New answers tagged sequence
1
Pyth, 30 bytes
Fdr2Kt*2QVr2hQI!%J*Nt^QdK/JK.q
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Uses a similar approach to my Python 2 answer. Note: I also used the upper bound of 2n-2 on d that @Bubbler mentioned in a comment.
(Q)
Implicitly initialize Q to be the input
Kt*2Q
Initialize K to be 2Q-1
Fdr2K
For d in range(2, 2Q-1):
Vr2hQ
- For N in range(2, Q+1):
...
0
Bash + bc, 103 bytes
for((m=1;2*m!=p;));do((m++));t=$(bc<<<"obase=$1;$m");p=$(bc<<<"ibase=$1;${t: -1}${t::-1}");done;echo $m
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1
Python 2, 78 bytes
Simple direct solution. Increments x and d until we obtain the appropriate answer.
n=input()
j=x=d=2;k=2*n-1
while j%k:j=x*n**d-x;x=[2,x+1][x<n];d+=x<3
print j/k
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0
Charcoal, 41 bytes
Ｎθ≔⊖⊗θηＩ÷⌊ΦＥ×θη∨‹ι⊗θ∨‹﹪ιθ²×﹪ιθ⊖Ｘθ÷ιθ¬﹪ιηη
Try it online! Link is to verbose version of code. Uses the OEIS quote plus the additional information provided by @Bubbler in a comment whereby d < 2n-1. Explanation:
Ｎθ
Input n.
≔⊖⊗θη
Calculate 2n-1.
Ｅ×θη
Loop nd + x from 0 to n(2n-1).
∨‹ι⊗θ
Ensure d > 1 by returning 1 if it is ...
3
05AB1E, 15 11 bytes
∞.ΔxsIвÁIβQ
-4 thanks to @Grimmy
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explanation:
∞ get all positive numbers
.Δ find the first number for which:
xs the number doubled and itself (e.g. 64, 32)
Iв convert to base input (e.g. [1, 0, 1, 2])
Á rotate right (e.g. [2, 1, 0, 1])
Iβ convert back from base input to a number (e.g. 64)
...
1
Japt, 11 bytes
_ѶZsU'é}a1
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_ѶZsU'é}a1 :Implicit input of integer U
_ :Function taking an integer Z as an argument
Ñ : Multiply by 2
¶ : Check for equality with
ZsU : Convert Z to a string in base U
'é : Rotate right (string is converted back to decimal afterwards)
...
3
JavaScript (ES7), 90 88 86 85 bytes
A recursive port of the Maple function.
n=>(d=1,x=n,g=(a=2**++d-1,b=q=n+~-n)=>b?g(b,a%b):(x=q/a)<n?(x+!~-x)*(n**d-1)/q:g())()
Try it online! (\$a(3)\$ to \$a(9)\$)
Or, for +5 bytes, a BigInt version:
n=>(d=1n,x=n,g=(a=2n**++d-1n,b=q=n+~-n)=>b?g(b,a%b):(x=q/a)<n?(~-x?x:2n)*~-(n**d)/q:g())()...
3
Python 2, 77 76 70 bytes
f=lambda n,m=1,i=1:i<m and f(n,m,i*n)or(m+m%n*i)/n-m*2and f(n,m+1)or m
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Input: base n
Output: The smallest integer m satisfies the requirement.
Saved 6 bytes thanks to @Bubbler!
This is a brute-force search that starts at m = 1 and works its way up. Will run out of recursion limit if the actual solution is too ...
0
cQuents, 10 bytes
D9*_+$)[0]
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Just another port of Grimmy's answer.
$ # Index
_+ # +1
9* # *9
D )[0] # First digit
0
cQuents, 10 bytes
K\rJ$)))^2
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First time using cQuents.
Explanation
$ # Take the index.
J ) # Convert to (default) base 2.
\r ) # Reverse it.
K ) # Convert from (default) base 2 to base 10.
^2 # And square it.
0
C (gcc), 38 bytes
r;f(n){for(r=n%2;n/=2;r+=r+n%2);r*=r;}
Reverse all the bits, square the result. Pretty standard.
-1 byte thanks to ceilingcat!
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2
MathGolf, 4 bytes
àxå²
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Explanation:
à # Convert the (implicit) input-integer to a binary-string
x # Reverse this string
å # Convert it from a binary-string back to an integer
² # Take the square of this integer
# (after which the entire stack joined together is output implicitly as result)
2
05AB1E, 4 bytes
bRCn
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Explanation
Implicit input
b Convert to binary
R Reverse
C Convert from binary
n Square
Implicit output
0
Pyth, 13 bytes
A(Z1)#HA(H+HG
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Explanation
A(Z1)#HA(H+HG
(Z1) : The tuple (0, 1)
A : Assign the first value of the tuple to G and the second to H
# : Loop until error statement
H : Print H
(H+HG : The tuple (H, H + G)
A : Assign the first value of the tuple to G ...
2
05AB1E, 48 47 bytes
g≠iā<DδmUεXøINǝ}Xšεā<sUœε©2.ÆíÆ.±Xε®Nèè}«P}O}ć÷
Sometimes 05AB1E's lack of almost all matrix builtins is pretty annoying.. ;)
Inspired by @Arnauld's JavaScript answer.
Try it online or verify almost all test cases (removed the last two largest ones, since they time out on TIO).
Explanation:
First handle the edge case of a ...
1
APL (Dyalog Unicode), 10 bytesSBCS
⊢⌹∘.*⍨∘⍳∘≢
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A port of Graham's APL+WIN solution into a modern APL, which happens to work exactly the same (and have the same byte count) as my own J solution.
How it works
⊢⌹∘.*⍨∘⍳∘≢ ⍝ Input: V, result of a polynomial evaluated at 0..m-1
⍳∘≢ ⍝ Generate 0..m-1
∘.*⍨∘ ⍝ Self outer product by * ...
1
Haskell, 77 bytes
h%(a:t)=h-a:a%t
h%_=[h]
f(h:t)=h:foldr(%)[](f$zipWith((/).(-h+))t[1..])
f e=e
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0
Jelly, 14 bytes
J’*þ`æ*-⁸æ×ær0
A monadic Link accepting a list of integers which yields a list of the exponents (floats and/or integers) with the lowest degree on the left of the same length as the input (with trailing zeros if need be).
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How?
J’*þ`æ*-⁸æ×ær0 - Link: list of integers, V
J - range of length (V)
’ - ...
3
J, 10 bytes
%.^/~@i.@#
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Obligatory J answer on a matrix-related challenge. Takes input as a vector of extended integers (otherwise the answer may have small floating-point errors), and gives the polynomial's coefficients in lowest-first order, possibly with some extra zeroes at the end.
How it works
%.^/~@i.@# NB. Input: a vector V of ...
2
Charcoal, 68 62 bytes
≔⟦¹⟧ηＦＬθ«⊞υ⁰≔÷⁻§θιΣＥυ×κＸιλ∨ΠＥι⊕κ¹ζＵＭυ⁺κ×ζ§ηλ⊞η⁰≔Ｅη⁻§η⊖λ×κιη»Ｉυ
Try it online! Link is to verbose version of previous version of code that excludes trailing zeros, but apparently it isn't necessary to do that, thus saving 6 bytes. Outputs the terms in power order i.e. the constant term is printed first. Explanation:
≔⟦¹⟧η
Start by ...
1
MATL, 12 bytes
n:qGyz3$ZQYo
The result is given with higher-order coefficients first, and may contain leading zeros.
Try it online! Or verify all test cases
Explanation
Consider input [-3, -1, 5, 15, 29] as an example.
n:q % Implicit input. Number of elements. Range. Subtract 1, element-wise
% STACK: [0, 1, 2, 3, 4]
G % Push input again
...
1
SageMath, 63 48 bytes
lambda v:QQ[x].lagrange_polynomial(enumerate(v))
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Outputs the polynomial as
$$a_k n^k + \cdots + a_3 n^3 + a_2 n^2 + a_1 n + a_0 $$
2
Python 3 + Numpy, 69 bytes
lambda x:polyfit(range(len(x)),x,len(x)-1).round()
from numpy import*
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May have leading zeros.
5
R, 55 52 bytes
-3 bytes thanks to Giuseppe.
round(solve(outer(n<-seq(a=u<-scan())-1,n,"^"))%*%u)
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Outputs \$(a_0, a_1,\ldots,)\$ with possible trailing zeros.
Let \$u\$ be the output sequence, and \$X\$ be the \$m\times m\$ matrix such that \$X_{i,j}=i^j\$ (0-indexed), i.e.
\$
X=\begin{pmatrix}
1&0&0&\ldots&0\\
1&...
2
Pari/GP, 38 bytes
a->Vecrev(polinterpolate([0..#a-1],a))
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4
Wolfram Language (Mathematica), 50 49 37 bytes
Returns a polynomial.
Mathematica is so awesome x+1 can be used as a variable in this context. Apart is a weird built-in that, quoting from the docs, seems to attempt to rewrite an expression as a sum of terms with minimal denominators, and also happens to expand polynomials (that are returned in a weird ...
8
JavaScript (ES7), 193 ... 154 145 bytes
Saved 9 bytes thanks to @Bubbler
Returns \$(a_0,a_1,...,a_k)\$ with some possible trailing zeros.
v=>v.map((_,i)=>(g=(i,m=v.map((n,y)=>v.map((_,x)=>x==i?n:y**x)))=>+m||m.reduce((s,[v],i)=>v*g(0,m.map(([,...r])=>r).filter(_=>i--))-s,0))(i)/g())
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(removed the ...
4
APL+WIN, 16 bytes
Index origin = 0
Prompts for input as a vector and outputs coefficients from a0 to an-1 where n is the length of the vector. The order of the polynomial can be obtained by summing the number of coefficients up to the last none zero coefficient:
0⍕n⌹m∘.*m←⍳⍴n←,⎕
Try it online! Courtesy of Dyalog Classic
0
R, 12 10 bytes
Producing two different sequences with the correct limiting frequencies whether \$s=0\$ or \$s\ne 0\$:
(!s):n%%10
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0
W, 2 bytes
(Port of the Charcoal answer.) How in the world can a 3-byter ever compress in the compressor‽
H≡
Uncompressed:
!)[
Explanation
! % Logical negate the first input.
) % Increment the negated input.
[ % Cyclic index the second input by that result.
1
MATLAB, 45 bytes
v=input('');rng(fix(~v(1)));randi(10,v(2),1)-1
randi(N,n,1) generates an n-length sequence of uniform random integers in the range 1,...,N. Taking two digits of the resulting sequence of realizations at a time, the required distribution follows naturally, which can be checked using the example code below.
N = 50000 ;
[h,x] = hist(10*(...
1
dc, 15 13 bytes
Saved 2 bytes thanks to Mitchell Spector!!!
[A/]sa?0=aA%p
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Input: \$n\$ and seed (\$s\$).
Output: \$n^{\text{th}}\$ term zero-indexed.
For \$s>0\$ the sequence is: \$0,1,2,\dots,9,\dots\$
For \$s=0\$ the sequence is: \$\underbrace{0,0,\dots,0}_{10},\underbrace{1,1,\dots,1}_{10},\underbrace{ 2,2,\dots,2}_{10},\dots,\...
1
Retina 0.8.2, 16 bytes
^(0,.+).
$1
!`.$
Try it online! Link includes test cases. Takes input as s,n and outputs the second last digit of n if it has one and s is zero otherwise the last digit of n. The individual sequences are 10% of each digit but the combined sequence only approaches 1% after the 10th term. (I have a 16 byte answer in Retina 1 for which ...
3
Bash, 25 23 bytes
Saved 2 bytes thanks to Mitchell Spector!!!
echo $[$2/($1?10:1)%10]
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Input: seed (\$s\$) and \$n\$.
Output: \$n^{\text{th}}\$ term zero-indexed.
For \$s=0\$ the sequence is: \$0,1,2,\dots,9,\dots\$
For \$s>0\$ the sequence is: \$\underbrace{0,0,\dots,0}_{10},\underbrace{1,1,\dots,1}_{10},\underbrace{ 2,2,\dots,2}_{...
2
JavaScript (Node.js), 15 bytes
s=>n=>n[!s+1]|0
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If seed is 0 it will take the 3rd digit of the number, else it will take the 2nd digit of the number
0
perl -apple, 18 bytes
$F[0]&&chop;$_%=10
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Prints the last digit of n for the special sequence (c = 0), otherwise, the penultimate digit of n. It doesn't quite work for single digit n's, but since we only have to tend towards equal distribution, it doesn't matter what it prints.
3
JavaScript (ES7), 19 bytes
Takes input as (seed)(n) and returns the \$n\$-th term. The special seed is \$0\$.
s=>n=>n/10**!s%10|0
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2
Charcoal, 9 5 bytes
§Ｓ⊕¬Ｎ
Try it online! Link is to verbose version of code. Takes input as n, s and outputs the (1-indexed) third digit of n if s is zero otherwise the second digit of n (so somewhat similar to @ExpiredData's answer, although this was unintentional). The two sequences separately always output an exact 10% frequency after a power of 10 ...
5
C (gcc), 24 bytes
f(s,n){s=(s?n/10:n)%10;}
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Input: seed (\$s\$) and \$n\$.
Output: \$n^{\text{th}}\$ term zero-indexed.
For \$s=0\$ the sequence is: \$0,1,2,\dots,9,\dots\$
For \$s>0\$ the sequence is: \$\underbrace{0,0,\dots,0}_{10},\underbrace{1,1,\dots,1}_{10},\underbrace{ 2,2,\dots,2}_{10},\dots,\underbrace{9,9,\dots ,9}_{10},\dots\...
4
05AB1E, 4 3 bytes
ΘÍè
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Takes the seed then n and outputs the nth term, outputs different sequence for seed of 1
Explanation
Θ - truthified (so 1 if input is 1, else 0)
Í - subtract 2 (so -1 if input is 1, else -2)
è - take this index of the nth term
Since 05AB1E uses modular ...
4
Wolfram Language (Mathematica), 38 bytes
Takes Input [n,seed] and outputs nth term
Specific seed is 0
If[#2>0,#~Mod~10,Floor[#~Mod~100/10]]&
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9
Python 2, 23 22 bytes
Thanks @xnor for figuring out a really cool way to convert the seed to -1 or -2, which saved 1 byte.
lambda s,n:`n+9`[2/~s]
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Input: The seed s (non-negative) and an index n (positive).
Output: The element at the nth index (the sequence is one-indexed)
If the seed is positive, the sequence is 0123456789 repeated ...
0
APL (Dyalog Unicode), 8 10 bytesSBCS
⊢↑∘∊⊣,/∘⍳⌈
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A dyadic train. Left arg = interval (n), right arg = number of terms (x). ⎕IO←1.
How it works
⊢↑∘∊⊣,/∘⍳⌈ ⍝ Left: interval n, Right: terms x
∘⍳⌈ ⍝ Generate range [1..max(n,x)]
⍝ (Too short array will complain at N-wise reduce)
⊣,/ ⍝ Extract length-n intervals (...
0
Io, 50 bytes
Port of Jonathan Allan's answer.
method(n,x,Range 0 to(x-1)map(v,(v/n)floor+v%n+1))
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0
APL (Dyalog Extended), 13 bytes
{⍺↑2-2|⍸⍵}⍣≡⍨
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Basically a port of 13-byte J code at the bottom of algorithmshark's J answer ($1+2|I.)^:_~.
Takes an integer n and gives first n terms.
How it works
{⍺↑2-2|⍸⍵}⍣≡⍨ ⍝ Input: n
{ } ⍝ At each step,
⍸⍵ ⍝ For each item a_i at index i, give a_i copies of i
2-2| ⍝ ...
0
dirt -v, 48 bytes
"122,11"|.*`,(1',.*(1'2|2'1)|2',.*(1"22"|2"11"))
Continuously outputs longer and longer Kolakoski sequences, with a , in the middle that moves right as the sequence is generated:
122,11
1221,12
12211,21
122112,122
1221121,221
12211212,2122
122112122,12211
1221121221,22112
12211212212,211211
122112122122,1121122
1221121221221,1211221
...
0
APL (Dyalog Extended), 17 bytes
{⍵⊃≢¨⊆⍨1+∊⊤¨⍳+⍨⍵}
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Gives nth term, 1-indexed.
How it works
{⍵⊃≢¨⊆⍨1+∊⊤¨⍳+⍨⍵}
{ } ⍝ ⍵←n
+⍨⍵ ⍝ Double of n
⍳ ⍝ 1 .. 2n, inclusive
∊⊤¨ ⍝ Convert each to binary and flatten
1+ ⍝ Add 1
⊆⍨ ⍝ Partition self into non-increasing segments
...
1
W x, 7 bytes
REALLY slow. The array is 1-indexed and it outputs all upto the input. (Glad that I tie with Husk BTW. Special bonus: it doesn't involve infinite lists!)
♫│x╤►U╟
Uncompressed:
^k2BLHkr
Explanation
^ % 10 ^ input. Make sure that enough items are calculated.
k % Find the length range of that.
2B % Convert every item to ...
0
Wolfram Language (Mathematica), 70 bytes
Tr[1^Join@@Partition[Split[Join@@IntegerDigits[Range[2#],2]],2][[#]]]&
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0
Zpr'(h, 369 bytes
s |> \
(g (foldr (op-> ++) () (map b |N)))
(e ())|>o
(e (S ()))|>z
(e (S (S .n)))|>(e n)
(h ())|>()
(h (S ()))|>()
(h (S (S .n)))|>(S (h n))
(b ())|>()
(b (S .n))|>((b (h (S n))) ++ (' (e n) ()))
(l (' z (' o .r)))|>1
(l (' z (' z .r)))|>(S (l (' z r)))
(l (' o (' o .r)))|>(S (l (' o r)))
(l (' o (' ...
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