4

Jelly, 13 bytes Thanks to @ovs for the fix. %Lċ0 *ṗ’ÇƤ€QL Try it online! 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}.


3

Python 2 (PyPy), 228 225 bytes This is based on the first PARI implementation on OEIS and computes terms up to \$n=6\$ on TIO. import math,itertools as I L=math.log R=range n=input() P=k=r=1 while k<n:k+=1;r*=k**int(P%k*L(n+.5)/L(k));P*=k*k print len({tuple(sum(x%o<1for x in s[:o])for o in R(n))for s in I.product(*[[i+1for i in R(r)if-1<r%~i]]*~-n)}...


1

Java (OpenJDK 8), 74 bytes a->{for(int c=0,i=-1,j=0;c<a.length;)a[c]=a[c++]==1?i+=2:(j+=2);return a;} Try it online!


1

Pyth, 10 bytes !%/.BQ"1"2 Try it online! Trying to learn Pyth, so this is my go at it. How it works: Convert the input to binary, count the number of "1"s, check whether they're even, invert the output (so it checks whether they're odd. In more detail: ! # negate output (0 -> 1, 1 -> 0) % # check ...


1

Vyxal, 5 bytes ǐ3v∴Π Try it Online! ǐ # Prime factors v∴ # Take max of each and... 3 # 3 Π # Take product


1

Japt, 20 bytes @TwXµY *!ZøX ªX+YÑ}h Try it


1

Python 2, 65 bytes n=a=0 v=[] exec'a+=[n,-n][a+n in v];print a;n-=1;v+=a,n;'*input() Try it online! Avoids negative numbers by adding them to the list of already visited numbers.


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