# Am I a Cullen Number?

A Cullen Number is any number that is contained in the sequence generated using the formula:

C(n) = (n*2^n)+1.

Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a Cullen Number.

## Input:

A non-negative integer between 0 and 10^9 (inclusive).

## Output:

A truthy/falsy value that indicates whether the input is a Cullen Number.

## Test Cases:

Input:    Output:
1   --->  truthy
3   --->  truthy
5   --->  falsy
9   --->  truthy
12  --->  falsy
25  --->  truthy


## Scoring:

This is , so the lowest score in bytes wins.

• What's the range of n? In particular, is 1 a Cullen Number?
– user62131
Jun 17, 2017 at 20:15
• @ais523 according to OEIS, it is. n seems to be 0-based. Jun 17, 2017 at 20:25
• Fair enough. Just needed to know whether my Jelly answer should have an Ḷ or R in it :-)
– user62131
Jun 17, 2017 at 20:25
• Related Jun 17, 2017 at 20:36
• Umm, what's with the downvote? Jun 20, 2017 at 13:47

## x86_64 machine code (System V ABI), 28 27 bytes

-1 byte thanks to @Cody Gray, thanks!

A constant-time algorithm!

_cullen:
0:   0f bd cf    bsrl    %edi, %ecx
3:   0f bd c1    bsrl    %ecx, %eax
6:   89 ca       movl    %ecx, %edx
8:   29 c2       subl    %eax, %edx
a:   0f bd c2    bsrl    %edx, %eax
d:   29 c1       subl    %eax, %ecx
f:   d3 e1       shll    %cl, %ecx
11:   ff c1       incl    %ecx
13:   31 c0       xorl    %eax, %eax
15:   39 f9       cmpl    %edi, %ecx
17:   0f 94 c0    sete    %al
1a:   c3          retq


Explanation:

Let y an integer and x=y*2^y + 1. Taking logs, we have y + log2(y) = log2(x-1), thus y=log2(x-1)-log2(y). Plugging back the value of y, we get y=log2(x-1)-log2(log2(x-1)-log2(y)). Doing this one more time, we obtain: y=log2(x-1)-log2[log2(x-1)-log2(log2(x-1)-log2(log2(x-1)-log2(y)))].

Let us remove the last terms (of the order of log2(log2(log2(log2(x)))), this should be safe!), and assume that x-1≈x, we obtain: y≈log2(x)-log2[log2(x)-log2(log2(x))]

Now, letting f(n) = floor(log2(n)), it can be verified manually that y can be exactly retrieved by: y=f(x)-f[f(x)-f(f(x))], for y < 26, and thus x ⩽ 10^9, as specified by the challenge(1).

The algorithm then simply consists of computing y given x, and verify that x == y*2^y + 1. The trick is that f(n) can simply be implemented as the bsr instruction (bit-scan reverse), which returns the index of the first 1-bit in n, and y*2^y as y << y.

Detailed code:

_cullen:                                 ; int cullen(int x) {
0:   0f bd cf    bsrl    %edi, %ecx   ;  int fx = f(x);
3:   0f bd c1    bsrl    %ecx, %eax   ;  int ffx = f(f(x));
6:   89 ca       movl    %ecx, %edx
8:   29 c2       subl    %eax, %edx   ;  int a = fx - ffx;
a:   0f bd c2    bsrl    %edx, %eax   ;  int ffxffx = f(a);
d:   29 c1       subl    %eax, %ecx   ;  int y = fx - ffxffx;
f:   d3 e1       shll    %cl, %ecx    ;  int x_ = y<<y;
11:   ff c1       incl    %ecx         ;  x_++;
13:   31 c0       xorl    %eax, %eax
15:   39 f9       cmpl    %edi, %ecx
17:   0f 94 c0    sete    %al
1a:   c3          retq                 ;  return (x_ == x);
; }


(1)In fact, this equality seems to hold for values of y up to 50000.

• Well, I'm pretty sure this qualifies as the most interesting code for this challenge so far. +1 Jun 18, 2017 at 14:30
• Pre-XORing eax would allow you to eliminate the movzbl, saving 1 byte. You'd need to do the XOR before the cmpl so it doesn't clobber the flags, of course, but that's totally fine because nothing after that depends on eax. Or, you could just decide that the method returns a Boolean in only the lower 8 bits, saving all 3 bytes! Jun 19, 2017 at 11:41
• @CodyGray Indeed, thanks a lot :) Jun 19, 2017 at 14:46

# Pyth, 6 5 bytes

/mh.<


try it online

# Jelly, 7 6 bytes

Ḷæ«i’


Try it online!

Takes input as a command-line argument. If given a Cullen number C(n), outputs n+1 (which is truthy in Jelly, being an nonzero integer; note that we have n≥0 because the input is an integer, and Cullen numbers with negative n are never integers). If given a non-Cullen number, returns 0, which is falsey in Jelly.

## Explanation

Ḷæ«i’
Ḷ        Form a range from 0 to (the input minus 1)
æ«      Left-shift each element in the range by
       itself
i’   Look for (the input minus 1) in the resulting array


Basically, form an array of Cullen numbers minus one, then look for the input minus one in it. If the input is a Cullen number, we'll find it, otherwise we won't. Note that the array is necessarily long enough to reach to the input, because C(n) is always greater than n.

## JavaScript (ES6), 37 35 bytes

Saved 2 bytes thanks to Neil

f=(n,k,x=k<<k^1)=>x<n?f(n,-~k):x==n


### Demo

f=(n,k,x=k<<k^1)=>x<n?f(n,-~k):x==n

console.log(JSON.stringify([...Array(1000).keys()].filter(n => f(n))))

• Does x<n?f(n,k+1):x==n work?
– Neil
Jun 17, 2017 at 20:57
• @Neil It sure does. :-) Jun 17, 2017 at 21:13
• Why does ~k work, while k+1 overloads the callstack? Jun 18, 2017 at 20:17
• @trlkly Basically, undefined+1===NaN but -~undefined===1. You can read more about this here. Jun 18, 2017 at 21:13

f n=or[x*2^x+1==n|x<-[0..n]]


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# Ohm, 8 bytes

@Dº*≥Dlε


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           Implicit input
@          Range [1,...,Input]
D         Duplicate
º        2^n each element
*       Multiply those two array
≥      Increment everything (now I have an array of all Cullen Numbers)
Dl    Push array length (= get input again, can't get again implicitly or using a function because it would be a string so I'd waste a byte again)
ε   Is input in array?


# PHP, 43 bytes

for(;$argn>$c=1+2**$n*$n++;);echo$argn==$c;


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• Is $argn a special variable? Changing it to $a would save 6 bytes: tio.run/##K8go@G9jX5BRwKWSaKtkaGaoZP0/… Jun 19, 2017 at 15:39
• @topher Yes $argn is available if you run PHP from the command line with the -R option Jun 19, 2017 at 16:26 # 05AB1E, 7 bytes ÝDo*¹<å  Try it online! Explanation: ÝDo*¹<å Example input: 9. Stack:  Ý Range 0-input. Stack: [[0,1,2,3,4,5,6,7,8,9]] D Duplicate. Stack: [[0,1,2,3,4,5,6,7,8,9],[0,1,2,3,4,5,6,7,8,9]] o 2** each item in the list. Stack: [[0,1,2,3,4,5,6,7,8,9], [1,2,4,8,16,32,64,128,256,512]] * Multiply the two lists. Stack: [[0, 2, 8, 24, 64, 160, 384, 896, 2048, 4608]] ¹ Push input again. Stack: [[0, 2, 8, 24, 64, 160, 384, 896, 2048, 4608],9] < Decrement. Stack: [[0, 2, 8, 24, 64, 160, 384, 896, 2048, 4608],8] å Is the first item of the stack in the second item? Stack:  Implicit print.  # R, 5351 46 bytes pryr::f(x%in%lapply(0:x,function(y)(y*2^y+1)))  Anonymous function. Checks if x is generated in the sequence C(n) for n in [0,x]. 3 bytes golfed by Giuseppe. Try it online! • use x%in%... instead of any(x==...); that'll drop you 4 bytes Jun 17, 2017 at 21:56 • So if I golf this by changing the lapply to simply checking the vector, and use scan instead of taking function arguments - I get @giuseppe 's answer. Thanks for posting it separately so I can see what I'm missing - I learn more by trying something on my own, even though I usually lose. – BLT Jun 18, 2017 at 1:09 ## C, C++, Java, C#, D : 70 bytes Due to the similarities between all these languages, this code works for each int c(int n){for(int i=0;i<30;++i)if((1<<i)*i+1==n)return 1;return 0;}  • I'll post the optimized D version this time, some pretty D-specific tricks can be used. Aug 3, 2017 at 21:55 • Suggest i=30;i--;)if(i<<i==n-1) instead of i=0;i<30;++i)if((1<<i)*i+1==n) Nov 28, 2018 at 4:24 # Python, 40 bytes lambda n:any(x<<x==n-1for x in range(n))  Try it online! # Python 2, 36 bytes f=lambda n,i=0:i<<i!=n-1and f(n,i+1)  Try it online! Outputs by not crashing / crashing, as currently allowed by this meta concensus. # Python 2, 42 bytes i=0 n=input()-1 while i<<i<n:i+=1 i<<i>n<c  Try it online! Outputs via exit code # R, 26 bytes a=0:26;scan()%in%(1+a*2^a)  Try it online! A slightly different approach than the other R answer; reads from stdin and since the input is guaranteed to be between 0 and 10^9, we just have to check n between 0 and 26. • I never remember scan(). Good work. – BLT Jun 17, 2017 at 22:10 # APL (Dyalog), 9 bytes To cover the case of n = 1, it requires ⎕IO←0 which is default on many systems. ⊢∊1+⍳×2*⍳  Try it online! ⊢ [is] n (the argument) ∊ a member of 1 one + plus ⍳ the integers 0 … (n-1) × times 2 two * to the power of ⍳ the integers 0 … (n-1) • So, "default on many systems" means that it just exists? Aug 3, 2017 at 22:03 • @Zacharý Yes, it would be wrong to call ⎕IO←0 non-standard, as many indeed always have it set like that, with no specification necessary each time. – Adám Aug 4, 2017 at 4:10 • Well. I'm going to use that trick in MY for sure (and MY can have non-0 and non-1 Index Origins) if I ever get the chance to. Aug 4, 2017 at 20:37 • @Zacharý Wouldn't that require an actual install base/versions where those values are default? E.g. in SAX and ngn, ⎕IO←0. – Adám Aug 5, 2017 at 22:04 • Yeah, I guess it would. And MY does have three iotas, so I think that wouldn't be used anyways. Aug 5, 2017 at 22:12 # Python 2, 32 bytes [n<<n|1for n in range(26)].count  Try it online! Creates the list of Cullen numbers up to 10^9, then counts how many times the input appears in it. Thanks to Vincent for pointing out n<<n|1 instead of (n<<n)+1, saving 2 bytes. • You can save two bytes using n<<n|1 (n<<n being even) ;) Jun 18, 2017 at 9:41 • This fails for 838860801. You need range(26), because the range isn't inclusive. Jun 19, 2017 at 16:52 • @mbomb007 Thanks. I've done this for a while and still sometimes forget ranges are right-exclusive. – xnor Jun 19, 2017 at 16:55 # TI-BASIC, 17 bytes max(Ans=seq(X2^X+1,X,0,25  Explanation seq(X2^X+1,X,0,25 Generate a list of Cullen numbers in the range Ans= Compare the input to each element in the list, returning a list of 0 or 1 max( Take the maximum of the list, which is 1 if any element matched  • You might want to add an explanation to this. Aug 1, 2017 at 23:58 • Done, thanks for the tip. Aug 2, 2017 at 0:12 • That works, but a command-by-command explanation usually helps garner most upvotes. I would reccomend doing something like the explanation on this answer. I don't know why somebody downvoted your post though. It's usually common courtesy to leave a comment when you do so, although that idea is often ignored. Aug 2, 2017 at 0:24 • Your welcome. I remember when I first joined the site, people told these types of things to me. Just passing on the favour. Aug 2, 2017 at 0:32 # D, 65 bytes This is a port of @HatsuPointerKun's algorithm to D (the original was already D code, but this is with D-specific tricks) T c(T)(T n){for(T i;i<30;++i)if((1<<i)*i+1==n)return 1;return 0;}  # How? (D specific tricks) D's template system is shorter than C++'s, and can infer types. And D also initializes its variables to the default upon declaration. # Nibbles, 5.5 bytes (11 nibbles) ?.,$+~*^2


Returns 1-based index in list of Cullen numbers (truthy), or zero (falsy) if not a Cullen number.

?.,$+~*^2 # full function ?.,$+~*^2$# (with implicit arguments shown): . # map over ,$           # 0..input-1
# with function:
~         #   1 (default)
#   to
^2      #   2 to the power of
$# input *$    #   multiplied by input
?           $# and get the index of the input in this # (or zero if not found) # Julia 1.0, 23 bytes ~n=n∈0:n.|>x->x*2^x+1  Try it online! # Mathematica, 30 bytes MemberQ[(r=Range@#-1)2^r+1,#]&  Pure function taking a nonnegative integer as input and returning True or False. If the input is n, then (r=Range@#-1) sets the variable r to be {0, 1, ..., n-1}, and then r2^r+1 vectorially computes the first n Cullen numbers. MemberQ[...,#] then checks whether n is an element of the list. # Mathematica, 32 bytes !Table[x*2^x+1,{x,0,#}]~FreeQ~#&  # Excel VBA, 45 Bytes Anonymous VBE immediate window function that takes input from cell [A1] and ouputs to the VBE immediate window Must be run in a clean module or have values for i,j be reset to default value of 0 between runs While j<[A1]:j=(i*2 ^ i)+1:i=i+1:Wend:?j=[A1]  ### Input / Output I/O as seen in VBE immediate window [A1]=25 While j<[A1]:j=(i*2 ^ i)+1:i=i+1:Wend:?j=[A1] True [A1]=1: i=0:j=0 ''# clearing module values While j<[A1]:j=(i*2 ^ i)+1:i=i+1:Wend:?j=[A1] True [A1]=5: i=0:j=0 ''# clearing module values While j<[A1]:j=(i*2 ^ i)+1:i=i+1:Wend:?j=[A1] False  ## Swi-Prolog, 69 bytes f(X) succeeds if it can find a value I where X = I*2^I+1. The range hint stops it running out of stack space, but it's enough for the range of Cullen numbers up to 10^9 in the question spec. :-use_module(library(clpfd)). f(X):-I in 0..30,X#=I*2^I+1,label([I]).  e.g. f(838860801). true  # cQuents, 9 bytes $0?1+$2^$


Try it online!

$0 n is 0-indexed ? Mode query. Given input n, output true/false for if n is in the sequence. 1+$2^$Each item in the sequence equals 1+index*2^index  # Arturo, 29 bytes $=>[in?&map 0..25'n->1+n*2^n]


Try it

# Raku, 30 bytes

{$_∈({1+$++*2**$++}...*>$_)}


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The parenthesized expression is a list of generated Cullen numbers, taken up to the point where a number is greater than the input argument (* > \$_). Then we just check that the input argument is a member of that list (∈).

# Japt-d, 7 bytes

ÑpU ¶NÉ


Try it

ÑpU ¶NÉ     :Implicit map of each U in the range [0,input)
Ñ           :Multiply by 2
pU         :Raised to the power of U
¶       :Is equal to
N      :Array of all inputs
É     :Minus 1
:Implicit output of whether or not any U returns true


# Vyxal, 6 bytes

ʀ:E*›c


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## Explained

ʀ:E*›c
ʀ:     # push 2 copies of the range [0, input]
E    # 2 ** x for x in ^
*   # vectorised multiply with the other copy
›  # + 1
c # contains the input?


# ThunnoD, $$\ 11 \log_{256}(96) \approx \$$ 9.05 bytes

RD2@z*s1-Aq


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#### Explanation

RD2@z*s1-Aq  # Implicit input
RD           # Push range(input) and duplicate
2@         # Pop one copy and push 2 ** each
z*       # Multiply elementwise
s1-    # Push the input decremented
Aq  # Is this in the list?


# Pyt, 8 bytes

Đ0⇹ŘĐ«⁺∈


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Đ             implicit input; Đuplicate
0⇹Ř          Řangify from 0 to n
Đ         Đuplicate
«        bitshift left
⁺       increment
∈      is input in list of Cullen numbers?; implicit print