26
Ruby, 26 bytes
->n{~-n*12-496/4**n%4+1/n}
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Revised version adding 1/n and subtracting 496/4**n%4 to get the +1,-3,-3,-1 offset for the first 4 terms.
Ruby, 32 bytes
->n{n>4?~-n*12:[0,1,9,21,35][n]}
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After 4, the sequence settles down to (n-1)*12. See diagram below (the equilateral triangles have been distorted into ...
9
JavaScript (ES6), 23 bytes
Based on Level River St's answer.
n=>[1,5,13,7][--n]^n*12
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How?
We compute \$(n-1)\times12\$ and adjust the first 4 values with a XOR.
$$\begin{array}{c|c}
n&1&2&3&4&5&6&7&8&9&10\\
\hline
(n-1)\times12&0&12&24&36&48&60&72&84&96&...
9
05AB1E, 9 bytes
<©12*3®cα
Try it online! or try a test suite.
< # input - 1
© # save to register
12* # multiply by 12
® # push the register
3 c # binomial coefficient(3, input - 1)
α # absolute difference
With 0-indexing, this would be 7 bytes:
12*3Icα
8
JavaScript (ES6), 91 ... 76 74 bytes
A recursive approach. This is very fast up to \$n=5\$. Computing \$a(6)\$ takes approximately 2 minutes on my laptop.
n=>(g=a=>1+(h=i=>i&&h(i-1)+(a[i]^i&&a[i-1]?g(b=[...a,0],b[i]++):0))(n))`1`
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How?
Instead of explicitly building the sequences, we just keep track of ...
8
Pyth, 12 bytes
f}hQ_XsjT;H1
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f: Count up from 1 to find the first number such that:
X ... H1: Increment or insert the value 1 into the dictionary H at index
sjT;: sum of base ten digits of current number
_: values of dictionary
}hQ: check whether input + 1 is contained within.
The first time this is true, exactly input numbers must have the ...
8
Jelly, 39 37 bytes
Soon to be crushed by MATL and, quite possibly, APL
p`œc⁸ḣ1ạ§ỊẸʋƇ@;QɗƬƊṪṢƊƑƇŒṬZṚŒṪƲƬṂ$€QL
Try it online! (It's pretty slow - a(6) took ~30 minutes locally!)
To see them instead try this (L -> ŒṬ€G€j⁾¶¶).
How?
Builds all index lists, where each represents the locations of 1s on a grid of 1s and 0s, up to those for a square ...
7
Pyth, 13 bytes
lfUI{T{.ysmRh
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Let's go through this step by step, starting at the end:
mRh: This autoexpands to mRhdQ. This means "Map the function m with right input hd over the input Q."
Q is an integer (e.g. 3) so it gets automatically cast to a range (e.g. [0, 1, 2]).
The m function is the map function.
d is defined to be the input to ...
7
Jelly, 8 bytes
’3cạ×ʋ12
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How?
’3cạ×ʋ12 - Link: integer, n
’ - decrement (n-1) 1 2 3 4 5 6 7 ...
12 - twelve 12 12 12 12 12 12 12 12 ...
ʋ - dyad:
3 - three 3 3 3 3 3 3 3 3 ...
c - ...
7
Python, 31 bytes
lambda n:n*12-11-(n>4or 5%-n%5)
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5
JavaScript (Node.js), 269 ... 250 243 bytes
Rather slow for \$n\ge7\$, but it does find \$a(8)=704\$ in a bit more than 2 minutes on my laptop.
f=(n,m=[...o=Array(w=n)],i=c=0)=>n?m.map((r,y)=>m.map((_,x,[...m])=>!i|1<<x&~r&(m[y+1]|r/2|r*2)&&f(n-1,m,m[y]|=1<<x)))|c:[0,0,0,0].some(_=>o[M=(m=m.map((_,y)=>...
4
Jelly, 18 bytes
x`€FŒ!QJƑ$Ƈ¹Ƥ€ẎQL‘
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An initial attempt at a Jelly answer. A monadic link taking an integer and returning an integer. Too slow for n>3 on TIO.
Explanation
x`€ | Repeat each of 1..n itself times
F | Flatten
Œ! | Permutations
$Ƈ | Keep those where the following ...
4
R, 113 106 98 bytes
g=function(n,N=1:n,B=N*0){for(e in N[c(1,B)[-n-1]>0&(F=all(B<=N))])F=F+g(n,,'[<-'(B,e,B[e]+1));+F}
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-9 bytes thanks to @RobinRyder
-2 bytes using @Arnauld idea of storing occurrences table only (I just noticed that basically I was using the same approach, but creating the actual list too)
4
Jelly, 9 bytes
ŻŒḂ€ḋṚ$HĊ
A monadic Link accepting a non-negative integer which yields a non-negative integer.
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How?
Counts all pairs without the \$a\leq b\$ restriction, halves and rounds up. Note that the halved count is only a fraction if \$\frac n 2\$ is a palindrome and in such cases we want to count this \$a=b\$ pair.
ŻŒḂ€ḋṚ$HĊ - ...
4
Jelly, 15 9 bytes
’3cạ×12$Ʋ
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A monadic link taking \$n\$ as its argument and returning \$a(n)\$.
Based on @LevelRiverSt’s clever Ruby answer so be sure to upvote that one too!
Thanks to @Grimmy for saving 6 bytes!
Explanation
’ | Subtract 1
Ʋ | Following as a monad
3c | - Number of ways of picking (n-1) items from 3
...
4
APL (Dyalog Unicode), 14 bytesSBCS
12(|×-3!⍨⊢)-∘1
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Direct translation of Jonathan Allan's Jelly answer. Even the code structure is the same. Jelly is a golfy descendant of APL; if you want to learn Jelly, learn APL first!
How it works
12(|×-3!⍨⊢)-∘1 ⍝ Monadic train, input: n
12( )-∘1 ⍝ Pass on to the inner function with left←12 and ...
4
brainfuck, 50 bytes
>>>>>>+<++<<----<+<,-[>++++++++++++[[>+<-]<]>>-]>.
Does i/o as raw byte values, like the others here.
4
Gaia, 13 10 bytes
1⟨┅Σ¦ṅC=⟩#
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Straightforward implementation.
Explanation:
1⟨ ⟩# | find the first 1 positive integers N where:
C | the count of
ṅ | the digital sum d(N)
┅Σ¦ | in the list [d(1)..d(N-1)]
= | is equal to the (implicit) input
3
Python 2, 73 70 63 bytes
lambda n:sum(`n-v`+`v`==(`v`+`n-v`)[::-1]for v in range(n/2+1))
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Note that:
(string_a == reverse(string_a)) and (string_b == reverse(string_b))
is equivalent to
reverse(string_a + string_b) == (string_b + string_a)
(where + is concatenation)
3
JavaScript (ES6), 74 73 bytes
n=>(g=a=>a>n-a?0:![a,n-a].some(n=>[...n+''].reverse().join``-n)+g(-~a))``
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3
APL (Dyalog Unicode), 62 bytesSBCS
Infinite printing full program, 63 bytes
s←⍎' '~⍨⍕
{⎕←s⊢a←⊃⍵⋄b←1+a⋄t←∪(b,⍨⊃a)(a,⊃⌽b)b,1↓⍵⋄t[⍋s¨t]}⍣≡⊂1 3
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This one is pretty fast. Theoretically can be made faster using a min heap, but the challenge is code golf after all.
How it works: the concept
If we denote each item in the sequence as a list of ...
3
Perl 6, 26 bytes
{$_*12-[-1,3,3,1][$_]}o*-1
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If this could be zero-indexed the o*-1 at the end can be removed. Returns (n-1)*12, offsetting the first 4 values.
3
Python 3, 40 36 bytes
lambda n:~-n*12-(*n*[0],1,3,3,-1)[4]
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3
Jelly, 10 bytes
D€§ċṪ$=ʋ1#
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A monadic link taking an integer \$n\$ and returning \$a(n)\$.
Explanation
ʋ1# | Find the first integer x where the following is true;
D€ | - Digits of 1..x
§ | - Sum each
ċṪ$ | - Count the number equal to the tail (after popping tail)
= | - Equal to (implicit original ...
3
05AB1E, 10 bytes
∞.Δ1Ÿ1öć¢Q
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3
APL (Dyalog Unicode), 27 bytesSBCS
{⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵}⍳+∘1
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How it works
{⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵}⍳+∘1 ⍝ Right argument: n
{ }⍳+∘1 ⍝ Find the first index of n+1 from...
⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵ ⍝ ... the list of cumulative count of own digit sum (sort of)
20×⍵ ⍝ Practical upper bound 20×...
2
05AB1E, 1 byte
η
Builtins ftw ¯\_(ツ)_/¯
Outputs as a list of prefixes.
Try it online or verify all test cases.
Explanation:
η # Push a list of prefixes of the (implicit) input-string
# (and output this list implicitly as result)
2
05AB1E, 19 18 17 bytes
ÝDÅΓœ€Œ€`ÙʒÙāQ}g>
Pretty slow (\$n=3\$ in about 5.5 0.5 seconds on TIO), but it works.
Inspired by @isaacg's Pyth answer.
I have the feeling this can definitely be golfed by at least a few bytes, though.
-1 byte thanks to @ExpiredData.
-1 byte and sped up by changing the order of some operations thanks to @Grimmy.
Try it online ...
2
Raku, 94 bytes
{1+set map ~*,grep {all .unique Z==^$_},.permutations>>[[\,] ^$_][*;*]}o{(^$_ Zxx 1..$_)[*;*]}
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Times out for \$n>3\$, mostly since it is running the same filter for all prefixes of all permutations of the largest list (i.e. [1,2,2,3,3,3,...n]). This results in a lot of duplicates that have to filtered out later.
2
Ruby, 100 bytes
Relatively simple answer similar to the Jelly solution, but it's able to calculate for n=4 without timing out on TIO (albeit taking 30 seconds to do so), which Jelly can't at time of writing.
->n{r=(1..n).flat_map{|j|[j]*j};i=0;r.sum{r.permutation(i+=1).uniq.count{|a|a.uniq==[*1..a.max]}}+1}
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2
Perl 6, 44 bytes
{+grep {.flip eq[R,] $_},(^$_ Z($_...$_/2))}
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Finds the number of pairs of numbers such that the reverse of the string representation is equal to the string representation of the reversed pair.
Only top voted, non community-wiki answers of a minimum length are eligible
