New answers tagged sequence
4
Python 3.8, 53 51 bytes (thanks @user for pointing out extra spaces)
def r(n):
i=n
while(a:=n**(1/i))%1:i-=1
return a
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3
05AB1E, 6 bytes
Inspired by SunnyMoon's answer.
LÀ.ΔBR
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L # push the range [1, 2, ..., n]
À # rotate the 1 to the back: [2, 3, ..., n, 1]
.Δ # find the first integer where ...
B # the input converted to that base
R # reversed
# implicit: is equal to 1 as a number
05AB1E, 8 bytes
LÀ.Δ.n.ï
...
2
05AB1E, 12 (10) bytes
ÓDā<Ør0š¿÷mP
Port of @UnrelatedString's Jelly answer.
The 0š shouldn't be necessary, but unfortunately there is a bug in 05AB1E for ¿ with empty lists.
Try it online or verify all test cases.
Explanation:
Ó # Get the prime exponents of the (implicit) input-integer
D # Duplicate this list of exponents
ā ...
5
R, 37 33 bytes
-4 bytes thanks to Dominic van Essen
match(T,!log(n<-scan(),1:n)%%1,1)
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A different (and shorter) approach than the one used in Giuseppe's and Dominic van Essen's R answers.
Finds the first integer k such that log(n,k) is an integer, or returns 1 if there is no such k (which corresponds to the special case n=1).
0
Perl 5 -p, 40 bytes
print$_,("@{[1-$_..0]}"=~/\d/g)[0..$_-1]
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1
PowerShell Core, 45 bytes
param($a)"$a$(($a-1)..0-join''|% su* 0 $a)$a"
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Explanation
($a-1)..0-join''|% su* 0 $a # builds an array from a-1 to 0, joins them with no separators
# and takes the `a` first chars
"$a ... $a" # surrounds the result with the parameter `...
0
Stax, 9 8 7 bytes
îσQçφ╠♦
Run and debug it
-1 byte from recursive.
Explanation (Unpacked)
rr$x%sy|S
r range 0..n-1
r reversed
$ converted to string
x( take n characters
y push n as string from register y
|S prepend and append it to the range
implicit output
3
Stax, 10 bytes
┌Pèó~JRå▲ï
Run and debug it
Explanation
R{xs|E|+1=}j
R range 1..n
{ }j get first number i where:
xs|E input(x) in base i digits
|+ summed
1= equals 1
3
Factor, 49 bytes
[ dup [1,b] 2dup '[ _ swap _ n^v member? ] find ]
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Slow for larger inputs because it tries to evaluate a large power.
[
dup [1,b] 2dup ! ( n 1..n n 1..n )
'[ ! Put stack items in the `_`s in the quotation
_(n) swap _(1..n) ! ( elt -- n elt 1..n )
n^v member? ! Test if n appears ...
1
Wolfram Language (Mathematica), 36 bytes
#^Last[#<2||1/GCD@@FactorInteger@#]&
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-11 bytes from @att
5
JavaScript (ES7), 36 bytes
Recursively looks for the highest \$i\le n\$ such that \$k=n^{1/i}\$ is an integer. Then returns this \$k\$.
f=(n,i=n)=>(k=n**(1/i))%1?f(n,i-1):k
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JavaScript (ES7), 37 bytes
A slightly longer version that performs more recursive calls but is not subject to rounding errors.
n=>(g=k=>k**i-n?g(k-1||i--|n):k)(...
4
R, 47 bytes
n=scan();match(n,sapply(0:n,"^",1:n))%/%n-(n<2)
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Struggled for ages trying to beat Giuseppe's answer, only to be totally outgolfed (seconds before posting) by Robin Ryder's comment (now an answer)...
2
Octave, 33 bytes
@(n)[~,j]=find((t=1:n)'.^t'==n,1)
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How it works
@(n) % anonymous function with input n
(t=1:n) % let t = [1, 2, ..., n] (row vector)
.^ % element-wise power with broadcast...
' % of t transposed...
...
7
Husk, 8 bytes
VȯεΣ`B¹ḣ
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V # index of first truthy element of
ȯ # applying 3 functions to
ḣ # 1...input
`B¹ # convert input to this base
Σ # sum of digits
ε # is at most 1
3
Japt -g, 10 bytes
I feel like I'm missing a trick here.
õ ï æ@¥Xrp
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õ ï æ@¥Xrp :Implicit input of integer U
õ :Range [1,U]
ï :Cartesian product with itself
æ :Get first pair that returns true
@ :When passed through the following function as X
¥ : Test U for equality with
Xr ...
2
Charcoal, 19 16 bytes
ＮθＩ⊕⌕Ｅθ⎇ιΣ↨θ⊕ιθ¹
Try it online! Link is to verbose version of code. Edit: Now back to a reformulation of my original answer. Works by converting n to each base 1..n and finding the first 1-indexed value with a digit sum of 1. Conveniently this automatically works for an input of 0 (the resulting list is empty, so the 1-indexed position ...
5
Retina 0.8.2, 60 bytes
.+
$*
(?<=(?=((?=((1*)(?=\5\3+$)1)(\2*$))\4)*1$)^(..+)).*
1
Try it online! Link includes test cases. Explanation:
.+
$*
Convert to unary.
(?<=(?=...$)^(..+)).*
Delete the earliest suffix leaving behind the smallest prefix \$r\$ (captured into \5) such that the \$n\$ matches the following:
((?=((1*)(?=\5\3+$)1)(\2*$))\4)*1
...
10
J, 14 bytes
(%+./)&.(_&q:)
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(%+./)&.(_&q:)
&.(_&q:) number to prime exponents
(%+./) divide them by their GCD
&.(_&q:) prime exponents to number
3
C (gcc), 65 60 58 bytes
Saved 5 bytes thanks to Sisyphus!!!
p;i;r;f(n){for(r=0;r++<n;)for(p=i=1;i++<n;p*=r)n=p-n?n:r;}
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5
k4, 26 24 21 bytes
-2 bytes by ignoring n=0 case
-3 bytes by applying
@caird coinheringaahing's logic
{(+/'(!x).q.vs\:x)?1}
Benefits from list ? value returning the count of the list if the value isn't present in it, and from convenient weirdness with the n=0 and n=1 edge cases.
2
Scala, 50 bytes
Back to 50 bytes because n=0 doesn't have to be handled anymore!
n=>1.to(n)find(r=>1.to(n)exists(n==math.pow(r,_)))
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7
Python 3, 55 \$\cdots\$ 59 57 bytes
Added 7 bytes to fix an error kindly pointed by user.
Saved 3 bytes thanks to user!!!
lambda n:{r**i:r for i in range(n)for r in range(n+1)}[n]
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0
05AB1E, 10 bytes
Î<ŸRJ¹£¹.ø
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Î<ŸRJ¹£¹.ø # full program
.ø # surround...
£ # first...
¹ # input...
£ # chars of...
J # joined elements of...
Ÿ # [...
Î # 0 ... input...
< # minus 1
Ÿ # ]...
R # reversed...
.ø # with...
¹ ...
8
Brachylog, 6 bytes
1|~^hℕ
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1|~^hℕ with the implicit input n
1 input and output is 1
| or
~^ find two numbers [r, i] so that r^i = n
h return r
ℕ to limit the search space: r must be positive
Search tries lowest i first, so we get the maximum r for free.
7
05AB1E, 8 11 8 bytes
-3 thanks to @ovs!
L¦BíCXkÌ
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I am trying to somehow implement a log function to check whether a number matches the regex 10*, but that is too mathematical for me...
Wait, how?
L # Push all numbers natural numbers up to input [1, 2, 3 ... I]
¦ # What is that 1 doing there? Remove it! ...
4
R, 49 44 bytes
n=scan();which(outer(1:n,n:1,"^")==n,T)[1,1]
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Thanks to Dominic van Essen for pointing out a bug.
n <- scan() # read input
arr <- outer(0:n,1:n,"^") # create the array of powers (0^(1:n), 1^(1:n), ... n^(1:n))
arr <- t(arr) # transpose, so the array is ((0:n)^1, (0:n)^2, .....
8
Jelly, 10 bytes
ÆEgƒ0:@ƊÆẸ
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ÆE:g/$ÆẸ errors given 1.
ÆE Take the exponents of the input's prime factorization.
:@Ɗ Divide each exponent by
gƒ0 the exponents' GCD (or 0 in the case that there are none).
ÆẸ Let the result be the exponents of the output's prime factorization.
5
Jelly, 8 6 5 bytes
bR§i1
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Uses the fact that n in base r has the format 100...000, meaning that the sum of the digits only equals 1 in base r
-1 byte (indirectly) thanks to Dominic van Essen's answer, make sure to give them an upvote
How it works
bR§i1 - Main link. Takes n on the left
R - [1, 2, 3, ..., n]
b - Convert n to each base 1, ...
2
Jelly, 10 bytes
ŒṗḟÐḟÆḞ€ṪP
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Times out for \$n > 60\$
How it works
ŒṗḟÐḟÆḞ€ṪP - Main link. Takes n on the left
Œṗ - Integer partitions of n
ÆḞ€ - First n Fibonacci numbers
Ðḟ - Keep the partitions p which yield an empty list under:
ḟ - Remove all elements of p which are in the first n Fibonacci numbers
...
0
Jelly, 11 bytes
ḶUDFṁ;D;@DḌ
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There is also this kinda cheaty 9 byter, which only works due to how the Y command in the Footer works, so I'm not counting that.
Jelly's dislike of strings really hurts it here, and CJam's stack is just helpful enough to beat Jelly.
How it works
ḶUDFṁ;D;@DḌ - Main link. Takes n on the left
Ḷ - [0, 1, 2, ....
0
Scala, 80 bytes
n=>Stream.iterate(1)(n*_)map(p=>p to p*n-1 find(x=>2*x==x%n*p+x/n))find(_!=None)
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Takes an integer n, returns an Option[Option[Int]] (really a Some[Some[Int]]).
0
SNOBOL4 (CSNOBOL4), 71 bytes
Y =1
R X =X + 1
Y =Y * X GT(X,Y) :S(O)
Y =Y / X
O OUTPUT =Y :(R)
END
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Prints the sequence forever.
1
Crystal, 41 bytes (-2 bytes thanks to Giuseppe)
def a(n)n<2?1: n>(f=a(n-1))?n*f: f//n end
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2
PowerShell Core, 65 bytes
Output the nth value:
$r=1
1..$args[0]|%{$r=([Math]::Floor($r/$_),($r*$_))[$_-gt$r]}
$r
Explanations
$r=1 # Initialises the result as 1
1..$args[0]|%{ # For 1 to the argument
$r=( # In an array calculate the two possible values
[Math]::Floor($r/$_), # ...
1
Jelly, 5 bytes
UQU⁻Q
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Just another port of xnor's answer, so go upvote that.
How it works
UQU⁻Q - Main link. Takes a list l on the left
U - rev(l)
Q - unique(rev(l))
U - rev(unique(rev(l)))
Q - unique(l)
⁻ - rev(unique(rev(l)) ≠ unique(l)
4
Java (JDK), 52 bytes
n->{int i,a=i=1;for(;i++<n;)a=i>a?i*a:a/i;return a;}
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Note: Thanks @RedwolfPrograms for -1 Byte and @user for -10(?) bytes.
1
R, 93 92 bytes
Edit: -1 byte thanks to Giuseppe
f=function(x)`if`(sum(x|1)<2,(-1)^(sum(x)!=2),`if`(!all(x%in%1:2),-1,(y=f(rle(x)$l))+(y>0)))
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Returns 1-based number of iterations for truthy input, and -1 for all falsy inputs. This needed quite careful input testing, especially for the last two test cases...
Commented:
f=function(x) ...
0
Scala 3, 460 bytes
This may be the longest answer here, but it's also the fastest one here...because it works at compile time.
import compiletime.ops.int._
type r[T,P]=T match{case h*:t=>r[t,h*:P]case E=>P}
type E=EmptyTuple
type F[K,R,L,S,P,I]=K match{case E=>L-S match{case 0=>R match{case E=>0 case _=>F[r[R,E],E,0,0,0,I+1]}case _=>0}...
4
Husk, 6 bytes
Ḟ§|*÷ṫ
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How?
Ḟ§|*÷ṫ - function: integer, n
ṫ - reversed range -> [n,n-1,...,2,1]
Ḟ - right-fold using this f(a,b):
§ - fork:
| - v logical-OR w
÷ - v: b integer-divide a
* - w: a multiplied by b
3
Hy, 54 bytes
(defn f[n](or(+(= n 1))(//(f(- n 1))n)(*(f(- n 1))n)))
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This function outputs the nth value of the sequence.
Ungolfed version:
(defn f [n]
(or (+ (= n 1))
(// (f (- n 1))
n)
(* (f (- n 1))
n)))
1
Stax, 26 bytes
äE⌐+É7∩ΦΓyr╔ßΣ·φƒÇe►ef%I»
Run and debug it
Tried with a generator, but it seems like a while loop is shorter.
3
APL (Dyalog Unicode), 56 54 49 bytes (SBCS)
Saved 2 7 bytes thanks to @ovs!
0∘{×⍵:r(,⍣d)(r,⍺)∇⍵-d←⍺∊⍨r←(⊃((≤∨⍺∊⍨-)⌷-,+)≢)⍺⋄⍬}
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Can be f k. Recursively builds up the first k duplicates.
⍺ holds Recamán's sequence (in reverse), and is set to 0 if no argument is given (at the start). If k (⍵) is 0, it returns an empty array (⍬). Otherwise, it ...
5
Forth (gforth), 51 bytes
: f 1+ 1 tuck ?do i 2dup <= if * else / then loop ;
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Code Explanation
: f \ start word definition
1+ \ add 1 to n
1 tuck \ set up accumulator and loop parameters
?do \ loop from 1 to n (if n > 1)
i 2dup \ set up top two stack values and duplicate
<= if \ if a(n-1) <= n
...
0
Jelly, 23 bytes
“J,1’D=þD§S×ẠƊ
0Ç<CɗƓ#Ṫ
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This takes h on the command line and n on STDIN
How it works
0Ç<CɗƓ#Ṫ - Main link. Takes h on the left
ɗ - Group the previous 3 links into a dyad f(k, h):
Ç - Call the helper link on k
< - Less than h?
C - Complement; 0 -> 1, 1 -> 0
0 Ɠ# - Read an ...
4
MathGolf, 11 9 bytes
1k{î`<¿*/
-2 bytes thanks to @ovs.
Outputs the \$n^{th}\$ value.
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Explanation:
1 # Push 1
k{ # Loop the input amount of times:
î # Push the 1-based loop index
` # Duplicate the top two items
<¿ # If the current value is smaller than the 1-based loop index: a(n-1)<n:
...
0
Perl 5 -p, 82 bytes
$;{1}++;map@;{$_*2+1,$_*3-1}=1,keys%;while$_*3>keys%;;$_=(sort{$a-$b}keys%;)[$_-1]
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2
Pyth, 20 bytes
J1V}2hQJ=J?>NJ*NJ/JN
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1-indexed, outputs the first n values.
0
Husk, 17 bytes
!OuΞ¡ṁ§eo←*3o→D;1
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There should be a shorter way using set difference.
This answer creates all the possibilities, merges them and deduplicates.
0
R, 122 bytes
function(n)while(n){T=T+1
m=T^2
t=0
while(m<-m-1){i=s=0;while((i=i+1)<m)if(!m%%i)s=s+i;t=t+(T==s)}
if(!t){print(T);n=n-1}}
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Anyone familiar with R programming will certainly wince when they look at this. Loops in general are horribly inefficient in R, and the fast and idiomatic way to tackle this kind of question would ...
1
Jelly, 16 bytes
ŒṖḌI’ḂƑƊƇḢ.ịr/ḟƲ
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Input as a list of digits. Outputs a singleton list [a] containing the missing element, or [] if there isn't one. +1 byte to output 0 instead, and +1 byte to input as a single integer.
How it works
ŒṖḌI’ḂƑƊƇḢ.ịr/ḟƲ - Main link. Takes a list of digits D on the left
ŒṖ - All partitions of D
Ḍ ...
Top 50 recent answers are included
