Shadow transform of a 0-based integer sequence \$a(n)\$ is another 0-based integer sequence \$s(n)\$ defined with the following equation:

$$ s(n) = \sum_{i=0}^{n-1}{(1 \text{ if } n \text{ divides } a(i), 0 \text{ otherwise})} $$

i.e. \$s(n)\$ is the number of terms in \$a(0), \cdots, a(n-1)\$ that are divisible by \$n\$.

\$s(0)\$ is always 0 because there are zero terms to consider, and \$s(1)\$ is always 1 because \$a(0)\$ is always divisible by 1. \$s(2)\$ may have a value of 0, 1, or 2, depending on how many terms out of \$a(0)\$ and \$a(1)\$ are even.


Given a non-negative integer \$n\$, compute the number of distinct shadow transforms of length \$n\$. This sequence is A226443.

The following is the list of first 11 terms for \$n = 0, \cdots, 10\$, as listed on the OEIS page.

1, 1, 1, 3, 12, 48, 288, 1356, 10848, 70896, 588480

Explanation: Let's call this sequence \$f(n)\$.

  • \$f(0)\$ counts the number of empty sequences, which is 1 (since [] counts).
  • \$f(1)\$ counts the possible number of [s(0)] which can only be [0].
  • \$f(2)\$ counts [s(0),s(1)]s which can only be [0,1].
  • Since s(2) can take any of 0, 1, or 2 independent of s(0) and s(1), \$f(3)\$ is 3.
  • s(3) is also independent of s(0) through s(2) (because 3 is relatively prime to 2) and take a value between 0 and 3 inclusive, so \$f(4) = 3 \cdot 4 = 12\$.
  • Finding \$f(5)\$ is slightly more complex because s(4) is tied with s(2). If s(4) == 4, all of a(0)..a(3) must be divisible by 4 (and therefore even), and s(2) can only be 2. If s(4) == 3, at least one of a(0) or a(1) must be even, and s(2) must be 1 or 2. Therefore, \$f(5) = 12 + 12 + 12 + 8 + 4 = 48\$.

Standard rules apply. I/O does NOT apply. The shortest code in bytes wins.

  • 3
    \$\begingroup\$ It is easy to generate all arrays with \$ n \$ elements each element in \$ 0 \dots \text{LCM}\left(1,\dots,n\right) \$ (to golf more bytes, I would choice \$1\dots n^n\$ instead), apply \$s\$ to it, and then count how many distinct results in all. But I don't know how to test it as it will not produce output in reasonable time even for \$n = 5\$. \$\endgroup\$
    – tsh
    Jul 9, 2021 at 6:27

2 Answers 2


Jelly, 13 bytes

Thanks to @ovs for the fix.


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}.

  • \$\begingroup\$ You're currently off by one, \$f(3)\$ should be \$3\$. µ -> would fix this at the cost of a bit of effiency. \$\endgroup\$
    – ovs
    Jul 9, 2021 at 13:01
  • \$\begingroup\$ @ovs fixed, thanks. \$\endgroup\$
    – Leaky Nun
    Jul 9, 2021 at 18:05
  • 2
    \$\begingroup\$ You can group the helper link with Ʋ and save a byte - *ṗ’%Lċ0ƲƤ€QL \$\endgroup\$ Jul 9, 2021 at 18:12

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
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)})

Try it online!

Naive bruteforcing is obviously a lot shorter (100 bytes):

lambda n,R=range:len({tuple(sum(s/n**(n*x)%n**n%o<1for x in R(o))for o in R(n))for s in R(n**n**2)})

Try it online! Crashes for \$n=4\$ with a MemoryError, when R(n**n**2) tries to create a list of \$2^{32}\$ integers.


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