ -- twoLines :: [Int] -> Stringon twoLines(xs)    script row        on |λ|(ns)            tab & intercalate(", ", ns)        end |λ|    end script    return unlines(map(row, chunksOf(16, xs)))end twoLines  ------------------------- GENERIC -------------------------- -- chunksOf :: Int -> [a] -> [[a]]on chunksOf(n, xs)    set lng to length of xs    script go        on |λ|(a, i)            set x to (i + n) - 1            if x ≥ lng then                a & {items i thru -1 of xs}            else                a & {items i thru x of xs}            end if        end |λ|    end script    foldl(go, {}, enumFromThenTo(1, n, lng))end chunksOf  -- enumFromThenTo :: Int -> Int -> Int -> [Int]on enumFromThenTo(x1, x2, y)    set xs to {}    set d to max(1, (x2 - x1))    repeat with i from x1 to y by d        set end of xs to i    end repeat    return xsend enumFromThenTo  -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl  -- intercalate :: String -> [String] -> Stringon intercalate(sep, xs)    set {dlm, my text item delimiters} to ¬        {my text item delimiters, sep}    set s to xs as text    set my text item delimiters to dlm    return send intercalate  -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b)on mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend mReturn  -- map :: (a -> b) -> [a] -> [b]on map(f, 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 tellend map  -- max :: Ord a => a -> a -> aon max(x, y)    if x > y then        x    else        y    end ifend max  -- take :: Int -> [a] -> [a]-- take :: Int -> String -> Stringon take(n, xs)    set c to class of xs    if list is c then        if 0 < n then            items 1 thru min(n, length of xs) 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 xs's |λ|()            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 ifend take  -- unlines :: [String] -> Stringon unlines(xs)    set {dlm, my text item delimiters} to ¬        {my text item delimiters, linefeed}    set str to xs as text    set my text item delimiters to dlm    strend unlines`
Output:
```First 25 Leonardo numbers:
1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973
1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049

First 25 Fibonacci numbers:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368```

AutoHotkey

`Leonardo(n, L0:=1, L1:=1, step:=1){	if n=0		return L0	if n=1		return L1	return Leonardo(n-1, L0, L1, step) + Leonardo(n-2, L0, L1, step) + step}`
Examples:
`output := "1st 25 Leonardo numbers, starting at L(0).`n"loop, 25	output .= Leonardo(A_Index-1) " "output .= "`n`n1st 25 Leonardo numbers, specifying 0 and 1 for L(0) and L(1), and 0 for the add number:`n"loop, 25	output .= Leonardo(A_Index-1, 0, 1, 0) " "MsgBox % outputreturn`
Output:
```1st 25 Leonardo numbers, starting at L(0).
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

1st 25 Leonardo numbers, specifying 0 and 1 for L(0) and L(1), and 0 for the add number:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368```

AWK

` # syntax: GAWK -f LEONARDO_NUMBERS.AWKBEGIN {    leonardo(1,1,1,"Leonardo")    leonardo(0,1,0,"Fibonacci")    exit(0)}function leonardo(L0,L1,step,text,  i,tmp) {    printf("%s numbers (%d,%d,%d):\n",text,L0,L1,step)    for (i=1; i<=25; i++) {      if (i == 1) {        printf("%d ",L0)      }      else if (i == 2) {        printf("%d ",L1)      }      else {        printf("%d ",L0+L1+step)        tmp = L0        L0 = L1        L1 = tmp + L1 + step      }    }    printf("\n")} `
Output:
```Leonardo numbers (1,1,1):
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Fibonacci numbers (0,1,0):
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Bash

` #!/bin/bash function leonardo_number () {    L0_value=\${2:-1}    L1_value=\${3:-1}    Add=\${4:-1}    leonardo_numbers=(\$L0_value \$L1_value)    for (( i = 2; i < \$1; ++i))    do       leonardo_numbers+=( \$((leonardo_numbers[i-1] + leonardo_numbers[i-2] + Add)) )    done    echo "\${leonardo_numbers[*]}"} `
Output:
```leonardo_number 25
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

leonardo_number 25 0 1 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

BASIC

BASIC256

` subroutine leonardo(L0, L1, suma, texto)	print "Numeros de " + texto + " (" + L0 + "," + L1 + "," + suma + "):"	for i = 1 to 25		if i = 1 then			print L0 + " ";		else			if i = 2 then				print L1 + " ";			else				print L0 + L1 + suma + " ";				tmp = L0				L0 = L1				L1 = tmp + L1 + suma			end if		end if	next i	print chr(10)end subroutine #--- Programa Principal ---call leonardo(1,1,1,"Leonardo")call leonardo(0,1,0,"Fibonacci")end `
Output:
```Numeros de Leonardo (1,1,1):
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

Numeros de Fibonacci (0,1,0):
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

IS-BASIC

`100 PROGRAM "Leonardo.bas"110 INPUT PROMPT "Enter values of L0, L1, and ADD, separated by comas: ":L0,L1,ADD120 PRINT L0;L1;130 FOR I=3 TO 25140   LET T=L1:LET L1=L1+L0+ADD:LET L0=T160   PRINT L1;170 NEXT180 PRINT`

Sinclair ZX81 BASIC

Runs on the 1k RAM model with room to spare; hence the long(ish) variable names. The parameters are read from the keyboard.

` 10 INPUT L0 20 INPUT L1 30 INPUT ADD 40 PRINT L0;" ";L1; 50 FOR I=3 TO 25 60 LET TEMP=L1 70 LET L1=L0+L1+ADD 80 LET L0=TEMP 90 PRINT " ";L1;100 NEXT I`
Input:
```1
1
1```
Output:
`1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049`
Input:
```0
1
0```
Output:
` 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368`

BBC BASIC

It's a shame when fonts don't make much of a distinction between l lower-case L and 1 the number One.

`REM >leonardo:PRINT "Enter values of L0, L1, and ADD, separated by commas:"INPUT l0%, l1%, add%PRINT l0% ' l1%FOR i% = 3 TO 25  temp% = l1%  l1% += l0% + add%  l0% = temp%  PRINT l1%NEXTPRINTEND`
Output:
```Enter values of L0, L1, and ADD, separated by commas:
?1, 1, 1
1
1
3
5
9
15
25
41
67
109
177
287
465
753
1219
1973
3193
5167
8361
13529
21891
35421
57313
92735
150049```
```Enter values of L0, L1, and ADD, separated by commas:
?0, 1, 0
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368```

Burlesque

`blsq ) 1 1 1{.+\/.+}\/+]23!CCLm]wdsh1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049 blsq ) 0 1 0{.+\/.+}\/+]23!CCLm]wdsh0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368`

C

This implementation fulfills the task requirements which state that the first 2 terms and the step increment should be specified. Many other implementations on this page only print out the first 25 numbers.

` #include<stdio.h> void leonardo(int a,int b,int step,int num){ 	int i,temp; 	printf("First 25 Leonardo numbers : \n"); 	for(i=1;i<=num;i++){		if(i==1)			printf(" %d",a);		else if(i==2)			printf(" %d",b);		else{			printf(" %d",a+b+step);			temp = a;			a = b;			b = temp+b+step;		}	}} int main(){	int a,b,step; 	printf("Enter first two Leonardo numbers and increment step : "); 	scanf("%d%d%d",&a,&b,&step); 	leonardo(a,b,step,25); 	return 0;} `

Output : Normal Leonardo Series :

```Enter first two Leonardo numbers and increment step : 1 1 1
First 25 Leonardo numbers :
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
```

Fibonacci Series :

```Enter first two Leonardo numbers and increment step : 0 1 0
First 25 Leonardo numbers :
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

C#

Works with: C sharp version 7
`using System;using System.Linq; public class Program{    public static void Main() {        Console.WriteLine(string.Join(" ", Leonardo().Take(25)));        Console.WriteLine(string.Join(" ", Leonardo(L0: 0, L1: 1, add: 0).Take(25)));    }     public static IEnumerable<int> Leonardo(int L0 = 1, int L1 = 1, int add = 1) {        while (true) {            yield return L0;            (L0, L1) = (L1, L0 + L1 + add);        }    }}`
Output:
```1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

C++

` #include <iostream> void leoN( int cnt, int l0 = 1, int l1 = 1, int add = 1 ) {    int t;    for( int i = 0; i < cnt; i++ ) {        std::cout << l0 << " ";        t = l0 + l1 + add; l0 = l1; l1 = t;    }}int main( int argc, char* argv[] ) {    std::cout << "Leonardo Numbers: "; leoN( 25 );    std::cout << "\n\nFibonacci Numbers: "; leoN( 25, 0, 1, 0 );    return 0;} `
Output:
```
Leonardo Numbers: 1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Fibonacci Numbers: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368

```

Common Lisp

` ;;;;;; leo - calculates the first n number from a leo sequence.;;; The first argument n is the number of values to return. The next three arguments a, b, add are optional.;;; Default values provide the "original" leonardo numbers as defined in the task.;;; a and b are the first and second element of the leonardo sequence.;;; add is the "add number" as defined in the task definition.;;; (defun leo (n &optional (a 1) (b 1) (add 1))  (labels ((iterate (n foo)             (if (zerop n) (reverse foo)                           (iterate (- n 1)                                    (cons (+ (first foo) (second foo) add) foo)))))     (cond ((= n 1) (list a))           (T       (iterate (- n 2) (list b a)))))) `
Output:
```> (leo 25)
(1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049)
> (leo 25 0 1 0)
(0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368)
```

Crystal

Translation of: Python
`def leonardo(l_zero, l_one, add, amount)    terms = [l_zero, l_one]    while terms.size < amount        new = terms[-1] + terms[-2]        new += add        terms << new    end    termsend puts "First 25 Leonardo numbers: \n#{ leonardo(1,1,1,25) }"puts "Leonardo numbers with fibonacci parameters:\n#{ leonardo(0,1,0,25) }" `
Output:
```First 25 Leonardo numbers:
[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]
Leonardo numbers with fibonacci parameters:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]
```

D

Translation of: C++
` import std.stdio; void main() {    write("Leonardo Numbers: ");    leonardoNumbers( 25 );     write("Fibonacci Numbers: ");    leonardoNumbers( 25, 0, 1, 0 );} void leonardoNumbers(int count, int l0=1, int l1=1, int add=1) {    int t;    for (int i=0; i<count; ++i) {        write(l0, " ");        t = l0 + l1 + add;        l0 = l1;        l1 = t;    }    writeln();} `
Output:
```
Leonardo Numbers: 1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Fibonacci Numbers: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368

```

F#

`open System let leo l0 l1 d =    Seq.unfold (fun (x, y) -> Some (x, (y, x + y + d))) (l0, l1) let leonardo = leo 1 1 1let fibonacci = leo 0 1 0 [<EntryPoint>]let main _ =     let leoNums = Seq.take 25 leonardo |> Seq.chunkBySize 16    printfn "First 25 of the (1, 1, 1) Leonardo numbers:\n%A" leoNums    Console.WriteLine()     let fibNums = Seq.take 25 fibonacci |> Seq.chunkBySize 16    printfn "First 25 of the (0, 1, 0) Leonardo numbers (= Fibonacci number):\n%A" fibNums     0 // return an integer exit code`
Output:
```First 25 of the (1, 1, 1) Leonardo numbers:
seq
[[|1; 1; 3; 5; 9; 15; 25; 41; 67; 109; 177; 287; 465; 753; 1219; 1973|];
[|3193; 5167; 8361; 13529; 21891; 35421; 57313; 92735; 150049|]]

First 25 of the (0, 1, 0) Leonardo numbers (= Fibonacci number):
seq
[[|0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; 233; 377; 610|];
[|987; 1597; 2584; 4181; 6765; 10946; 17711; 28657; 46368|]]```

Factor

`USING: fry io kernel math prettyprint sequences ;IN: rosetta-code.leonardo-numbers : first25-leonardo ( vector add -- seq )    23 swap '[ dup 2 tail* sum _ + over push ] times ; : print-leo ( seq -- ) [ pprint bl ] each nl ; "First 25 Leonardo numbers:" printV{ 1 1 } 1 first25-leonardo print-leo "First 25 Leonardo numbers with L(0)=0, L(1)=1, add=0:" printV{ 0 1 } 0 first25-leonardo print-leo`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers with L(0)=0, L(1)=1, add=0:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Fortran

Happily, no monster values result for the trial run, so ordinary 32-bit integers suffice. The source style uses the F90 facilities only to name the subroutine being ended (i.e. `END SUBROUTINE LEONARDO` rather than just `END`) and the I0 format code that shows an integer without a fixed space allowance, convenient in produced well-formed messages. The "\$" format code signifies that the end of output from its WRITE statement should not trigger the starting of a new line for the next WRITE statement, convenient when rolling a sequence of values to a line of output one-by-one as they are concocted. Otherwise, the values would have to be accumulated in a suitable array and then written in one go.

Many versions of Fortran have enabled parameters to be optionally supplied and F90 has standardised a protocol, also introducing a declaration syntax that can specify multiple attributes in one statement which in this case would be `INTEGER, OPTIONAL:: AF` rather than two statements concerning AF. However, in a test run with `CALL LEONARDO(25,1,1)` the Compaq F90/95 compiler rejected this attempt because there was another invocation with four parameters, not three, in the same program unit. By adding the rigmarole for declaring a MODULE containing the subroutine LEONARDO, its worries would be assuaged. Many compilers (and linkers, for separately-compiled routines) would check neither the number nor the type of parameters so no such complaint would be made - but when run, the code might produce wrong results or crash.

The method relies on producing a sequence of values, rather than calculating L(n) from the start each time a value from the sequence is required.
`      SUBROUTINE LEONARDO(LAST,L0,L1,AF)	!Show the first LAST values of the sequence.       INTEGER LAST	!Limit to show.       INTEGER L0,L1	!Starting values.       INTEGER AF	!The "Add factor" to deviate from Fibonacci numbers.       OPTIONAL AF	!Indicate that this parameter may be omitted.       INTEGER EMBOLISM	!The bloat to employ.       INTEGER N,LN,LNL1,LNL2	!Assistants to the calculation.        IF (PRESENT(AF)) THEN	!Perhaps the last parameter has not been given.          EMBOLISM = AF			!It has. Take its value.         ELSE			!But if not,          EMBOLISM = 1			!This is the specified default.        END IF			!Perhaps there should be some report on this?        WRITE (6,1) LAST,L0,L1,EMBOLISM	!Announce.    1   FORMAT ("The first ",I0,	!The I0 format code avoids excessive spacing.     1   " numbers in the Leonardo sequence defined by L(0) = ",I0,     2   " and L(1) = ",I0," with L(n) = L(n - 1) + L(n - 2) + ",I0)        IF (LAST .GE. 1) WRITE (6,2) L0	!In principle, LAST may be small.        IF (LAST .GE. 2) WRITE (6,2) L1	!!So, suspicion rules.    2   FORMAT (I0,", ",\$)	!Obviously, the \$ sez "don't finish the line".        LNL1 = L0	!Syncopation for the sequence's initial values.        LN = L1		!Since the parameters ought not be damaged.        DO N = 3,LAST	!Step away.          LNL2 = LNL1		!Advance the two state variables one step.          LNL1 = LN		!Ready to make a step forward.          LN = LNL1 + LNL2 + EMBOLISM	!Thus.          WRITE (6,2) LN	!Reveal the value. Overflow is distant...        END DO		!On to the next step.        WRITE (6,*)	!Finish the line.      END SUBROUTINE LEONARDO	!Only speedy for the sequential production of values.       PROGRAM POKE       CALL LEONARDO(25,1,1,1)	!The first 25 Leonardo numbers.      CALL LEONARDO(25,0,1,0)	!Deviates to give the Fibonacci sequence.      END `

Output:

```The first 25 numbers in the Leonardo sequence defined by L(0) = 1 and L(1) = 1 with L(n) = L(n - 1) + L(n - 2) + 1
1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049,
The first 25 numbers in the Leonardo sequence defined by L(0) = 0 and L(1) = 1 with L(n) = L(n - 1) + L(n - 2) + 0
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368,
```

FreeBASIC

` Sub leonardo(L0 As Integer, L1 As Integer, suma As Integer, texto As String)    Dim As Integer i, tmp    Print "Numeros de " &texto &" (" &L0 &"," &L1 &"," &suma &"):"    For i = 1 To 25        If i = 1 Then            Print L0;        Elseif i = 2 Then            Print L1;        Else            Print L0 + L1 + suma;            tmp = L0            L0 = L1            L1 = tmp + L1 + suma        End If    Next i    Print Chr(10)End Sub '--- Programa Principal ---leonardo(1,1,1,"Leonardo")leonardo(0,1,0,"Fibonacci")End `
Output:
```Numeros de Leonardo (1,1,1):
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

Numeros de Fibonacci (0,1,0):
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Go

`package main import "fmt" func leonardo(n, l0, l1, add int) []int {    leo := make([]int, n)    leo[0] = l0    leo[1] = l1    for i := 2; i < n; i++ {        leo[i] = leo[i - 1] + leo[i - 2] + add    }    return leo} func main() {    fmt.Println("The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:")    fmt.Println(leonardo(25, 1, 1, 1))    fmt.Println("\nThe first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:")    fmt.Println(leonardo(25, 0, 1, 0))}`
Output:
```The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:
[1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049]

The first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:
[0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368]
```

`import Data.List (intercalate, unfoldr)import Data.List.Split (chunksOf) --------------------- LEONARDO NUMBERS ----------------------- L0 -> L1 -> Add number -> Series (infinite)leo :: Integer -> Integer -> Integer -> [Integer]leo l0 l1 d = unfoldr (\(x, y) -> Just (x, (y, x + y + d))) (l0, l1) leonardo :: [Integer]leonardo = leo 1 1 1 fibonacci :: [Integer]fibonacci = leo 0 1 0 --------------------------- TEST ---------------------------main :: IO ()main =  (putStrLn . unlines)    [ "First 25 default (1, 1, 1) Leonardo numbers:\n"    , f \$ take 25 leonardo    , "First 25 of the (0, 1, 0) Leonardo numbers (= Fibonacci numbers):\n"    , f \$ take 25 fibonacci    ]  where    f = unlines . fmap (('\t' :) . intercalate ",") . chunksOf 16 . fmap show`
Output:
```First 25 default (1, 1, 1) Leonardo numbers:

1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973
3193,5167,8361,13529,21891,35421,57313,92735,150049

First 25 of the (0, 1, 0) Leonardo numbers (= Fibonacci numbers):

0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610
987,1597,2584,4181,6765,10946,17711,28657,46368```

Alternately, defining the list self-referentially instead of using unfoldr:

`leo :: Integer -> Integer -> Integer -> [Integer]leo l0 l1 d = s where  s = l0 : l1 : zipWith (\x y -> x + y + d) s (tail s)`

J

` leo =:  (] , {[email protected][ + _2&{@] + {:@])^:(_2&[email protected]{:@[) `
Output:
``` 1 25 leo 1 1
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

0 25 leo 0 1
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Java

Translation of: Kotlin
`import java.util.Arrays;import java.util.List; @SuppressWarnings("SameParameterValue")public class LeonardoNumbers {    private static List<Integer> leonardo(int n) {        return leonardo(n, 1, 1, 1);    }     private static List<Integer> leonardo(int n, int l0, int l1, int add) {        Integer[] leo = new Integer[n];        leo[0] = l0;        leo[1] = l1;        for (int i = 2; i < n; i++) {            leo[i] = leo[i - 1] + leo[i - 2] + add;        }        return Arrays.asList(leo);    }     public static void main(String[] args) {        System.out.println("The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:");        System.out.println(leonardo(25));        System.out.println("\nThe first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:");        System.out.println(leonardo(25, 0, 1, 0));    }}`
Output:
```The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:
[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]

The first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]```

JavaScript

ES6

`const leoNum = (c, l0 = 1, l1 = 1, add = 1) =>    new Array(c).fill(add).reduce(        (p, c, i) => i > 1 ? (            p.push(p[i - 1] + p[i - 2] + c) && p        ) : p, [l0, l1]    ); console.log(leoNum(25));console.log(leoNum(25, 0, 1, 0));`
```[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]```

Or, taking N terms from a non-finite Javascript generator:

Translation of: Python
`(() => {    'use strict';     // leo :: Int -> Int -> Int -> Generator [Int]    function* leo(L0, L1, delta) {        let [x, y] = [L0, L1];        while (true) {            yield x;            [x, y] = [y, delta + x + y];        }    }     // ----------------------- TEST ------------------------    // main :: IO ()    const main = () => {        const            leonardo = leo(1, 1, 1),            fibonacci = leo(0, 1, 0);         return unlines([            'First 25 Leonardo numbers:',            indentWrapped(take(25)(leonardo)),            '',            'First 25 Fibonacci numbers:',            indentWrapped(take(25)(fibonacci))        ]);    };     // -------------------- FORMATTING ---------------------     // indentWrapped :: [Int] -> String    const indentWrapped = xs =>        unlines(            map(x => '\t' + x.join(','))(                chunksOf(16)(                    map(str)(xs)                )            )        );     // ----------------- GENERIC FUNCTIONS -----------------     // chunksOf :: Int -> [a] -> [[a]]    const chunksOf = n =>        xs => enumFromThenTo(0)(n)(            xs.length - 1        ).reduce(            (a, i) => a.concat([xs.slice(i, (n + i))]),            []        );     // enumFromThenTo :: Int -> Int -> Int -> [Int]    const enumFromThenTo = x1 =>        x2 => y => {            const d = x2 - x1;            return Array.from({                length: Math.floor(y - x2) / d + 2            }, (_, i) => x1 + (d * i));        };     // map :: (a -> b) -> [a] -> [b]    const map = f =>        // The list obtained by applying f        // to each element of xs.        // (The image of xs under f).        xs => [...xs].map(f);     // str :: a -> String    const str = x =>        x.toString();     // 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.apply([], Array.from({            length: n        }, () => {            const x = xs.next();            return x.done ? [] : [x.value];        }));     // unlines :: [String] -> String    const unlines = xs => xs.join('\n');     // MAIN ---    return main();})();`
Output:
```First 25 Leonardo numbers:
1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973
3193,5167,8361,13529,21891,35421,57313,92735,150049

First 25 Fibonacci numbers:
0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610
987,1597,2584,4181,6765,10946,17711,28657,46368```

jq

Naive Implementation

`def Leonardo(zero; one; incr):  def leo:    if . == 0 then zero    elif . == 1 then one    else ((.-1) |leo) + ((.-2) | leo) +  incr    end;  leo;`

Implementation with Caching

An array is used for caching, with `.[n]` storing the value L(n).

`def Leonardo(zero; one; incr):  def leo(n):    if .[n] then .    else leo(n-1)   # optimization of leo(n-2)|leo(n-1)      | .[n] = .[n-1] + .[n-2] +  incr    end;  . as \$n | [zero,one] | leo(\$n) | .[\$n];`

(To compute the sequence of Leonardo numbers L(1) ... L(n) without redundant computation, the last element of the pipeline in the last line of the function above should be dropped.)

Examples

`[range(0;25) | Leonardo(1;1;1)]`
Output:
`[1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973,3193,5167,8361,13529,21891,35421,57313,92735,150049]`
`[range(0;25) | Leonardo(0;1;0)]`
Output:
`[0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368]`

Julia

Works with: Julia version 0.6
`function L(n, add::Int=1, firsts::Vector=[1, 1])    l = max(maximum(n) .+ 1, length(firsts))    r = Vector{Int}(l)    r[1:length(firsts)] = firsts    for i in 3:l        r[i] = r[i - 1] + r[i - 2] + add    end    return r[n .+ 1]end # Task 1println("First 25 Leonardo numbers: ", join(L(0:24), ", ")) # Task 2@show L(0) L(1) # Task 4println("First 25 Leonardo numbers starting with [0, 1]: ", join(L(0:24, 0, [0, 1]), ", "))`
Output:
```First 25 Leonardo numbers: 1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049
L(0) = 1
L(1) = 1
First 25 Leonardo numbers starting with 0, 1: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368```

Kotlin

`// version 1.1.2 fun leonardo(n: Int, l0: Int = 1, l1: Int = 1, add: Int = 1): IntArray {    val leo = IntArray(n)    leo[0] = l0    leo[1] = l1    for (i in 2 until n) leo[i] = leo[i - 1] + leo[i - 2] + add    return leo} fun main(args: Array<String>) {    println("The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:")    println(leonardo(25).joinToString(" "))    println("\nThe first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:")    println(leonardo(25, 0, 1, 0).joinToString(" "))}`
Output:
```The first 25 Leonardo numbers with L[0] = 1, L[1] = 1 and add number = 1 are:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

The first 25 Leonardo numbers with L[0] = 0, L[1] = 1 and add number = 0 are:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Lua

`function leoNums (n, L0, L1, add)  local L0, L1, add = L0 or 1, L1 or 1, add or 1  local lNums, nextNum = {L0, L1}  while #lNums < n do    nextNum = lNums[#lNums] + lNums[#lNums - 1] + add    table.insert(lNums, nextNum)  end  return lNumsend function show (msg, t)  print(msg .. ":")  for i, x in ipairs(t) do    io.write(x .. " ")  end  print("\n")end show("Leonardo numbers", leoNums(25))show("Fibonacci numbers", leoNums(25, 0, 1, 0))`
Output:
```Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

Fibonacci numbers:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368```

Maple

`L := proc(n, L_0, L_1, add) if n = 0 then   return L_0; elif n = 1 then   return L_1; else   return L(n - 1) + L(n - 2) + add; end if; end proc: Leonardo := n -> (L(1, 1, 1),[seq(0..n - 1)]) Fibonacci := n -> (L(0, 1, 0), [seq(0..n - 1)])`
```[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]```

Mathematica

`L[0,L0_:1,___]:=L0L[1,L0_:1,L1_:1,___]:=L1L[n_,L0_:1,L1_:1,add_:1]:=L[n-1,L0,L1,add]+L[n-2,L0,L1,add]+add L/@(Range[25]-1)L[#,0,1,0]&/@(Range[25]-1)`
```{1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049}
{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368}```

min

Works with: min version 0.19.3
`(over over + rolldown pop pick +) :next(('print dip " " print! next) 25 times newline) :leo "First 25 Leonardo numbers:" puts!1 1 1 leo"First 25 Leonardo numbers with add=0, L(0)=0, L(1)=1:" puts!0 0 1 leo`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers with add=0, L(0)=0, L(1)=1:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Modula-2

`MODULE Leonardo;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE leonardo(a,b,step,num : INTEGER);VAR    buf : ARRAY[0..63] OF CHAR;    i,temp : INTEGER;BEGIN    FOR i:=1 TO num DO        IF i=1 THEN            FormatString(" %i", buf, a);            WriteString(buf)        ELSIF i=2 THEN            FormatString(" %i", buf, b);            WriteString(buf)        ELSE            FormatString(" %i", buf, a+b+step);            WriteString(buf);             temp := a;            a := b;            b := temp + b + step        END    END;    WriteLnEND leonardo; BEGIN    leonardo(1,1,1,25);    leonardo(0,1,0,25);     ReadCharEND Leonardo.`

Nim

`import strformat proc leonardoNumbers(count: int, L0: int = 1,                      L1: int = 1, ADD: int = 1) =  var t = 0  var (L0_loc, L1_loc) = (L0, L1)  for i in 0..<count:    write(stdout, fmt"{L0_loc:7}")    t = L0_loc + L1_loc + ADD    L0_loc = L1_loc    L1_loc = t    if i mod 5 == 4:      write(stdout, "\n")  write(stdout, "\n") echo "Leonardo Numbers:"leonardoNumbers(25)echo "Fibonacci Numbers: "leonardoNumbers(25, 0, 1, 0)`
Output:
```Leonardo Numbers:
1      1      3      5      9
15     25     41     67    109
177    287    465    753   1219
1973   3193   5167   8361  13529
21891  35421  57313  92735 150049

Fibonacci Numbers:
0      1      1      2      3
5      8     13     21     34
55     89    144    233    377
610    987   1597   2584   4181
6765  10946  17711  28657  46368
```

Perl

`no warnings 'experimental::signatures';use feature 'signatures'; sub leonardo (\$n, \$l0 = 1, \$l1 = 1, \$add = 1) {  (\$l0, \$l1) = (\$l1, \$l0+\$l1+\$add)  for 1..\$n;  \$l0;} my @L = map { leonardo(\$_) } 0..24;print "Leonardo[1,1,1]: @L\n";my @F = map { leonardo(\$_,0,1,0) } 0..24;print "Leonardo[0,1,0]: @F\n";`
Output:
```Leonardo[1,1,1]: 1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Leonardo[0,1,0]: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Phix

`function leonardo(integer n, l1=1, l2=1, step=1)--return the first n leonardo numbers, starting {l1,l2}, with step as the add numbersequence res = {l1,l2}    while length(res)<n do        res = append(res,res[\$]+res[\$-1]+step)    end while    return resend function?{"Leonardo",leonardo(25)}?{"Fibonacci",leonardo(25,0,1,0)}`
Output:
```{"Leonardo",{1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973,3193,5167,8361,13529,21891,35421,57313,92735,150049}}
{"Fibonacci",{0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368}}
```

PicoLisp

`(de leo (A B C)   (default A 1  B 1  C 1)   (make      (do 25         (inc            'B            (+ (link (swap 'A B)) C) ) ) ) ) (println 'Leonardo (leo))(println 'Fibonacci (leo 0 1 0))`
Output:
```Leonardo (1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049)
Fibonacci (0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368)```

Plain English

`To run:Start up.Write "First 25 Leonardo numbers:" on the console.Show 25 of the Leonardo numbers starting with 1 and 1 and using 1 for the add number.Write "First 25 Leonardo numbers with L(0)=0, L(1)=1, add=0:" on the console.Show 25 of the Leonardo numbers starting with 0 and 1 and using 0 for the add number.Wait for the escape key.Shut down. To show a number of the Leonardo numbers starting with a first number and a second number and using an add number for the add number:If the number is less than 2, exit.Privatize the number.Privatize the first number.Privatize the second number.Subtract 2 from the number.Write the first number then " " then the second number on the console without advancing.Loop.If a counter is past the number, write "" on the console; exit.Swap the first number with the second number.Put the first number plus the second number plus the add number into the second number.Write the second number then " " on the console without advancing.Repeat.`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers with L(0)=0, L(1)=1, add=0:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

PureBasic

` EnableExplicit #N = 25 Procedure leon_R(a.i, b.i, s.i = 1, n.i = #N)   If n>2    Print(Space(1) + Str(a + b + s))        ProcedureReturn leon_R(b, a + b + s, s, n-1)  EndIf   EndProcedure If OpenConsole()   Define r\$   Print("Enter first two Leonardo numbers and increment step (separated by space) : ")  r\$ = Input()  PrintN("First " + Str(#N) + " Leonardo numbers : ")    Print(StringField(r\$, 1, Chr(32)) + Space(1) +         StringField(r\$, 2, Chr(32)))   leon_R(Val(StringField(r\$, 1, Chr(32))),         Val(StringField(r\$, 2, Chr(32))),                    Val(StringField(r\$, 3, Chr(32))))   r\$ = Input()EndIf `
Output:
```Enter first two Leonardo numbers and increment step (separated by space) : 1 1 1
First 25 Leonardo numbers :
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Enter first two Leonardo numbers and increment step (separated by space) : 0 1 0
First 25 Leonardo numbers :
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Python

Finite iteration

`def Leonardo(L_Zero, L_One, Add, Amount):    terms = [L_Zero,L_One]    while len(terms) < Amount:        new = terms[-1] + terms[-2]        new += Add        terms.append(new)    return terms out = ""print "First 25 Leonardo numbers:"for term in Leonardo(1,1,1,25):    out += str(term) + " "print out out = ""print "Leonardo numbers with fibonacci parameters:"for term in Leonardo(0,1,0,25):    out += str(term) + " "print out `
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Leonardo numbers with fibonacci parameters:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Non-finite generation

Or, for a non-finite stream of Leonardos, we can use a Python generator:

Works with: Python version 3
`'''Leonardo numbers''' from functools import (reduce)from itertools import (islice)  # leo :: Int -> Int -> Int -> Generator [Int]def leo(L0, L1, delta):    '''A number series of the       Leonardo and Fibonacci pattern,       where L0 and L1 are the first two terms,       and delta = 1 for (L0, L1) == (1, 1)       yields the Leonardo series, while       delta = 0 defines the Fibonacci series.'''    (x, y) = (L0, L1)    while True:        yield x        (x, y) = (y, x + y + delta)  # main :: IO()def main():    '''Tests.'''     print('\n'.join([        'First 25 Leonardo numbers:',        folded(16)(take(25)(            leo(1, 1, 1)        )),        '',        'First 25 Fibonacci numbers:',        folded(16)(take(25)(            leo(0, 1, 0)        ))    ]))  # FORMATTING ---------------------------------------------- # folded :: Int -> [a] -> Stringdef folded(n):    '''Long list folded to rows of n terms each.'''    return lambda xs: '[' + ('\n '.join(        str(ns)[1:-1] for ns in chunksOf(n)(xs)    ) + ']')  # GENERIC ------------------------------------------------- # chunksOf :: Int -> [a] -> [[a]]def chunksOf(n):    '''A series of lists of length n,       subdividing the contents of xs.       Where the length of xs is not evenly divible,       the final list will be shorter than n.'''    return lambda xs: reduce(        lambda a, i: a + [xs[i:n + i]],        range(0, len(xs), n), []    ) if 0 < n else []  # take :: Int -> [a] -> [a]# take :: Int -> String -> Stringdef take(n):    '''The prefix of xs of length n,       or xs itself if n > length xs.'''    return lambda xs: (        xs[0:n]        if isinstance(xs, list)        else list(islice(xs, n))    )  # MAIN ---if __name__ == '__main__':    main()`
Output:
```First 25 Leonardo numbers:
[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973
3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]

First 25 Fibonacci numbers:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]```

R

` leonardo_numbers <- function(add = 1, l0 = 1, l1 = 1, how_many = 25) {	result <- c(l0, l1)	for (i in 3:how_many) 		result <- append(result, result[[i - 1]] + result[[i - 2]] + add)	result}cat("First 25 Leonardo numbers\n")cat(leonardo_numbers(), "\n") cat("First 25 Leonardo numbers from 0, 1 with add number = 0\n")cat(leonardo_numbers(0, 0, 1), "\n") `
Output:
```First 25 Leonardo numbers
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers from 0, 1 with add number = 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Racket

`#lang racket(define (Leonardo n #:L0 (L0 1) #:L1 (L1 1) #:1+ (1+ 1))  (cond [(= n 0) L0]        [(= n 1) L1]        [else         (let inr ((n (- n 2)) (L_n-2 L0) (L_n-1 L1))           (let ((L_n (+ L_n-1 L_n-2 1+)))             (if (zero? n) L_n (inr (sub1 n) L_n-1 L_n))))])) (module+ main  (map Leonardo (range 25))  (map (curry Leonardo #:L0 0 #:L1 1 #:1+ 0) (range 25))) (module+ test  (require rackunit)  (check-equal? (Leonardo 0) 1)  (check-equal? (Leonardo 1) 1)  (check-equal? (Leonardo 2) 3)  (check-equal? (Leonardo 3) 5))`
Output:
```'(1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049)
'(0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368)```

Raku

(formerly Perl 6)

`sub 𝑳 ( \$𝑳0 = 1, \$𝑳1 = 1, \$𝑳add = 1 ) { \$𝑳0, \$𝑳1, { \$^n2 + \$^n1 + \$𝑳add } ... * } # Part 1say "The first 25 Leonardo numbers:";put 𝑳()[^25]; # Part 2say "\nThe first 25 numbers using 𝑳0 of 0, 𝑳1 of 1, and adder of 0:";put 𝑳( 0, 1, 0 )[^25];`
Output:
```The first 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

The first 25 numbers using 𝑳0 of 0, 𝑳1 of 1, and adder of 0:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368```

REXX

`/*REXX pgm computes Leonardo numbers, allowing the specification of L(0), L(1), and ADD#*/numeric digits 500                               /*just in case the user gets ka-razy.  */@.=1                                             /*define the default for the  @. array.*/parse arg N L0 L1 a# .                           /*obtain optional arguments from the CL*/if  N =='' |  N ==","  then    N= 25             /*Not specified?  Then use the default.*/if L0\=='' & L0\==","  then  @.0= L0             /*Was     "         "   "   "   value. */if L1\=='' & L1\==","  then  @.1= L1             /* "      "         "   "   "     "    */if a#\=='' & a#\==","  then  @.a= a#             /* "      "         "   "   "     "    */say 'The first '   N   " Leonardo numbers are:"  /*display a title for the output series*/if @.0\==1 | @.1\==1  then say 'using '     @.0     " for L(0)"if @.0\==1 | @.1\==1  then say 'using '     @.1     " for L(1)"if @.a\==1            then say 'using '     @.a     " for the  add  number"say                                              /*display blank line before the output.*/\$=                                               /*initialize the output line to "null".*/             do j=0  for N                       /*construct a list of Leonardo numbers.*/             if j<2  then [email protected].j                  /*for the 1st two numbers, use the fiat*/                     else do                     /*··· otherwise, compute the Leonardo #*/                          [email protected].0                  /*save the old primary Leonardo number.*/                          @.[email protected].1                /*store the new primary number in old. */                          @.[email protected].0  +  _  +  @.a  /*compute the next Leonardo number.    */                          [email protected].1                  /*store the next Leonardo number in Z. */                          end                    /* [↑]  only 2 Leonardo #s are stored. */             \$=\$ z                               /*append the just computed # to \$ list.*/             end   /*j*/                         /* [↓]  elide the leading blank in  \$. */say strip(\$)                                     /*stick a fork in it,  we're all done. */`
output   when using the default input:
```The first  25  Leonardo numbers are:

1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
```
output   when using the input of:     12   0   1   0
```The first  25  Leonardo numbers are:
using  0  for L(0)
using  1  for L(1)
using  0  for the  add  number

0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Ring

` # Project : Leanardo numbers n0 = 1n1 = 1add = 1see "First 25 Leonardo numbers:" + nlleonardo()n0 = 1n1 = 1add = 0see "First 25 Leonardo numbers with L(0) = 0, L(1) = 1, step = 0 (fibonacci numbers):" + nlsee "" + add + " "leonardo() func leonardo()        see "" + n0 + " " + n1        for i=3 to 25              temp=n1              n1=n0+n1+add              n0=temp             see " "+ n1        next         see nl `

Output:

```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Leonardo numbers with L(0) = 0, L(1) = 1, step = 0 (fibonacci numbers):
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025
```

Ruby

Enumerators are nice for this.

`def leonardo(l0=1, l1=1, add=1)  return to_enum(__method__,l0,l1,add) unless block_given?  loop do      yield l0    l0, l1 = l1, l0+l1+add  endend p leonardo.take(25)p leonardo(0,1,0).take(25) `
Output:
```[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]
```

Run BASIC

`sqliteconnect #mem, ":memory:"#mem execute("CREATE TABLE lno (name,L0,L1,ad)")#mem execute("INSERT INTO lno VALUES('Leonardo',1,1,1),('Fibonacci',0,1,0);")#mem execute("SELECT * FROM lno")for j = 1 to 2#row  = #mem #nextrow()name\$ = #row name\$()L0    = #row L0()L1    = #row L1()ad    = #row ad()print :print name\$;" add=";ad :print" ";L0;" ";L1;" ";for i = 3 to 25  temp  = L1  L1    = L0 + L1 + ad  L0    = temp  print L1;" ";next inext jend`
```Leonardo add=1
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Rust

`fn leonardo(mut n0: u32, mut n1: u32, add: u32) -> impl std::iter::Iterator<Item = u32> {    std::iter::from_fn(move || {        let n = n0;        n0 = n1;        n1 += n + add;        Some(n)    })} fn main() {    println!("First 25 Leonardo numbers:");    for i in leonardo(1, 1, 1).take(25) {        print!("{} ", i);    }    println!();    println!("First 25 Fibonacci numbers:");    for i in leonardo(0, 1, 0).take(25) {        print!("{} ", i);    }    println!();}`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Fibonacci numbers:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Scala

`def leo( n:Int, n1:Int=1, n2:Int=1, addnum:Int=1 ) : BigInt = n match {  case 0 => n1  case 1 => n2  case n => leo(n - 1, n1, n2, addnum) + leo(n - 2, n1, n2, addnum) + addnum} {println( "The first 25 Leonardo Numbers:")(0 until 25) foreach { n => print( leo(n) + " " ) } println( "\n\nThe first 25 Fibonacci Numbers:")(0 until 25) foreach { n => print( leo(n, n1=0, n2=1, addnum=0) + " " ) }} `
Output:
```The first 25 Leonardo Numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

The first 25 Fibonacci Numbers:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Seed7

`\$ include "seed7_05.s7i"; const proc: leonardo (in var integer: l0, in var integer: l1, in integer: add, in integer: count) is func  local    var integer: temp is 0;  begin    for count do      write(" " <& l0);      temp := l0 + l1 + add;      l0 := l1;      l1 := temp;    end for;    writeln;  end func; const proc: main is func  begin    write("Leonardo Numbers:");    leonardo(1, 1, 1, 25);    write("Fibonacci Numbers:");    leonardo(0, 1, 0, 25);  end func;`
Output:
```Leonardo Numbers: 1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Fibonacci Numbers: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

Sidef

`func 𝑳(n, 𝑳0 = 1, 𝑳1 = 1, 𝑳add = 1) {    { (𝑳0, 𝑳1) = (𝑳1, 𝑳0 + 𝑳1 + 𝑳add) } * n    return 𝑳0} say "The first 25 Leonardo numbers:"say 25.of { 𝑳(_) } say "\nThe first 25 numbers using 𝑳0 of 0, 𝑳1 of 1, and adder of 0:"say 25.of { 𝑳(_, 0, 1, 0) }`
Output:
```The first 25 Leonardo numbers:
[1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049]

The first 25 numbers using 𝑳0 of 0, 𝑳1 of 1, and adder of 0:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]
```

Swift

`struct Leonardo: Sequence, IteratorProtocol {    private let add : Int    private var n0: Int    private var n1: Int     init(n0: Int = 1, n1: Int = 1, add: Int = 1) {        self.n0 = n0        self.n1 = n1        self.add = add    }     mutating func next() -> Int? {        let n = n0        n0 = n1        n1 += n + add        return n    }} print("First 25 Leonardo numbers:")print(Leonardo().prefix(25).map{String(\$0)}.joined(separator: " ")) print("First 25 Fibonacci numbers:")print(Leonardo(n0: 0, add: 0).prefix(25).map{String(\$0)}.joined(separator: " "))`
Output:
```First 25 Leonardo numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
First 25 Fibonacci numbers:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

VBA

` Option Explicit Private Sub LeonardoNumbers()Dim L, MyString As String    Debug.Print "First 25 Leonardo numbers :"    L = Leo_Numbers(25, 1, 1, 1)    MyString = Join(L, "; ")    Debug.Print MyString    Debug.Print "First 25 Leonardo numbers from 0, 1 with add number = 0"    L = Leo_Numbers(25, 0, 1, 0)    MyString = Join(L, "; ")    Debug.Print MyString    Debug.Print "If the first prarameter is too small :"    L = Leo_Numbers(1, 0, 1, 0)    MyString = Join(L, "; ")    Debug.Print MyStringEnd Sub Public Function Leo_Numbers(HowMany As Long, L_0 As Long, L_1 As Long, Add_Nb As Long)Dim N As Long, Ltemp     If HowMany > 1 Then        ReDim Ltemp(HowMany - 1)        Ltemp(0) = L_0: Ltemp(1) = L_1        For N = 2 To HowMany - 1             Ltemp(N) = Ltemp(N - 1) + Ltemp(N - 2) + Add_Nb        Next N    Else        ReDim Ltemp(0)        Ltemp(0) = "The first parameter is too small"    End If    Leo_Numbers = LtempEnd Function `
Output:
```First 25 Leonardo numbers :
1; 1; 3; 5; 9; 15; 25; 41; 67; 109; 177; 287; 465; 753; 1219; 1973; 3193; 5167; 8361; 13529; 21891; 35421; 57313; 92735; 150049
First 25 Leonardo numbers from 0, 1 with add number = 0
0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; 233; 377; 610; 987; 1597; 2584; 4181; 6765; 10946; 17711; 28657; 46368
If the first prarameter is too small :
The first parameter is too small```

Visual Basic .NET

Translation of: C#
`Module Module1     Iterator Function Leonardo(Optional L0 = 1, Optional L1 = 1, Optional add = 1) As IEnumerable(Of Integer)        While True            Yield L0            Dim t = L0 + L1 + add            L0 = L1            L1 = t        End While    End Function     Sub Main()        Console.WriteLine(String.Join(" ", Leonardo().Take(25)))        Console.WriteLine(String.Join(" ", Leonardo(0, 1, 0).Take(25)))    End Sub End Module`
Output:
```1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368```

Wren

`var leonardo = Fn.new { |first, add, limit|    var leo = List.filled(limit, 0)    leo[0] = first[0]    leo[1] = first[1]    for (i in 2...limit) leo[i] = leo[i-1] + leo[i-2] + add    return leo} System.print("The first 25 Leonardo numbers with L(0) = 1, L(1) = 1 and Add = 1 are:")for (l in leonardo.call([1, 1], 1, 25)) System.write("%(l) ") System.print("\n\nThe first 25 Leonardo numbers with L(0) = 0, L(1) = 1 and Add = 0 are:")for (l in leonardo.call([0, 1], 0, 25)) System.write("%(l) ")System.print()`
Output:
```The first 25 Leonardo numbers with L(0) = 1, L(1) = 1 and Add = 1 are:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

The first 25 Leonardo numbers with L(0) = 0, L(1) = 1 and Add = 0 are:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

XPL0

`int N, L, L0, L1, Add;[Text(0, "Enter L(0), L(1), Add: ");L0:= IntIn(0);L1:= IntIn(0);Add:= IntIn(0);IntOut(0, L0);  ChOut(0, ^ );IntOut(0, L1);  ChOut(0, ^ );for N:= 3 to 25 do        [L:= L1 + L0 + Add;        IntOut(0, L);  ChOut(0, ^ );        L0:= L1;        L1:= L;        ];]`
Output:
```Enter L(0), L(1), Add: 1 1 1
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049
Enter L(0), L(1), Add: 0 1 0
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```

zkl

`fcn leonardoNumber(n, n1=1,n2=1,addnum=1){   if(n==0) return(n1);   if(n==1) return(n2);   self.fcn(n-1,n1,n2,addnum) + self.fcn(n-2,n1,n2,addnum) + addnum}`
`println("The first 25 Leonardo Numbers:");foreach n in (25){ print(leonardoNumber(n)," ") }println("\n"); println("The first 25 Fibonacci Numbers:");foreach n in (25){ print(leonardoNumber(n, 0,1,0)," ") }println();`
Output:
```The first 25 Leonardo Numbers:
1 1 3 5 9 15 25 41 67 109 177 287 465 753 1219 1973 3193 5167 8361 13529 21891 35421 57313 92735 150049

The first 25 Fibonacci Numbers:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368
```