Hofstadter Q-sequence

Question

Definition

1. a(1) = 1
2. a(2) = 1
3. a(n) = a(n-a(n-1)) + a(n-a(n-2)) for n> 2 where n is an integer

Task

Given positive integer n, generate a(n).

Testcases

n a(n)
1 1
2 1
3 2
4 3
5 3
6 4
7 5
8 5
9 6
10 6
11 6
12 8
13 8
14 8
15 10
16 9
17 10
18 11
19 11
20 12

Reference

• Obligatory OEIS A005185

• Leaky Nun: Related.
• Dennis: Can we return True in languages where it can be used as 1?
• Leaky Nun: @Dennis If in that language true is equivalent to 1 then yes.
• Martin Ender: Apart from the OEIS link it might be good to reference GEB where the sequence first appeared.
• Martin Ender: Completing the list of GEB-related sequence challenges.

Answer 1

Haskell, (削除) 35 (削除ここまで) 33 bytes

a n|n<3=1|b<-a.(-)n=b(b 1)+b(b 2)

Defines a function a.

• xnor: Nice trick with the bind! Wouldn't something like (b.b)1+(b.b)2 be shorter than the sum?

Answer 2

Retina, (削除) 84 (削除ここまで) (削除) 83 (削除ここまで) (削除) 79 (削除ここまで) 74 bytes

Byte count assumes ISO 8859-1 encoding.

.+
$*;1¶1¶
+`;(?=(1)+¶(1)+)(?=(?<-1>(1+)¶)+)(?=(?<-2>(1+)¶)+)
3ドル4ドル¶
G3=`
1

Try it online! (The first line enables a linefeed-separated test suite.)

I'll have to golf this some more later.

Answer 3

Julia, 29 bytes

!n=n<3||!(n-!~-n)+!(n-!~-~-n)

Try it online!

How it works

We redefine the unary operator ! for our purposes.

If n is 1 or 2, n<3 returns true and this is our return value.

If n larger than 2, n<3 returns false and the || branch gets executed. This is a straightforward implementation of the definition, where ~-n yields n - 1 and ~-~-n yields n - 2.

Answer 4

Sesos, 54 bytes

0000000: eefb5b 04f83a a75dc2 36f8d7 cf6dd0 af7b3b 3ef8d7 ..[..:.].6...m..{;>..
0000015: cfed12 f661f0 ae9d83 ee63e6 065df7 ce6183 af7383 ....a.....c..]..a..s.
000002a: 76ef3c 3f6383 7eff9c b9e37f v.<?c.~.....

Try it online

Disassembled

set numin
set numout
add 1
fwd 1
add 1
fwd 6
get
sub 1
jmp
 jmp
 sub 1
 fwd 1
 add 1
 rwd 1
 jnz
 fwd 1
 sub 1
 rwd 2
 add 2
 jmp
 rwd 4
 jmp
 sub 1
 fwd 3
 add 1
 rwd 3
 jnz
 fwd 4
 jmp
 sub 1
 rwd 3
 add 1
 rwd 1
 add 1
 fwd 4
 jnz
 rwd 3
 jmp
 sub 1
 fwd 3
 add 1
 rwd 3
 jnz
 fwd 4
 add 2
 jmp
 rwd 5
 jmp
 rwd 1
 jmp
 sub 1
 fwd 2
 add 1
 rwd 2
 jnz
 fwd 1
 jmp
 sub 1
 rwd 1
 add 1
 fwd 1
 jnz
 rwd 1
 sub 1
 jnz
 fwd 2
 jmp
 sub 1
 rwd 1
 add 1
 rwd 1
 add 1
 fwd 2
 jnz
 fwd 1
 jmp
 rwd 2
 jmp
 sub 1
 fwd 1
 add 1
 rwd 1
 jnz
 fwd 2
 jmp
 sub 1
 rwd 2
 add 1
 fwd 2
 jnz
 fwd 1
 jnz
 fwd 3
 sub 1
 jnz
 rwd 2
 jmp
 sub 1
 rwd 3
 add 1
 fwd 3
 jnz
 fwd 1
 sub 1
 jnz
 fwd 2
jnz
rwd 7
put

Or in Brainfuck notation:

+>+>>>>>>,-[[->+<]>-<<++[<<<<[->>>+<<<]>>>>[-<<<+<+>>>>]<<<[->>>+<<<]>>>>++[<<<<<[<
[->>+<<]>[-<+>]<-]>>[-<+<+>>]>[<<[->+<]>>[-<<+>>]>]>>>-]<<[-<<<+>>>]>-]>>]<<<<<<<.

Answer 5

C, (削除) 43 (削除ここまで) 42 bytes

Saved 1 byte thanks to @Dennis

Every answer is the same, I must do something different!

Try it online!

a(n){return n<3?:a(n-a(n-2))+a(n---a(n));}

Explanation: it's basically a(n-a(n-2))+a(n-a(n-1)) but with swaggy^undefinedbehavior (works on my phone (gcc) and ideone).

• Dennis: 1. You should also mention the compiler; your "swag" is undefined behavior. 2. With GCC, you don't need the 1 between ? and :.
• AdmBorkBork: @Dennis Interestingly, that same formulation works in my iterative PowerShell answer... $b+=$b[$_-$b[$_-2]]+$b[$_---$b[$_]]
• betseg: @TimmyD some compilers may compile the a(n) before the n--, and there isn't a standart (or defined) behavior for that. Thus, undefined behavior.
• AdmBorkBork: @betseg Yep, I agree. Just pointing out that it's not necessarily unique to C.
• betseg: @TimmyD Oh I misunderstood that. I just wanted to change the function that everybody uses, so mine would be different and swaggy :D

Answer 6

Mathematica, 36 bytes

Byte count assumes ISO 8859-1 encoding and Mathematica's $CharacterEncoding set to WindowsANSI (the default on Windows; other settings might work as well, but some like UTF-8 definitely don't).

±1=±2=1
±n_:=±(n-±(n-1))+±(n-±(n-2))

Defines ± as a unary operator.

I tried getting rid of the duplication, but ended up with the same byte count:

±1=±2=1
±n_:=Tr[±(n-±(n-#))&/@{1,2}]

• Leaky Nun: I may give you a +200 bounty if you do it in Retina
• Martin Ender: @LeakyNun okay? :)
• Leaky Nun: Two days later.
• mbomb007: @LeakyNun Soon you're going to have no rep left if you give out bounties so easily.
• LegionMammal978: Let us continue this discussion in chat.

Answer 7

Jelly, (削除) 15 (削除ここまで) 14 bytes

2Rạ8߀$+Sμ1>?2

Try it online! or verify all test cases (takes a few seconds).

How it works

2Rạ8߀$+Sμ1>?2 Main link. Argument: n (integer)
2R Yield [1, 2].
 $ Combine the previous three links into a monadic chain.
 8 Yield n.
 ạ Take the absolute difference of the return value and n.
 ߀ Recursively call the main link on each result.
 + Duplicate the chain.
 The first copy maps [1, 2] to [a(n - 1), a(n - 2)].
 The second copy maps [a(n - 1), a(n - 2)] to
 [a(n - a(n - 1)), a(n - a(n - 2))].
 S Take the sum.
 μ Combine all links to the left into a chain.
 ? If...
 > 2 n is greater than 2, call the chain.
 1 Else, return 1.

• Leaky Nun: I may give you a +400 bounty if you do it in Sesos.
• Yytsi: @LeakyNun There seems to be an Sesos answer. It came out one day after your comment.

Answer 8

JavaScript (ES6), (削除) 45 Bytes (削除ここまで) 34 Bytes

A recursive solution in ES6. Any golfing tips much appreciated.

a=n=>n>2?a(n-a(n-1))+a(n-a(n-2)):1

Thank you to /u/ismillo for shortening even further.

Answer 9

Jelly, (削除) 14 (削除ここまで) (削除) 12 (削除ここまで) 11 bytes

ịḣ2S;
1Ç8¡2ị

This is an iterative approach.

Try it online! or verify all test cases.

How it works

1Ç¡2ị Main link. Argument: n
1 Set the return value to 1.
 Ç¡ Call the helper link n times, updating the return value after each call.
 2ị Extract the second element of the resulting array.
ịḣ2S; Helper link. Argument: A (array)
ị At-index; retrieve the elements of A at the values of A.
 ḣ2 Head 2; extract the first two results.
 S Take the sum of the result.
 ; Prepend the sum to A.

Answer 10

Python, (削除) 45 (削除ここまで) 40 bytes

a=lambda n:n<3or a(n-a(n-1))+a(n-a(n-2))

Simple naïve interpretation of the challenge.

Saved 5 bytes thanks to @LeakyNun!

Answer 11

Haskell, (削除) 39 (削除ここまで) 37 Bytes

h n|n<3=1|n>2=h(n-h(n-1))+h(n-h(n-2))

exactly like described in the challenge, using guards

• Laikoni: Sorry, I didn't saw your solution before posting my (identical) haskell solution. However isn't the byte count 38 as the new-line has to be taken into account?
• Laikoni: And the guard has to be n<3 for h 2 to be 1.
• KarlKastor: @Laikoni It's 37 according to Pythons len feature with a multiline (""") string, unless you count newline as two bytes. Yeah, I noticed the other thing it's fixed now.
• Laikoni: TIL notepad++ counts newline as two character.
• KarlKastor: @Laikoni got rid of the newline it's indisputably 37 bytes now.

Answer 12

R, 50 bytes

a=function(n)ifelse(n<3,1,a(n-a(n-1))+a(n-a(n-2)))

Usage:

> a(1)
 1
> a(20)
 12

Answer 13

CJam, (削除) 19 (削除ここまで) 18 bytes

XXri{_(2ドル$$+}*]-3=

Try it online!

Uses the iterative approach.

Answer 14

C#, (削除) 51 (削除ここまで) 44 bytes

int a(int n)=>n<3?1:a(n-a(n-1))+a(n-a(n-2));

(削除) i wonder if this can be shortened by making it anonymous (削除ここまで) thanks pinkfloydx33!

• pinkfloydx33: c# 6 expression bodied function int a(int n)=>n<3?1:a(n-a(n-a))+a(n-a(n-2));
• pinkfloydx33: Seems I typod while typing that on my phone. The inner most -a in the first set of parens should be -1
• downrep_nation: I didn't notice it either, ill fix it rq

Answer 15

Golfscript, 29 bytes

~[1 1]{..0=(=\..1=(=@+\+}@*2=

Try it online!

Answer 16

05AB1E, 29 bytes

An iterative solution.

XˆXˆÍL>v ̄¤ys-è ̄y ̄yÍè-è+ˆ} ̄1<è

Try it online

Answer 17

APL, 20 bytes

{⍵≤2:1⋄+/∇ ̈⍵-∇ ̈⍵-⍳2}

Explanation:

{⍵≤2:1⋄+/∇ ̈⍵-∇ ̈⍵-⍳2}
 ⍵≤2:1 If argument is 2 or less, return 1
 ⋄ Otherwise:
 ⍵-⍳2 Subtract [1, 2] from the argument
 ∇ ̈ Recursive call on both
 ⍵- Subtract both results from the argument 
 ∇ ̈ Recursive call on both again
 +/ Sum 

Answer 18

VBA Excel 87 bytes

Non-recursive, since I want this to work for n=100000, say:

Function A(N):ReDim B(N):For i=3 To N:B(i)=B(i-B(i-1)-1)+B(i-B(i-2)-1)+1:Next:A=B(N)+1

... and press return (byte #87) at the end of the line to get the End Function statement for "free". Note that B values are offset by -1 to avoid initializing for n=1 and 2.

Invoke in spreadsheet as normal, eg =A(100000) to get 48157

The recursive version, 61 bytes,

Function Y(N):If N<3 Then Y=1 Else Y=Y(N-Y(N-1))+Y(N-Y(N-2))

starts to get unreasonably slow for n>30, and couldn't be said to work at all for n>40.

• mbomb007: We don't care about performance. We care about code length. You should move your shorter solution to the top of your answer.
• Joffan: @mbomb007 Since I'm nowhere near winning the golf, I'll make my own choices on what constitutes a working program. Not able to handle even single byte integers is not good enough as far as I'm concerned, when there is a solution which can do so easily.

Answer 19

Ruby, 36 bytes

A direct implementation. Any golfing suggestions are welcome.

a=->n{n<3?1:a[n-a[n-1]]+a[n-a[n-2]]}

• Seims: Afaik, you can get rid of the a=. If you post it here, it suffices when your code starts with the ->. It counts as an anonymous function then.
• Sherlock9: @Seims Unfortunately, as the function calls itself with a[n-1] and such, the function needs to be named.

Answer 20

Java 7, (削除) 68 (削除ここまで) (削除) 61 (削除ここまで) 51 bytes

17 saved thanks to Leaky Nun.

int a(int n){return n<3?1:a(n-a(n-1))+a(n-a(n-2));}

• AdmBorkBork: Welcome to PPCG!
• Kevin Cruijssen: Welcome to PPCG! You might like Tips for Golfing in Java. An alternate form would be: int a(int n){return n<3?1:a(n-a(n-2))+a(n---a(n));}, but unfortunately it uses the same amount of bytes as the answer you already have.. Also, I would specify that your answer is in Java 7, since the Java 8 answer would be shorter: n->return n<3?1:a(n-a(n-1))+a(n-a(n-2)) (39 bytes).
• Justin: Thanks for the welcomes guys, and thanks for the tip on Java8 - I didn't realize lambdas were allowed like that - although they're allowed like that in Python, so I guess I just never thought about it. Does the lambda need a semi-colon?
• Kevin Cruijssen: @JustinTervay I don't use Java 8 a lot, but from what I've heard the semi-colon isn't counted on single-line expressions, according to a comment by @DavidConrad and @CAD97 in one of my own Java answers.

Answer 21

Oasis, (削除) 9 (削除ここまで) (削除) 7 (削除ここまで) 5 bytes

(削除) Non-competing, since the language postdates the challenge. (削除ここまで) Thanks to Kenny Lau for saving 4 bytes. Code:

ece+V

Expanded form (V is short for 11):

a(n) = ece+
a(0) = 1
a(1) = 1

Code:

e # Stack is empty, so a(n - 1) is used, and it calculates a(n - a(n - 1))
 c # Calculate a(n - 2)
 e # Calculate a(n - a(n - 2))
 + # Add up

Try it online!. Calculates n = 1000 in 0.1 seconds.

Answer 22

PowerShell v2+, (削除) 85 (削除ここまで) (削除) 79 (削除ここまで) 69 bytes

param($n)$b=1,1;2..$n|%{$b+=$b[$_-$b[$_-1]]+$b[$_-$b[$_-2]]};$b[$n-1]

Takes input $n, sets $b to be an array of @(1, 1), then enters a loop from 2 .. $n. Each iteration we tack onto $b the latest calculation in the sequence with a simple += and the definition of the sequence. We then output the appropriate number from $b (with a -1 because arrays in PowerShell are zero-indexed). This works if $n is 1 or 2 because both of those values are pre-populated into the lower indices of $b from the start, so even if the loop tacks on junk, it's ignored anyway.

---

Recursive solution (削除) 78 (削除ここまで) 76 bytes

$a={param($k)if($k-lt3){1}else{(&$a($k-(&$a($k-1))))+(&$a($k-(&$a($k-2))))}}

First time I've used the equivalent of a lambda as the answer, as usually an iterative solution is shorter (as you can see from all the nested parens). (削除) But, in this case, the nested parens are almost duplicated in the iterative solution with the nested array calls, so the recursive solution is shorter. (削除ここまで) Nope, the iterative solution is indeed shorter (see above).

Call it via the execution-operator, like &$a 20. Just a straight-up recursive call.

Answer 23

JavaScript (ES6), 66 bytes

n=>[...Array(n+1)].reduce((p,_,i,a)=>a[i]=i<3||a[i-p]+a[i-a[i-2]])

Non-recursive version for speed; recursive version is probably shorter but I'll leave it for someone else to write. I always like it when I get to use reduce. Note: 1 byte saved by returning true (which casts to 1 when used in an integer context) for of a(1) and a(2).

Answer 24

Pyth, 16 bytes

L|<b3smy-bytdtBb
L def y(b):
 |<b3 b < 3 or ...
 m tBb map for d in [b - 1, b]:
 y-bytd y(b - y(d - 1))
 s sum

Defines a function y.

Try it online (added yMS20 to print the first 20 values)

Answer 25

Forth, 76 bytes

I finally got it working!

: Q recursive dup dup 3 < if - 1+ else 2dup 2 - Q - Q -rot 1- Q - Q + then ;

Try it online

Explanation:

: Q recursive \ Define a recursive function Q
 dup dup 3 < \ I moved a dup here to golf 2 bytes
 if \ If n < 3, return 1
 - 1 \ Golf: n-n is zero, add one. Same as 2drop 1+
 else
 2dup 2 - Q - Q \ Copy n until 4 on stack, find Q(n-Q(n-2))
 -rot \ Move the result below 2 copies of n
 1- Q - Q + \ Find Q(n-Q(n-2)), then add to previous ^
 then ;

Try it online (slightly un-golfed from above)

Unfortunately, mutual recursion is a bit too wordy to use for golfing.

Answer 26

Lua, 59 bytes

function a(n)return n<3 and 1 or a(n-a(n-1))+a(n-a(n-2))end

Answer 27

Maple, (削除) 43 (削除ここまで) 41 bytes

a:=n->`if`(n>2,a(n-a(n-1))+a(n-a(n-2)),1)

Usage:

> a(1);
 1
> a(20);
 12

This problem is certainly a good candidate for memoization. Using option cache, the run times are cut down significantly:

aC := proc(n) 
 option cache; 
 ifelse( n > 2, aC( n - aC(n-1) ) + aC( n - aC(n-2) ), 1 ); 
end proc:

This can be seen using:

CodeTools:-Usage( aC(50) );

Answer 28

Husk, 17 bytes

!¡λṁ!1m≠→L1↑_2)ḋ3

Try it online!

Feels too long. I'll post something with fix soon.

Answer 29

Japt, 20 bytes

No big shenanigans, just recursively calculates the result.

§2?1:ßU-ßUÉ)+ßU-ßU-2 
§2 // If the input is less than or equal to 2
 ?1 // return 1,
 : // otherwise
 ß // recursively recall with
 U-ßUÉ // input minus recursive recall input minus one
 )+ // plus
 ßU-ßU-2 // same with input minus two.

Try it here.

• Shaggy: 15 bytes

Answer 30

J, (削除) 29 (削除ここまで) 28 bytes

1:`(+&$:/@:-$:@-&1 2)@.(2&<)

Uses the recursive definition.

Usage

Extra commands are used for formatting multiple input/output.

 f =: 1:`(+&$:/@:-$:@-&1 2)@.(2&<)
 (,:f"0) >: i. 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1 1 2 3 3 4 5 5 6 6 6 8 8 8 10 9 10 11 11 12

Explanation

1:`(+&$:/@:-$:@-&1 2)@.(2&<) Input: n
 2&< If n < 2
1: Return 1
 Else
 -&1 2 Subtract [1, 2] from n to get [n-1, n-2]
 $:@ Call recursively on n-1 and n-2
 - Subtract each of the results from n
 /@: Reduce using
 $: A recursive call on each
 +& Then summation
 Return that value as the result