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11:00 AM
@Cowsquack better?
 
btw how long is your solution?
 
@Cowsquack 6 bytes, same as yours
 
@cairdcoinheringaahing What is the correct order?
No
I figured it out
@cairdcoinheringaahing WHy doesn't this work?
Ʋ‘8×
ÇÐfḶ
 
@Cowsquack for the one which demonstrates the ÇÐf method, 11 bytes
@Mr.Xcoder "times by 8, increment, is square?"
Yours is "is square, increment, times by 8"
 
@cairdcoinheringaahing This doesn't work either:
8בƲ
ÇÐfḶ
Nor does this:
×8‘Ʋ
ÇÐfḶ
 
11:05 AM
@Mr.Xcoder "8 times 1 + is square?" for the first one, and the second link is wrong for the second one
 
@cairdcoinheringaahing Hooray, thanks:
×8‘Ʋ
çÐfḶ
@cairdcoinheringaahing Is there a more golfy way?
 
@Mr.Xcoder yes. Quicks are the way to reduce it
 
@cairdcoinheringaahing I am on the right track though :)
 
@Mr.Xcoder you should be able to remove 4 bytes
 
@Mr.Xcoder what is this for?
 
11:09 AM
@cairdcoinheringaahing 3 bytes... Give me the very last character
 
@Mr.Xcoder sorry, the solution should be 6 bytes
 
@LeakyNun Finding triangular numbers below the input
 
@Mr.Xcoder hint: what is a triangular number?
 
@LeakyNun can you beat 6 bytes?
 
@cairdcoinheringaahing of course
 
11:10 AM
@LeakyNun A number that can form an equilateral triangle.
I have an idea anyway
 
@Mr.Xcoder that's rather vague; can you make it more concrete?
 
On other thoughts, a new post for OEIS answer chain has been posted
 
@LeakyNun a(n) = a(n-1)+(a(n-1)-a(n-2)+1)
 
@Mr.Xcoder simpler?
 
@LeakyNun a(n)=2a(n-1)-a(n-2)+1
 
11:12 AM
@Mr.Xcoder simpler
 
@cairdcoinheringaahing Even more simple?
 
@Mr.Xcoder yep
 
@cairdcoinheringaahing If the challenge was about finding triangulars less than or equal to n, then I have an obfuscated beast of 10 bytes
 
@Cowsquack I don't think it works: [1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5]
 
@Cowsquack I have literally no idea what that does
 
11:15 AM
*less than or equal to
@Mr.Xcoder oh, it finds the number of triangular numbers less than or equal to n for all numbers from 1 to the argument
 
@cairdcoinheringaahing @LeakyNun I know a very tiny bit of Jelly since today, so can you give a more concrete hint?
 
@Mr.Xcoder do you want the simplest formula?
 
@cairdcoinheringaahing yes
 
@Mr.Xcoder a(n) = n + a(n-1)
 
@cairdcoinheringaahing Ah, of course, didn't notice
Wait, I now must write something recursive in... Jelly?!
 
11:18 AM
@Mr.Xcoder my 6 byte answer doesn't use recursion, so no.
 
@Mr.Xcoder \
 
@LeakyNun I was just about to post a solution when you've posted the Cheddar
 
@Mr.Xcoder sorry :p
 
@cairdcoinheringaahing the formula for the nth triangular number can be described as $\frac{n(n+1)}{2}$, this can be simplified into $\frac{(k+0.5)(k-0.5)}{2}=\frac{k^2-0.25}{2}$ where $k$ is $n+0.5$. The Jelly program finds $k$ by doing $\sqrt{2*x+0.25}-0.5$ and then floors the result.
 
what language did you use? @Mr.Xcoder
 
11:22 AM
@LeakyNun Swift
 
@Cowsquack meanwhile I can do it in 6 bytes
@Mr.Xcoder maybe you can do the next sequence swiftly :P
 
do what, or <?
 
@LeakyNun No, I've read it, I cannot
 
@LeakyNun well, I did say "obfuscated"
 
Can you guys give me a simpler task in Jelly?
 
11:24 AM
find the nth triangular number
 
@cairdcoinheringaahing - or +?
 
@Mr.Xcoder do Cows Quack's
@Cowsquack I got 3 bytes
 
@Cowsquack Jelly, 4 bytes: ב:2
 
5-byte spoiler does not use non-printables
 
I got a different three bytes @cairdcoinheringaahing
 
11:26 AM
There is H for Halve
 
@Mr.Xcoder can you get 3 bytes?
 
Jelly, 3 bytes: בH
 
@Cowsquack what was yours?
 
using only ASCII characters?
 
3 mins ago, by Cows quack
find the nth triangular number
2 bytes please
 
11:28 AM
@LeakyNun come on, you can get that to 3 bytes easily
@cairdcoinheringaahing a 3-byter using only ascii characters
 
@Cowsquack no, that was the previous challenge
for your challenge, I can do it in 2 bytes
 
@LeakyNun Uses a built-in we haven't used?
 
@Mr.Xcoder no such builtin
 
@Mr.Xcoder it isn't a built-in
 
Ok,ok, I forgot we are talking about Jelly, sorry :)
 
11:30 AM
oh obviously 2 bytes
 
@Cowsquack Is there something for n+n/2?
 
so @Mr.Xcoder try doing it in 2 bytes of Jelly
@Mr.Xcoder There is an easier way
 
@Cowsquack Does it use another formula?
 
@LeakyNun how do you remove duplicates from a list?
 
@cairdcoinheringaahing Q
 
11:33 AM
Does it use Æ?
 
@Mr.Xcoder +H$
@Mr.Xcoder I said it isn't a built-in
 
only two atoms
 
@LeakyNun @Cowsquack +Ḥ
No, does not Work
 
@Mr.Xcoder its 0-indexed
 
H+Ḥ
 
11:35 AM
visualise the problem in another way
 
2 easy bytes: RS
 
well done
you can use this to do caird's triangle CMC as well
 
Lol, abuse range built-ins
 
@LeakyNun at that time I forgot +/ becomes S lol
 
What's jelly's map?
Is it ?
 
11:38 AM
@Mr.Xcoder yes
 
@LeakyNun how do you apply 2 atoms to one argument? e.g. x => [min(x),max(x)]
 
@cairdcoinheringaahing there's no short way
 
How do you apply a monad over a range? Like S mapped to ẆR?
 
@cairdcoinheringaahing Ṣ.ịU
@Mr.Xcoder what do you mean?
ẆR doesn't make much sense
 
@LeakyNun I noticed, I missed the point
 
11:42 AM
but you would append S€€
@Mr.Xcoder you mean I missed the point?
 
@LeakyNun No, I missed tthe point
 
@Mr.Xcoder what point?
 
I wanted to solve it in a way, but picked the wrong things
Really, NVM
 
@Qwerp-Derp sa-zaap-go hun-diong...
 
I will have lunch now, I'll be back shortly
 
11:45 AM
Hey, if I make a meta post that holds deleted posts from this OEIS , so people can at least see what someone spent 30 minutes on, would that be in line?
or would it be a main post
 
@StepHen it would probably be closed as off-topic
You could create a github repository though
 
that's what I was afraid of
you're right
brb
 
@StepHen i said "close to impossible"
 
@cairdcoinheringaahing if the sequence only goes to 10, why not hardcode it?
 
@StepHen hardcoding works, but there's no formula for it so
@LeakyNun do you think Mr. Xcoder should apply for Hypertraining?
 
11:54 AM
Guys I found a flame war :P
 
Hi again
 
@Mr.XCoder Hello
 
@Mr.Xcoder apply for hypertraining
 
@cairdcoinheringaahing Sure, hopefully I won't be kicked because I am too lazy
 
@cairdcoinheringaahing What's hypertraining?
 
11:55 AM
@cairdcoinheringaahing Applied

 Jelly Hypertraining

Practice your Jelly :) Rules and stuff are here: golfingsucces...
 
@Mr.Xcoder I'm pretty sure they don't do that @Qwerp-Derp Jelly training
 
@cairdcoinheringaahing I like rule #6
> Dennis has to be a student too, because, well, if he just knows Jelly the best he isn't afraid of proving it, is he?
 
@Mr.Xcoder genuinely made me laugh
 
@cairdcoinheringaahing You're already level 1, wow
@cairdcoinheringaahing Only Leaky and Erik can grant me access, right?
CMC: Given a list and a positive integer, return all the combinations of the list with the length equal to the number. 1 ≤ integer ≤ length of the list
 
I may or may not maintain it, but I'll accept PRs
 
12:18 PM
Are there while loops in Jelly?
 
@Qwerp-Derp Yes, ¿
Use it as such: <body><condition>¿
 
@Mr.Xcoder use this:
func bit_sum_parity(n: Int) -> Int {
    var res = 0;
    var test = n;
    while test > 0 {
        res ^= test%2;
        test /= 2;
    }
    return res;
}

func A000028(n: Int) -> Int {
    var res = 1;
    var iter = n;
    while iter > 0 {
        var clone = res;
        var p = 2;
        var parity = 0;
        while clone > 1 {
            var count = 0;
            while clone % p == 0 {
                clone /= p;
                count += 1;
            }
            parity ^= bit_sum_parity(n: count);
 
@LeakyNun For what? The next OEIS?
 
@Mr.Xcoder yes
 
@LeakyNun Doing it now
 
12:23 PM
@Mr.Xcoder I should have used prime sieve to optimize it first though :p
just post it anyway
 
@LeakyNun Wait, that's Swift? How many languages do you know?
 
@Mr.Xcoder I don't know swift
I just searched for "function" and "loops" here
 
@LeakyNun OK
 
@Mr.Xcoder btw, link
 
Hardcoding in a language as dedicated to computation as rust... you earned my downvote. — Leaky Nun 1 hour ago
my "scorched earth" tactics go unappreciated >◡>;
 
12:27 PM
@Lynn :p
 
@LeakyNun Posted
0
A: One OEIS after another

Mr. Xcoder43. Swift, 702 bytes, A000028 func bit_sum_parity(n: Int) -> Int { var res = 0; var test = n; while test > 0 { res ^= test%2; test /= 2; } return res; } func A000028(n: Int) -> Int { var res = 1; var iter = n; while iter > 0 { var clone ...

 
@Mr.Xcoder did you count the main function into the byte length?
 
@LeakyNun yes, I think
 
@Mr.Xcoder I mean, the footer
 
This is it:
func bit_sum_parity(n: Int) -> Int {
    var res = 0;
    var test = n;
    while test > 0 {
        res ^= test%2;
        test /= 2;
    }
    return res;
}

func A000028(n: Int) -> Int {
    var res = 1;
    var iter = n;
    while iter > 0 {
        var clone = res;
        var p = 2;
        var parity = 0;
        while clone > 1 {
            var count = 0;
            while clone % p == 0 {
                clone /= p;
                count += 1;
            }
            parity ^= bit_sum_parity(n: count);
 
12:29 PM
You shouldn't count the last part in
 
@LeakyNun Ah yeah, didn't really look at that
@LeakyNun Fixed
0
A: One OEIS after another

Mr. Xcoder43. Swift, 651 bytes, A000028 func bit_sum_parity(n: Int) -> Int { var res = 0; var test = n; while test > 0 { res ^= test%2; test /= 2; } return res; } func A000028(n: Int) -> Int { var res = 1; var iter = n; while iter > 0 { var clone ...

 
trying to do this in SOGL, 21 bytes for the gray node creation..
 
1:05 PM
Why partitions
I give up on this challenge
 
@totallyhuman I did it for you
The next sequence is so easy
@totallyhuman @Mr.Xcoder @Qwerp-Derp
 
@LeakyNun I cannot post it yet
 
@totallyhuman please
 
@LeakyNun is that the answer to A190?
 
@cairdcoinheringaahing yes
post it for me please
 
1:17 PM
Sure why not?
 
thanks
 
@LeakyNun Like, half the challenges answers there are yours
 
@Mr.Xcoder yep
 
@LeakyNun Look on Hypertraining
 
I've gained rep from 4 posts all on OEIS chain
 
1:21 PM
@cairdcoinheringaahing Me too
 
@Mr.Xcoder I thought you've done more than 4
 
@cairdcoinheringaahing Nope
 
who wants to post the next answer?
 
@cairdcoinheringaahing I actually have only 3
 
(Not that I have already written it)
 
1:22 PM
@LeakyNun Leaky Nun.
 
@LeakyNun none of us can
 
@LeakyNun I'll post it
 
@BlackCap ok, wait a minute
 
Leaky's just giving us the answers, then we're claiming them as our own. Seems right
 
@cairdcoinheringaahing you can always give me credit :)
 
1:23 PM
@LeakyNun I've added courtesy of Leaky Nun once
then, it's self-evident
 
@dzaima do you want to take a crack at this with SOGL?
 
@LeakyNun I beat you to it
 
@BlackCap nice
@BlackCap where's your answer?
 
dupes, huh?
 
@LeakyNun what dupes?
 
1:46 PM
@cairdcoinheringaahing maybe, the next sequence seems easy enough. So the AWK answer is valid, right?
 
@dzaima yeah I think so
 
@dzaima yes it is
 
Does anyone want to post I1+4*3-£1+2/o in my language √ å ı ¥ ® Ï Ø ¿?
 
√ doesn't work in the online interpreter..
 
@dzaima for SOGL?
 
1:49 PM
@cairdcoinheringaahing yeah; that's what you get for trying to make a javas bigDecimal replacement :p
 
AAAH! I have 30 minutes before I can post!
 
I still have 13 minutes before I can post my answer
 
Should I just delete the draft of my next answer?
 
@cairdcoinheringaahing or just give it to us
 
1:55 PM
@cairdcoinheringaahing it doesn't work, they say
 
# 48. [√ å ı ¥ ® Ï Ø ¿](https://github.com/SatansSon/UnprintableName), 13 bytes, [A000194](https://oeis.org/A000194)

I1+4*3-£1+2/o

[Next sequence](https://oeis.org/A000013)
 
@cairdcoinheringaahing I couldn't get your √ å ı ¥ ® Ï Ø ¿ code to work
 
It works on my version here. Ill check the differences
 
omg if I post mine, the next sequence will be Fib
 
1:56 PM
@LeakyNun we've had that
 
@cairdcoinheringaahing right
 
I have a (3-byte) SOGL answer for A000194 too
 
5
A: One OEIS after another

HyperNeutrino30. Python 1, 1112 bytes, A000046 def rotations(array): rotations = [] for divider_index in range(len(array)): rotations.append(array[divider_index:] + array[:divider_index]) return rotations def next(array): for index in range(len(array) - 1, -1, -1): array[index] = 1 - array[index] ...

underrated answer (5 upvotes is not enough!)
you're gonna kill me if I make it 47 bytes
 
@LeakyNun why?
 
@cairdcoinheringaahing because it's one of the x^2 and y^2 challenges again
bye, you guys can post or not post it, I can't post it now anyway
 
1:58 PM
@LeakyNun try to avoid it then
 
just add some spaces somewhere
 
so.. should I post mine?
 
@BlackCap should be fixed now
 
@cairdcoinheringaahing Can you please join me on Hypertraining?
 
No output
 
2:04 PM
Is there an o at the end of the code?
 
oh, I can't use a file, right
 
@BlackCap what do you mean?
 
I assumed that the first argument was a file with code
 
python <name>.py code input is how to use it from command line
 
You should print that if the program is run without arguments
 
2:13 PM
gg there's no way I'll solve the next sequence
 
the Number of Primes of the Form n^2 + a
 
I really want to reach 50 answers :)
 
2:33 PM
CMC: Print 9999999999
 
@Mr.Xcoder +[--------->+<]>..........
 
@LeakyNun What language, Brainf***?
 
@Mr.Xcoder yes
 
4 bytes: *T`9
Your turn
 
,,,, 5 bytes: '910×
 
print(int(1e10-1)) also :(
 
Also 4 bytes: t^TT
Now, who else writes another CMC?
 
2:37 PM
@Mr.Xcoder no, we keep on doing it in different languages
 
v['9.]v*'9<^9999999999
 
Dyalog APL, 6 bytes, 10⍴'9'
 
I do it in Jelly
 
9<insert quote character>10x
 
2:38 PM
who wants to take turns with me?
 
sure
 
@LeakyNun turns?
 
(though I know I will run out of langs soon)
 
@Mr.Xcoder I post one language, you post one language
 
Oh, I do
 
2:39 PM
@Mr.Xcoder your turn
 
malbolge: D'`%ML[Z}Y9WDCwRRQra);L9lH#G4ggU#Ay>w<^)\rwvutm3210/.-,+*u
 
cQuents: "#10::9
 
@LeakyNun Without the ones above, right?
 
@Mr.Xcoder sure
 
@LeakyNun Swift: print("9999999999")
 
2:41 PM
@LeakyNun 9dddddddddO
 
Röda, {["9"*10]}
 
@Mr.Xcoder 3 4-byte SOGL answers: LL^H, UO⁹‘, ` 9L*`
 
@cairdcoinheringaahing does the name have a meaning?
 
@LeakyNun ”9ẋ10
 
2:42 PM
Processing, print("999999999");
 
@LeakyNun they're "special" commands in the language
 
10.times{$><<9}
$><<999999999
¯\_(ツ)_/¯
 
$><<'9'*10
 
2:43 PM
@LeakyNun Ha get golfed, and I don't even know Nim: echo(int(1E10-1))
 
@StepHen neither do I
 
@LeakyNun I'm doing a trio with you and Mr. XCoder btw
 
Dear comment section, how to say 9999999999.
Today I learned how to do say 9999999999!
I sang "9999999999"!
That's all about how to do say 9999999999.
Your faithful student, BlackCap.
PS: FiM++
 
2:44 PM
V, 10é9.
 
@Mr.Xcoder why the unevaluated form?
 
@LeakyNun print10**10-1
(Remove the space)
 
@Phoenix tried, doesn't work
 
Really?
Huh
 
yes
 
2:45 PM
@LeakyNun unevaluated?
 
@LeakyNun HODOR: Walder
Hodor Hodor Hodor Hodor Hodor Hodor Hodor Hodor Hodor
HODOR!! HODOR!! HODOR!! HODOR!! HODOR!! HODOR!! HODOR!! HODOR!! HODOR!! HODOR!!
HODOR!!!
 
@Phoenix otherwise it could be a variable name
 
@StepHen ?
That version is above
 
since variables can contain numbers
 
5 mins ago, by Mr. Xcoder
@LeakyNun Swift: print("9999999999")
 
2:46 PM
it's Leaky's turn now @LeakyNun
 
right
 
Hey, someone solved A03
 
> Number of odd integers <= 2^n of form x^2 + y^2.
 
baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet
baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaa
 
2:50 PM
ok then
 
@Mr.Xcoder @Cowsquack
 
JS, a=>1e10-1
 
2 mins ago, by Mr. Xcoder
@LeakyNun R: print("9999999999")
And
50 secs ago, by Leaky Nun
baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet
baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa bleeeeeeeeeet baaaa
Lol
 
@Downgoat would be proud
 
2:52 PM
@LeakyNun @Cowsquack 9999999999
 
wait is it leaky's or my turn now?
 
@Cowsquack Leaky's
 
@Mr.Xcoder T°<
 
@LeakyNun You beat me to that
 
Vim, 10i9.
 
2:53 PM
2sable, 9999999999
C++:
#include<iostream>
int main(){std::cout<<"9999999999";}
Your turns
 
I gtg so, dc 10 10^1-p. bc print 10^10-1. Bash echo 9999999999. Carrot 9^*9 ><> '9999999999'>o<. Fission 2, R"9999999999". sed s/$/9999999999/. ///, 9999999999
 
hold on, I'm trying to think of interesting solutions
 
C++:
 
@Downgoat I managed to solve the os is not defined issue earlier, you were missing a require('os');. Why it worked on other systems I have no idea, but after making the change and rebuilding, ./vsl -dast put me into this weird prompt thing I don't understand.
 
#include<stdio.h>
int main(){puts("9999999999");}
 
2:57 PM
yes 9|head|tr -d '\n'
 
Wow, that's hard: oeis.org/A000074
 
@Mr.Xcoder main(){puts("9999999999");} is sufficient, in GCC at least. Try it online!
 
I really got bored of 9999999999, so let's pick another CMC…
 
@Mr.Xcoder how is that hard
 
2:59 PM
@LeakyNun It's hard when all the languages I know are gone
@LeakyNun Your CMC turn
 
@Mr.Xcoder I mean, the sequence you linked to
@Mr.Xcoder do A74 :p
 
@LeakyNun Don't.
 
CMC: Given a directory name, recursively remove all files in it, but keep the directory structure.
 
@Phoenix This is 100% made for bash and sed
 

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