One OEIS after another
22. FiM++, 982 bytes, A000024
Note: if you are reading this, you might want to sort by "oldest".
Dear PPCG: I solved A000024!
I learned how to party to get a number using the number x and the number y.
Did you know that the number beers was x?
For every number chug from 1 to y,
beers became beers times x!
That's what I did.
Then you get beers!
That's all about how to party.
Today I learned how to do math to get a number using the number n.
Did you know that the number answer was 0?
For every number x from 1 to n,
For every number y from 1 to n,
Did you know that the number tmp1 was how to party using x and 2?
Did you know that the number tmp2 was how to party using y and 2?
Did you know that the number max was how to party using 2 and n?
tmp2 became tmp2 times 10!
tmp1 became tmp1 plus tmp2!
If tmp1 is more than max then: answer got one more.
That's what I did.
That's what I did.
Then you get answer!
That's all about how to do math.
Your faithful student, BlackCap.
PS: This is the best answer
PPS: This really is the best answer
Next sequence
73. Starry, 363 bytes, A000252
, + + * '. `
+ + + + * * * +
+` +* + `
+ + + + + + * '
+ ' ####` + + +
+ + #### +* + *
' ##### + + '
` ######+ + + +
+ + + ######### * '
+ + + #####+ + +
* + + * + * * +
+ * + + + + * *
+ + + * + ` + +
+ + + + *' + +.
Try it online!
Next sequence
Uses the formula "a(n) = n^4 * product p^(-3)(p^2 - 1)*(p - 1)
where the product is over all the primes p that divide n" from OEIS.
The moon's a no-op, but hey, this isn't code-golf.
1. Triangular, 10 bytes, A000217
$\:_%i/2*<
Try it online!
Next Sequence
How it works
The code formats into this triangle
$
\ :
_ % i
/ 2 * <
with the IP starting at the $
and moving South East (SE), works like this:
$ Take a numerical input (n); STACK = [n]
: Duplicate it; STACK = [n, n]
i Increment the ToS; STACK = [n, n+1]
< Set IP to W; STACK = [n, n+1]
* Multiply ToS and 2ndTos; STACK = [n(n+1)]
2 Push 2; STACK = [n(n+1), 2]
/ Set IP to NE; STACK = [n(n+1), 2]
_ Divide ToS by 2ndToS; STACK = [n(n+1)/2]
\ Set IP to SE; STACK = [n(n+1)/2]
% Output ToS as number; STACK = [n(n+1)/2]
* Multiply ToS by 2ndToS (no op); STACK = [n(n+1)/2]