13
Jelly, 6 bytes
R×\⁸¡Ṫ
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APL (Dyalog Unicode), 10 bytes
{⊃⌽×\⍣⍵⍳⍵}
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The main trick is to observe how the computation of x!y progresses as y increases.
1!0=1 2!0=2 3!0=3 4!0=4 ...
1!1=1 2!1=1*2 3!1=1*2*3 4!1=1*2*3*4 ...
1!2=1!1 2!2=1!1*2!1 3!2=1!1*2!1*3!1 4!2=1!1*2!1*3!...
11
Jelly, 4 bytes
R¡FP
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(Starting from [n],)
R Recursively replace each x with [1..x]
¡ n times
F Flatten
P Product
For example R¡ for the input 4 yields
[
[[[1]]],
[[[1]], [[1], [1, 2]]],
[[[1]], [[1], [1, 2]], [[1], [1, 2], [1, 2, 3]]],
[[[1]], [[1], [1, 2]], [[1], [1, 2], [1, 2, 3]], [[1], [1, 2], [1,...
11
Jelly, 6 bytes
3Ḋ*)SI
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-1 byte thanks to Jonathan Allan
1-indexed sequence without leading zeros. Uses the formula \$\sum_{k=1}^{n} 3^k - 2^k\$.
Explanation
3Ḋ*)SI Main monadic link
) For each k in the range [1..n]:
3Ḋ Remove the first element from the range [1..3]
* Raise each to the power of k
S Sum (producing [sum(2^k)...
11
Python, 20 bytes
lambda x,y:x*x>y*y-y
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Checks whether \$x^2>y^2-y \$.
Outputs True/False for whether x comes after y in the listing. Note that we're not asked handle the case x==y.
To see that this works, note that each of the x*x value in the table is greater than all the y*y-y values to the left of it, but none to the right of it.
...
9
Haskell, 43 bytes
f x=x!x
x!0=x
x!n=product$map(!(n-1))[1..x]
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+8 bytes due to the problem needing a function that only takes a single argument. Thanks to @Lynn for pointing that out
Thanks to OVS for pointing out map is shorter than a list comprehension and taking 1 byte off.
My first time ever submitting a haskell answer. I am a beginner so ...
8
Python 3 48 bytes
f=lambda x,y:f(x,y-1)*f(x-1,y)if x*y else(y>0)+x
TIO
55 bytes
if a more conventional input is required. I don't know how to initialize the default 2nd argument to equal the first, so I've rewritten everything as t=x-y without much more thought.
f=lambda x,t=0:f(x,t+1)*f(x-1,t-1)if(x-t)*x else(x>t)+x
Thanks to Arnauld for saving 3 ...
8
JavaScript, 58 bytes
a=>a.some((r,y)=>r.some((c,x)=>(r/(r=c)==a[x][x])/-y--<c))
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When c==0
r/(r=c) is NaN or Infinity;
(r/(r=c)==a[x][x]) is false
(r/(r=c)==a[x][x])/-y-- is 0 or NaN
(r/(r=c)==a[x][x])/-y--<c is false
When y==0 (cells on main diagonal) and c>=1
(r/(r=c)==a[x][x])/-y-- is NaN or Infinity
(r/(r=c)==a[x][x])...
8
R, 23 bytes
3^(n=scan()+2)/2-2^n+.5
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Outputs without the leading zeros, 0-indexed.
Uses formula from OEIS page:
$$ a(n) = 9 \cdot 3^n/2 - 4 \cdot 2^n + 1/2 $$
8
Jelly, 7 bytes
cþæ*2FS
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Now that xigoi has outgolfed me, I thought I'd share my answer.
This outputs the \$n\$ term of the sequence without leading zeroes.
How it works
We generate the matrix
$$\left[\begin{matrix}
\binom 1 1 & \binom 2 1 & \cdots & \binom n 1 \\
\binom 1 2 & \binom 2 2 & \cdots & \binom n 2 \\
\vdots &...
7
Python 3, 70 bytes
f=lambda a,b:b and f(b,a-b*((t:=a/b+.5+.5j).real//1+t.imag//1*1j))or a
7
Jelly, 14 13 bytes
ŒDµḢ=Ɲo@ƑḢ>FẠ
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-1 because I actually thought to check if the input can contain negative integers
ŒDµ Consider the diagonals of the input matrix.
Ḣ Pop the main diagonal;
Ɲ for each pair of its adjacent elements
= are they equal?
(Especially note ...
7
Alchemist, 76 bytes
_+0j->2c
b+0j->3d
0j+0_+0b->j
c+j+x->_+j
d+j->x+b+j
j+0c+0d->Out_x+Out_" "!b
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Outputs nonzero values infinitely using the formula \$S(n, 3) = \sum_{k=1}^{n-2} (3^k - 2^k)\$.
On TIO, this computes up to the 13th element (2375101) before timing out.
6
JavaScript (Node.js), 43 bytes
Expects and returns a BigInt.
f=(x,n=x,g=_=>x?f(x--,~-n)*g():1n)=>n?g():x
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This actually simplifies down to ...
JavaScript (Node.js), 37 bytes
f=(x,n=x)=>n?x?f(x,~-n)*f(~-x,n):1n:x
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... which is essentially the same as Alwin's answer.
6
Python 2, 24 bytes
lambda n:3**n/6-~-2**n/2
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Formula from OEIS for \$n>0\$:
$$ f(n) = \frac{1}{2}(3^{n-1}-2^n+1)$$
6
APL (Dyalog Unicode), 14 bytes
Anonymous tacit prefix function using xnor's formula. 1-indexed.
2÷⍨1-2∘*-3*-∘1
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6
Jelly, 9 bytes
Port of xnor's answer, made/fixed and golfed with hyper-neutrino's and caird coinheringaahing's help.
’3*_2*$‘H
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This doesn't handle leading zeroes.
’3*_2*$‘H
’ n-1
3* 3^(n-1)
_ From that, subtract
2*$ 2^n
‘ Increment that
H Halve it
6
Jelly, 12 bytes
3ẹ@þṗ¥ẠƇṢ€QL
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A monadic link that takes \$n\$ as its argument and returns \$a(n)\$. This is longer and less efficient than the answers based on the OEIS formulae, but should work for any value of \$k\$ by varying the first number in the link from 3.
Also, removing the L at the end yields the actual sets.
Explanation
3 ¥ ...
6
Jelly, 9 bytes
Surely there is an eight or less out there...
1×Ɱ3$¡FQS
A full program that accepts a non-negative integer from STDIN and prints the result using the 0-indexed, no leading zeros option.
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How?
1×Ɱ3$¡FQS - Main Link: no arguments
1 - start with x=1
¡ - repeat this (read from STDIN) times:
$ - last two ...
6
Desmos, 21 bytes
f(x,y)=\{xx-x<yy,0\}
(Newline is required)
Edit: Just noticed that xnor's answer is very similar to mines, but I found this independently.
Function is \$f(x,y)\$, and it returns \$0\$ if \$x\$ has a higher index, otherwise returns \$1\$.
This seems to work, but I just randomly stumbled upon it, so I can't explain how it works.
Try It On ...
6
J, 8 bytes
>&(-3^|)
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Or verbose: (x - 3 ^ abs(x)) > (y - 3 ^ abs(y)).
This maps the inputs to the following sequence to compare them in:
0 1 _1 2 _2 3 _3 4 _4 5 _5
_1 _2 _4 _7 _11 _24 _30 _77 _85 _238 _248
If sorted input as output is allowed, then 7 byte /:*.@%: is another fun way to sort /: the input list: *. ...
5
Community Wiki answer for trivial answers
Please add your language to this if the builtin GCD function correctly handles Gaussian integers.
J, 2 bytes
+.
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Wolfram Language (Mathematica), 3 bytes
GCD
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Pari/GP, 3 bytes
gcd
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5
Factor, 53 45 41 bytes
[ dup [1,b] [ cum-product ] repeat last ]
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A port of @Bubbler's answers; take the product scan (cumulative product) of 1..x x times and then return the last element.
5
Rattle, 111 bytes
|sI^>s[[PgI#1I#s2=#-#1[^0[^1g2[^0=q]][1g2[^0[^1=q]P=#4<s<=#3-sg0>I~<I~s_3P=#4+<s<=#3sg0>I~<I~<[^~=q]]]]g1]`]`=1
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Needless to say, Rattle is not built to handle matrices and this approach is pretty brute-force. However, this code really shows off most of Rattle's features!
Explanation
| ...
5
Scala, 132 bytes
n=>(for{i<-1 to n-1
j<-i+1 to n-1
p<-1.to(n).permutations}yield{Set(p.slice(0,i),p.slice(i,j),p.slice(j,n))map(_.toSet)}).toSet.size
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Is it short? No. Is it efficient? No. Is it clever? No. Why did I make it? I...don't know. I'll try golfing it later, if possible.
5
Husk, 7 bytes
F>m€Θݱ
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Returns 1 if the second is larger than the first, 0 otherwise
4
Jelly, 13 bytes
Thanks to @ovs for the fix.
%Lċ0
*ṗ’ÇƤ€QL
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The first line computes the shadow transform.
The second line looks at the shadow transform of all sequences of length n with elements in {1, 2, ..., n^n}.
4
Wolfram Language (Mathematica), 27 bytes
1##&@@#&//@Nest[Range,#,#]&
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Port of Lynn's solution.
Wolfram Language (Mathematica), 33 bytes
Nest[f1##&@@f~Array~#&,D,#]@#&
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Builds \$x!0...x!x\$:
D x => x!0
f1##&@@f~Array~#& (x => x!n) => (x ...
4
JavaScript (ES6), 46 bytes
x=>(g=p=>x>1n?x--**p*g(p*y++/i++):x)(i=1n,y=x)
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Use
$$ f\left(n\right) = \prod_{i=1}^n i^{{2n-i-1}\choose{n-1}} $$
with exception
$$ f\left(0\right) = 0 $$
4
Haskell, 41 bytes
f n=product$iterate(>>= \x->[1..x])[n]!!n
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Same idea as my Jelly answer: starting from [n], repeatedly replace each x by 1..x, n times, then take the product.
[4]
→ [1,2,3,4]
→ [1,1,2,1,2,3,1,2,3,4]
→ [1,1,1,2,1,1,2,1,2,3,1,1,2,1,2,3,1,2,3,4]
→ [1,1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,1,2,1,1,2,1,2,3,1,1,2,1,2,3,1,2,3,4]...
4
Jelly, 8 7 bytes
Ṭ€+\I’Ȧ
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-1 byte thanks to Jonathan Allan
Outputs an empty array (falsey) if it can be the result, and a non-empty array (truthy) if not.
How it works
Ṭ€+\I’Ȧ - Main link. Takes a list L on the left
€ - For each integer I in L:
Ṭ - Untruth; Generate a list of zeroes of length I, replace the last with 1
\ - ...
Only top voted, non community-wiki answers of a minimum length are eligible
