Recaman's sequence
You are encouraged to solve this task according to the task description, using any language you may know.
The Recamán's sequence generates Natural numbers.
Starting from zero, the n'th term a(n) is the previous term minus n i.e a(n) = a(n-1) - n but only if this is both positive and has not been previousely generated.
If the conditions don't hold then a(n) = a(n-1) + n.
- Task
- Generate and show here the first 15 members of the sequence.
- Find and show here, the first duplicated number in the sequence.
- Optionally: Find and show here, How many terms of the sequence are needed until all the integers 0..1000, inclusive, are generated.
- References
- A005132, The On-Line Encyclopedia of Integer Sequences.
- The Slightly Spooky Recamán Sequence, Numberphile video.
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- define TRUE 1
- define FALSE 0
typedef int bool;
bool used[1001];
bool contains(int a[], int b, int len) {
int i;
for (i = 0; i < len; ++i) {
if (a[i] == b) return TRUE;
}
return FALSE;
}
int firstFalse(int start) {
int i;
for (i = start; i <= 1000; ++i) {
if (!used[i]) return i;
}
return -1;
}
void init() {
int i; used[0] = TRUE; for (i = 1; i <= 1000; ++i) used[i] = FALSE;
}
int main() {
int i, n, k = 0, next, *a;
bool foundDup = FALSE;
init();
a = malloc(400000 * sizeof(int));
a[0] = 0;
for (n = 1; n <= 15 || !foundDup || k != -1; ++n) {
next = a[n - 1] - n;
if (next < 1 || contains(a, next, n)) {
next += 2 * n;
}
a[n] = next;
if (next >= 0 && next <= 1000) used[next] = TRUE;
if (n == 14) {
printf("The first 15 terms of the Recaman's sequence are: ");
printf("[");
for (i = 0; i < 15; ++i) printf("%d ", a[i]);
printf("\b]\n");
}
if (!foundDup && contains(a, next, n)) {
printf("The first duplicated term is a[%d] = %d\n", n, next);
foundDup = TRUE;
}
k = firstFalse(k);
if (k == -1) {
printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n);
}
}
free(a);
return 0;
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Go
<lang go>package main
import "fmt"
func contains(a []int, b int) bool {
for _, j := range a {
if j == b {
return true
}
}
return false
}
func main() {
a := []int{0}
used := make(map[int]bool, 1001)
used[0] = true
for n, foundDup := 1, false; n <= 15 || !foundDup || len(used) < 1001; n++ {
next := a[n-1] - n
if next < 1 || contains(a, next) {
next += 2 * n
}
a = append(a, next)
if next >= 0 && next <= 1000 {
used[next] = true
}
if n == 14 {
fmt.Println("The first 15 terms of the Recaman's sequence are:", a)
}
if !foundDup && contains(a[:n], next) {
fmt.Printf("The first duplicated term is a[%d] = %d\n", n, next)
foundDup = true
}
if len(used) == 1001 {
fmt.Printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n)
}
}
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Haskell
A basic recursive function for the first N terms, <lang haskell>import Data.List (find) import Data.Maybe (isNothing)
recaman :: Int -> [Int] recaman n = fst <$> reverse (go n)
where
go x
| 1 > x = []
| 1 == x = [(0, 1)]
| otherwise =
let xs@((r, i):_) = go (pred x)
back = r - i
in ( if 0 < back && isNothing (find ((back ==) . fst) xs)
then back
else r + i
, succ i) :
xs
main :: IO () main = print $ recaman 15</lang>
- Output:
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]
Or, a little more flexibly, a recamanUpto (predicate) function. <lang haskell>import Data.Set (Set, fromList, insert, isSubsetOf, member, size)
recamanUpto :: (([Int], Int, Set Int) -> Bool) -> ([Int], Int, Set Int) recamanUpto p =
until
p
(\(rs@(r:_), i, set) ->
let nxt = nextRecaman set i r
in (nxt : rs, succ i, insert nxt set))
([0], 1, fromList [0])
nextRecaman :: Set Int -> Int -> Int -> Int nextRecaman set i nRec =
let back = nRec - i
in if 0 < back && not (member back set)
then back
else nRec + i
firstNRecamans :: Int -> [Int] firstNRecamans n =
let (rs, _, _) = recamanUpto (\(_, i, _) -> n == i) in reverse rs
firstDuplicateR :: Int firstDuplicateR =
let (rs, _, _) = recamanUpto (\(rs, _, set) -> size set /= length rs) in head rs
recamanSuperSet :: Set Int -> [Int] recamanSuperSet setInts =
let (rs, _, _) = recamanUpto (\(_, _, setR) -> isSubsetOf setInts setR) in tail rs
-- TEST --------------------------------------------------------------- main :: IO () main =
putStrLn $ unlines [ "First 15 Recamans:" , show $ firstNRecamans 15 , "\nFirst duplicated Recaman:" , show firstDuplicateR , "\nLength of Recaman series required to include [0..1000]:" , (show . length . recamanSuperSet) $ fromList [0 .. 1000] ]</lang>
- Output:
First 15 Recamans: [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9] First duplicated Recaman: 42 Length of Recaman series required to include [0..1000]: 328002
JavaScript
<lang javascript>(() => {
const main = () => {
console.log(
'First 15 Recaman:\n' +
recamanUpto(i => i === 15)
);
console.log(
'\n\nFirst duplicated Recaman:\n' +
last(recamanUpto(
(_, set, rs) => set.size !== rs.length
))
);
const setK = new Set(enumFromToInt(0, 1000));
console.log(
'\n\nNumber of Recaman terms needed to generate' +
'\nall integers from [0..1000]:\n' +
(recamanUpto(
(_, setR) => isSubSetOf(setK, setR)
).length - 1)
);
};
// RECAMAN --------------------------------------------
// nextRecaman :: Set Int -> Int -> Int
const nextRecaman = (setSeen, i, nRec) => {
const back = nRec - i;
return (0 < back && !setSeen.has(back)) ? (
back
) : nRec + i;
};
// recamanUpto :: (Int -> Set Int > [Int] -> Bool) -> [Int]
const recamanUpto = p => {
let
i = 1,
intR = 0,
rs = [intR];
const setR = new Set(rs);
while (!p(i, setR, rs)) {
intR = nextRecaman(setR, i, intR);
setR.add(intR);
rs.push(intR);
i++;
}
return rs;
}
// GENERIC --------------------------------------------
// enumFromToInt :: Int -> Int -> [Int]
const enumFromToInt = (m, n) =>
m <= n ? iterateUntil(
x => n <= x,
x => 1 + x,
m
) : [];
// isSubsetOf :: Ord a => Set a -> Set a -> Bool
const isSubSetOf = (a, b) => {
for (let x of a) {
if (!b.has(x)) return false;
}
return true;
};
// iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
const iterateUntil = (p, f, x) => {
const vs = [x];
let h = x;
while (!p(h))(h = f(h), vs.push(h));
return vs;
};
// last :: [a] -> a
const last = xs =>
0 < xs.length ? xs.slice(-1)[0] : undefined;
// MAIN ------------------------------------------------ return main();
})();</lang>
- Output:
First 15 Recaman: 0,1,3,6,2,7,13,20,12,21,11,22,10,23,9 First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
Kotlin
<lang scala>// Version 1.2.60
fun main(args: Array<String>) {
val a = mutableListOf(0)
val used = mutableSetOf(0)
var foundDup = false
var n = 1
while (n <= 15 || !foundDup || used.size < 1001) {
var next = a[n - 1] - n
if (next < 1 || a.contains(next)) next += 2 * n
a.add(next)
if (next in 0..1000) used.add(next)
if (n == 14) {
println("The first 15 terms of the Recaman's sequence are: $a")
}
if (!foundDup && a.subList(0, n).contains(next)) {
println("The first duplicated term is a[$n] = $next")
foundDup = true
}
if (used.size == 1001) {
println("Terms up to a[$n] are needed to generate 0 to 1000")
}
n++
}
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Microsoft Small Basic
<lang smallbasic>' Recaman's sequence - smallbasic - 05/08/2015
nn=15
TextWindow.WriteLine("Recaman's sequence for the first " + nn + " numbers:")
recaman()
TextWindow.WriteLine(Text.GetSubTextToEnd(recaman,2))
nn="firstdup"
recaman()
TextWindow.WriteLine("The first duplicated term is a["+n+"]="+firstdup)
Sub recaman
recaman=""
firstdup=""
If nn="firstdup" Then
nn=1000
firstdup="True"
EndIf
For n=0 To nn-1
ap=a[n-1]+n
If a[n-1]<=n Then
a[n]=ap 'a[n]=a[n-1]+n
b[ap]=1
Else
am=a[n-1]-n
If b[am]=1 Then
a[n]=ap 'a[n]=a[n-1]+n
b[ap]=1
Else
a[n]=am 'a[n]=a[n-1]-n
EndIf
EndIf
If firstdup Then
If dup[a[n]]=1 Then
firstdup=a[n]
Goto exitsub
EndIf
dup[a[n]]=1
EndIf
recaman=recaman+","+a[n]
EndFor
exitsub:
EndSub </lang>
- Output:
Recaman's sequence for the first 15 numbers: 0,1,3,6,2,7,13,20,12,21,11,22,10,23,9 The first duplicated term is a[24]=42
Perl 6
<lang perl6>my @recamans = 0, {
state %s;
state $n;
$n++;
my $k = $^m - $n;
$k = $m + $n unless ($k > 0) && !%s{$k};
%s{$k} = True;
$k
} … *;
put "First fifteen term of Recaman's sequence: ", @recamans[^15];
say "First duplicate at term: a[{ @recamans.first({@recamans[^$_].Bag.values.max == 2})-1 }]";
my @seen = 0 xx 1001; for ^Inf {
@seen[@recamans[$_]] = 1 if @recamans[$_] <= 1000; say "Range 0..1000 covered by terms up to a[$_]" and last if so (sum(@seen[^1001]) == 1001);
}</lang>
- Output:
First fifteen term of Recaman's sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate at term: a[24] Range 0..1000 covered by terms up to a[328002]
Python
<lang python>from itertools import islice
class Recamans():
"Recamán's sequence generator callable class"
def __init__(self):
self.a = None # Set of results so far
self.n = None # n'th term (counting from zero)
def __call__(self):
"Recamán's sequence generator"
nxt = 0
a, n = {nxt}, 0
self.a = a
self.n = n
yield nxt
while True:
an1, n = nxt, n + 1
nxt = an1 - n
if nxt < 0 or nxt in a:
nxt = an1 + n
a.add(nxt)
self.n = n
yield nxt
if __name__ == '__main__':
recamans = Recamans()
print("First fifteen members of Recamans sequence:",
list(islice(recamans(), 15)))
so_far = set()
for term in recamans():
if term in so_far:
print(f"First duplicate number in series is: a({recamans.n}) = {term}")
break
so_far.add(term)
n = 1_000
setn = set(range(n + 1)) # The target set of numbers to be covered
for _ in recamans():
if setn.issubset(recamans.a):
print(f"Range 0 ..{n} is covered by terms up to a({recamans.n})")
break</lang>
- Output:
First fifteen members of Recamans sequence: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] First duplicate number in series is: a(24) = 42 Range 0 ..1000 is covered by terms up to a(328002)
REXX
<lang rexx>/*REXX program computes & displays the Recaman sequence (also known as Recamán sequence)*/ parse arg N h . /*obtain optional arguments from the CL*/ if N== | N=="," then N= 15 /*Not specified? Then use the default.*/ say "Recamán's sequence for the first " N " numbers:" say recaman(N) say say "The first duplicate number in the Recamán's sequence is: " recaman(0) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ recaman: procedure; parse arg y 1 oy,,$ !. d.; u=0 /*U: a unique flag. */
if y==0 then do; y=1e8; u=1; end /*for duplicate stuff.*/
@.=0 /*initialize @ array*/
do j=0 for y; jm=j-1; p=@.jm
_=p+j
if p<=j then do; @.j=_; !._=.; end /*for pository values.*/
else do; m=p-j
@.j=m /*for negatory values.*/
if !.m==. then do; @.j=_; !._=. /*was defined before? */
end
end
z=@.j /*get the @.J value.*/
if u then do; if d.z==. then return z; d.z=.; iterate /*j*/; end
$=$ z /*append Z to list. */
end /*j*/
return strip($) /*return the $ list.*/</lang>
- output when using the default input:
Recamán's sequence for the first 15 numbers: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 The first duplicate number in the Recamán's sequence is: 42
zkl
<lang zkl>fcn recamanW{ // -->iterator -->(n,a,True if a is a dup)
Walker.tweak(fcn(rn,rp,d){
n,p,a := rn.value, rp.value, p - n;
if(a<=0 or d.find(a)) a+=2*n;
d.incV(a); rp.set(a);
return(rn.inc(),a,d[a]>1);
}.fp(Ref(0),Ref(0),Dictionary()) )
}</lang> <lang zkl>print("First 15 members of Recaman's sequence: "); recamanW().walk(15).apply("get",1).println();
n,a := recamanW().filter1("get",2); // ie filter(a[n].dup) println("First duplicate number in series is: a(%d) = %d".fmt(n,a));
rw,ns,n,a,dup := recamanW(),1000,0,0,0; do{ n,a,dup=rw.next(); if(not dup and a<1000) ns-=1; }while(ns); println("Range 0..1000 is covered by terms up to a(%,d)".fmt(n));</lang>
- Output:
First 15 members of Recamans sequence: L(0,1,3,6,2,7,13,20,12,21,11,22,10,23,9) First duplicate number in series is: a(24) = 42 Range 0..1000 is covered by terms up to a(328,002)