Am I a Cullen Number?

Question

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

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

Your Task:

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 code-golf, so the lowest score in bytes wins.

Answer 1

Pyth, 6 5 bytes

/mh.<

try it online

Answer 2

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.

Answer 3

Jelly, 7 6 bytes

Ḷæ«`i’

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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.

Answer 4

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))))

Answer 5

Haskell, 28 bytes

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

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Answer 6

R, 53 51 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.

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Answer 7

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?

Answer 8

PHP, 43 bytes

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

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Answer 9

05AB1E, 7 bytes

ÝDo*¹<å

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Explanation:

ÝDo*¹<å Example input: 9. Stack: [9]
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: [1]
        Implicit print.

Answer 10

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;}

Answer 11

Python 2, 36 bytes

f=lambda n,i=0:i<<i!=n-1and f(n,i+1)

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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

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Outputs via exit code

Answer 12

Python, 40 bytes

lambda n:any(x<<x==n-1for x in range(n))

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Answer 13

R, 26 bytes

a=0:26;scan()%in%(1+a*2^a)

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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.

Answer 14

APL (Dyalog), 9 bytes

To cover the case of n = 1, it requires ⎕IO←0 which is default on many systems.

⊢∊1+⍳×2*⍳

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⊢ [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)

Answer 15

Python 2, 32 bytes

[n<<n|1for n in range(26)].count

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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.

Answer 16

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.

Answer 17

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.

Answer 18

Mathematica, 32 bytes

!Table[x*2^x+1,{x,0,#}]~FreeQ~#&

Answer 19

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 

Answer 20

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

Answer 21

cQuents, 9 bytes

$0?1+$2^$

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Explanation

$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`

Answer 22

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

Answer 23

QBIC, 24 bytes

[0,:|~a*(2^a)+1=b|_Xq}?0

Explanation

[0,:|           FOR a = 0 to b (input from cmd line)
~a*(2^a)+1=b    IF calculating this a results in b
|_Xq            THEN quit, printing 1
}               NEXT a
?0              We haven't quit early, so print 0 and end.

Answer 24

k, 19 bytes

{&1=x-{x**/x#2}'!x}

Try it online. Truthy is an array with a number in it: ,3 or ,0 et cetera. Falsey is an empty array: () or !0 depending on your interpreter.

Answer 25

Java (OpenJDK 8), 56 bytes

i->{int n,c;for(n=0;(c=n*(2<<n++-1)+1)<i;);return c==i;}

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Answer 26

Pari/GP, 25 bytes

n->!prod(i=0,n,n-i*2^i-1)

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