# A curious valuation of this sequence

The sequence $a_n$ given by $$a_n=\sum_{k=0}^n\frac{n!}{k!}$$ is found at A000522 on OEIS with a description: total number of arrangements of a set with $n$ elements. Let $\nu_2(x)$ denote the $2$-adic valuation of the integer $x$. My first question:

Is it true that $\nu_2(a_n)=\nu_2(n+13)$?

Well, that is what experiments suggest to me. My second question (curiosity) is:

What is special about the presence of the number $13$ above?

Thanks to darij grinberg the above claim has failed, which prompts me to ask:

What is the $2$-adic valuation of $a_n$ then?

• Well, $\nu_2(n+13) = \nu_2((n+1) + 12)$. And 12 is special :) – Marty Aug 8 '18 at 21:00
• This might be in arxiv.org/abs/1508.01987v1 : your sequence $a_n$ fits into both classes; in particular it is $\mathbf{a}\left(X, 1, 1\right)$. So I'd look at Theorem 5 in particular. – darij grinberg Aug 8 '18 at 21:13

To your first question: No, it is false. For $n = 256 - 13$, the number $a_n$ has $\nu_2\left(a_n\right) = 7 < 8 = \nu_2 \left(n+13\right)$.

HOWEVER, it is almost correct: namely, it is correct whenever $128 \nmid n+13$ (so the smallest counterexample is $n = 128 - 13 = 115$). This is the following theorem:

Theorem 1. We have $\nu_2\left(a_n\right) = \nu_2\left(n+13\right)$ for all $n \in \mathbb{N}$ that don't satisfy $128 \mid n+13$.

Proof of Theorem 1 (sketched). We observe the following facts:

1. Each nonnegative integer $n$ and each positive integer $k$ satisfy $a_{n+k} \equiv a_n \mod k$. (This is proven by induction on $n$. The base case boils down to $a_k \equiv 1 \mod k$, which follows from $a_k = k a_{k-1} + 1$. The induction step uses the recursion $a_n = na_{n-1} + 1$.)

2. Each $n \in \mathbb{N}$ satisfies $a_{n+128} \equiv a_n \mod 128$. (This follows by applying fact 1 to $k = 128$.)

3. Each $n \in \mathbb{N}$ satisfies $128 \nmid a_n$ unless $128 \mid n+13$. (This needs only to be checked for the first $128$ values of $n$, due to fact 2 above.)

4. Combining facts 2 and 3, obtain $\nu_2\left(a_{n+128}\right) = \nu_2\left(a_n\right)$ for each $n \in \mathbb{N}$ unless $128 \mid n+13$.

Using fact 4, the Theorem 1 can be proven by induction (with $128$ base cases, unfortunately). $\blacksquare$

So what about the cases when $128 \mid n+13$ ? Their behavior is weird. The smallest such case is $n = 115$, in which $\nu_2 \left(a_n\right) = 9 > 7 = \nu_2 \left( n + 13 \right)$. However, for all larger $n$'s of the form $2^k - 13$, the sign flips:

Theorem 2. We have $\nu_2\left(a_n\right) = 7 < \nu_2\left(n+13\right)$ for all $n$ of the form $n = 2^k - 13$ with $k \geq 8$.

Proof of Theorem 2 (sketched). We claim that $a_{2^k - 13} \equiv 2^7 \mod 2^8$ for each $k \geq 8$. This is proven by induction on $k$. The induction base ($k = 8$) is verified by computer; the induction step relies on fact 1 from the proof of Theorem 1. $\blacksquare$

Theorem 2 should not suggest that $\nu_2\left(a_n\right)$ is always at most $7$; it is not. Higher values of $\nu_2\left(a_n\right)$ appear when $n+13$ is a multiple of $128$ but not itself a power of $2$. For example, $\nu_2\left(a_{2675}\right) = 15$.

Note that the same inductive argument that we used to prove fact 1 can be used to show the following stronger result:

Proposition 3. For any nonnegative integer $n$ and any positive integer $k$, we have \begin{align*} a_{n+k}-a_{n} & \equiv% \begin{cases} k, & \text{if }n+k\text{ or }n\text{ is odd};\\ 0, & \text{if neither }n+k\text{ nor }n\text{ is odd}% \end{cases} \mod 2k. \end{align*}

This might come useful.

Code: It is not immediately obvious how to compute $a_n$ for large values of $n$ efficiently. After all, $a_n \geq n!$, so we'd at least need a large integer type. Fortunately, all we care about are the $2$-adic valuations $\nu_2\left(a_n\right)$ and the remainders of $a_n$ modulo powers of $2$. The former valuations can easily be obtained from the latter remainders, so we only need to care about the remainders. And we can easily compute the remainders of $a_n$ modulo any given integer $k$ by recursion, because the recursive relation $a_n = na_{n-1} + 1$ descends to $\mathbb{Z}/k\mathbb{Z}$. Here, for example, is some simple Sage (or Python2) code that recursively $a_n$ modulo $2^k$ for given $n$ and $k$:

def a(n, k): # This gives a_n modulo 2^k.
if n == 0:
return 1
return (n * a(n-1) + 1) % (2 ** k)


This is already quite useful (e.g., you can let it compute "a(256-13, 8)" to check that $\nu_2\left(a_n\right) = 7$ for $n = 256-13$), but not really optimal, because it's recursive and eventually (around $n = 1000$) exceeds Python's maximum recursion depth. So let us replace the recursion by dynamic programming.

Here is some better Sage (or Python2) code, which tabulates $\nu_2(a_n)$ for $n = 1, 2, \ldots, m$ given a positive integer $m$:

def val2(k):
# Return the 2-adic valuation \nu_2(k) of the integer k.
if k == 0:
return # returns None, standing for -\infty.
res = 0
kk = k
while kk % 2 == 0:
res += 1
kk = kk // 2
return res

def aas(n, k):
# Return the list [a_0, a_1, \ldots, a_{n-1}] modulo 2^k.
pwk = 2 ** k
res =  * n
for i in range(1, n):
res[i] = (i * res[i-1] + 1) % pwk
return res

def output(m):
# Return a LaTeX table of \nu_2(a_n) for n = 1, 2, \ldots, m.
k = 3
# Starting with modulo 2^3, will later replace by better 2-adic precision.
nums = aas(m+1, k)
while (0 in nums):
k += 1
nums = aas(m+1, k)
vs = [val2(i) for i in nums]
res = r"\begin{array}{|c|c|} \hline" + "\n" + r"n & \nu_2(a_n) \\ \hline "
for i in range(1, m+1):
res += "\n "
res += str(i) + r"&" + str(vs[i]) + r" \\"
res += " \n" + r"\hline \end{array}"
return res


If I let it execute "print output(25)", it returns me the LaTeX code for a table of values of $\nu_2(a_n)$... which I'm not quoting here, because Theorem 1 renders it obsolete. A better idea is to tabulate only those values that Theorem 1 does not cover:

def output2(m):
# Return a LaTeX table of \nu_2(a_{128n - 13}) for n = 1, 2, \ldots, m.
k = 3
# Starting with modulo 2^3, will later replace by better 2-adic precision.
M = 128*m - 13
nums = aas(M+1, k)
while (0 in nums):
k += 1
nums = aas(M+1, k)
vs = [val2(i) for i in nums]
res = r"\begin{array}{|c|c|c|} \hline" + "\n" + r"n & \nu_2(a_{128n-13}) & \nu_2(128n) \\ \hline "
for j in range(1, m+1):
i = 128 * j - 13
res += "\n "
res += str(j) + r"&" + str(vs[i]) + r"&" + str(val2(128 * j)) + r" \\"
res += " \n" + r"\hline \end{array}"
return res


Now, executing "print output2(25)" yields

\begin{equation} \begin{array}{|c|c|c|} \hline n & \nu_2(a_{128n-13}) & \nu_2(128n) \\ \hline 1&9&7 \\ 2&7&8 \\ 3&8&7 \\ 4&7&9 \\ 5&11&7 \\ 6&7&8 \\ 7&8&7 \\ 8&7&10 \\ 9&9&7 \\ 10&7&8 \\ 11&8&7 \\ 12&7&9 \\ 13&10&7 \\ 14&7&8 \\ 15&8&7 \\ 16&7&11 \\ 17&9&7 \\ 18&7&8 \\ 19&8&7 \\ 20&7&9 \\ 21&15&7 \\ 22&7&8 \\ 23&8&7 \\ 24&7&10 \\ 25&9&7 \\ \hline \end{array} . \end{equation}

• This should go to the "eventual counterexamples" question. – Gerry Myerson Aug 10 '18 at 5:38