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\begin{center}
\vskip 1cm{\LARGE\bf Some $2$-adic Integers Related \\
\vskip .1in
to the Odd Part of $2^e!$
}
\vskip 1cm
\large
Donald M.\ Davis\\
Department of Mathematics\\
Lehigh University\\
Bethlehem, PA 18015
USA\\
\href{mailto:dmd1@lehigh.edu}{\tt dmd1@lehigh.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
The odd part of $2^e!$ as $e\to\infty$ leads to a $2$-adic integer
$z$. The bits of $z$ were publicized in OEIS sequence \seqnum{A359349},
where two
conjectures were made, relevant to computing $z$. We prove both of those
conjectures. A second $2$-adic integer, the limit of $((2^e-1)!!-1)/2^e$,
plays a key role in one proof.
\end{abstract}




\def\on{\operatorname}
\def\a{\alpha}
\def\Si{\Sigma}
\def\b{\beta}
\def\ni{\noindent}
\def\od{\operatorname{od}}
\def\odprod{\on{odpr}}
\def\uns{\on{uns}}
\def\stab{\on{stab}}
\def\Z{\mathbb{Z}}
\def\sihat{\widehat{\sigma}}

\section{Introduction}\label{intro}
In \cite{MAA}, the author noted that the odd part of $2^e!$ and of $2^{e-1}!$ agree mod $2^e$, and so the 2-adic limit as $e$ approaches $\infty$ is a 2-adic integer, which we call $z$. In sequence \seqnum{A359349} of the
{\it On-Line Encyclopedia of Integer Sequences} (OEIS), 
the author and Schoenfield publicized the sequence of bits of $z$ and made two conjectures. One involved a relationship between the bits of $z$ and some of the unstable bits in the odd part of $2^e!$, while the other leads to a more efficient way of computing $z$. In this paper we prove both conjectures and some generalizations.

In this introductory section, we review the two conjectures, stating them as theorems.
In Sections \ref{sec2} and \ref{sec3}, we prove generalizations of both.

Let $\nu(n)$ denote the exponent of 2 in the prime factorization of $n$, and $\od(n)=2/2^{\nu(n)}$, the odd part of $n$. Then $\od(2^e!)=2^e!/2^{2^e-1}$. In Table 1 we tabulate the first 40 bits in the backward binary expansion (BBE) of $\od(2^e!)$ for $2\le e\le30$. The first 31 bits of row 30 of Table 1 are the first 31 bits of the BBE of the 2-adic integer $z$ defined above. For example, the binary expansion of $z$ ends
\begin{equation} \cdots011010001011.\label{z12}\end{equation}
In \seqnum{A359349}, a larger table (64 bits for $e\le40$) is presented.


\begin{table}[H]
\centering
\begin{tabular}{r|l}
$e$&\\
\hline
2&110$\ $0000000000000000000000000000000000000\\
3&1101$\ $110010000000000000000000000000000000\\
4&11010$\ $11101110111011100000110010000000000\\
5&110100$\ $1011001110100011000001101010001001\\
6&1101000$\ $000000001101000110100010110011110\\
7&11010001$\ $10010101001010001010001101011100\\
8&110100010$\ $0111100111001000110000010011101\\
9&1101000101$\ $000110101110110000001011000011\\
10&11010001011$\ $10101101100010111001010101001\\
11&110100010110$\ $0010011111110101111010110111\\
12&1101000101101$\ $110111110000001100111011011\\
13&11010001011010$\ $11100001101100011101011000\\
14&110100010110100$\ $1011001111011110110100001\\
15&1101000101101000$\ $000011110100000100110000\\
16&11010001011010001$\ $10000001011001001111011\\
17&110100010110100010$\ $0101001000011101001001\\
18&1101000101101000101$\ $011101001000000000010\\
19&11010001011010001011$\ $00000101101001010101\\
20&110100010110100010111$\ $1001101111101111101\\
21&1101000101101000101110$\ $011011110000010111\\
22&11010001011010001011101$\ $00100001101011100\\
23&110100010110100010111011$\ $1101001111111011\\
24&1101000101101000101110110$\ $111101110010100\\
25&11010001011010001011101100$\ $10000001111101\\
26&110100010110100010111011000$\ $0101001100111\\
27&1101000101101000101110110001$\ $011101101100\\
28&11010001011010001011101100011$\ $00000011011\\
29&110100010110100010111011000111$\ $1001011000\\
30&1101000101101000101110110001110$\ $011111001
\end{tabular}
\caption{first 40 bits in BBE of $\od(2^e!)$.}
\label{table}
\end{table}


Bits $0$ through $e$ of $\od(2^e!)$ are {\it stable}; they agree with those of $z$. The {\it unstable} bits of $\od(2^e!)$ are those in position $\ge e+1$; they appear to the right of  the space in Table \ref{table}. Note that, for $e>d$, the first $d$ unstable bits of $\od(2^e!)$ occur in the same positions as the last $d$ stable bits of $\od(2^{e+d}!)$. The latter are the stable bits of $z$ in position $e+1$ through $e+d$. Let $\uns(e,d)$ and $\stab(e+1,d)$ denote the numbers whose BBE's are these sequences of $d$ bits. The following theorem is the first conjecture of \seqnum{A359349}.

\begin{theorem}\label{thm1} There is a 2-adic integer $K$ such that, for all $d$ and $e>d$, we have
$$\uns(e,d)+K\equiv \stab(e+1,d)\pmod{2^d}.$$
\end{theorem}
\begin{example} The BBE of $K$ begins $1011011$. This will be seen in Example \ref{expl3}. The BBE's of $\uns(17,7)$ and $\stab(18,7)$
are $0101001$ and $1110110$, respectively. Note that the BBE of stab(18,7) is the 7 bits to the left of the space in line 24 of Table \ref{table}. We verify Theorem \ref{thm1} with $d=7$ and $e=17$ by reversing the order of the bits and checking the binary addition in Table \ref{addn}.
\end{example}


\begin{table}[h]
\centering
\begin{tabular}{c|r}
$\uns(17,7)$&$\cdots1001010$\\
$K$&$\cdots1101101$\\
\hline
$\stab(18,7)$&$\cdots0110111$
\end{tabular}
\caption{Binary addition check.}
\label{addn}
\end{table}

\bigskip

This relationship between the stable and unstable parts is, at least, a curiosity. It could be useful in calculations. We will see in Section \ref{sec2} that the first $d$ bits of $K$ can be determined from $\od(2^d!)$ mod $2^d$ and $(2^d-1)!!$ mod $2^{2d}$. Computing $\uns(e,d)$ requires $\od(2^e!)$ mod $2^{e+d}$. The desired $\stab(e+1,d)$, which can be obtained as the sum of these,  would, if computed directly, require $\od(2^{e+d}!)$ mod $2^{e+d}$, which is a much larger calculation, requiring the multiplication of much larger numbers.

Let $\odprod(\ell,m)$ denote the product of all odd integers $j$ satisfying $\ell\le j\le m$, and let \begin{equation}\label{hdef}h(m)=\odprod(2^{m-1}+1,2^m-1).\end{equation}
We begin with the following elementary proposition.
\begin{proposition} For any $e\ge1$, we have
$$\od(2^e!)=\prod_{m=2}^eh(m)^{e+1-m}.$$\label{hprop}
\end{proposition}
\begin{proof} Each factor $j$ of $h(m)$ occurs with coefficient $2^i$ for $0\le i\le e-m$ in $2^e!$.
\end{proof}
This yields a method of computing $\od(2^e!)$ mod $2^B$, reducing mod $2^B$ at each step. The following theorem, which is the second conjecture of \seqnum{A359349}, makes it more efficient.
\begin{theorem}\label{thm2} If $2\le m-1\le B\le 3m-7$, and $d=2+\lfloor \frac{B-m}2\rfloor$, then
$$h(m)\equiv \odprod(2^{m-1}+1,2^{m-1}+2^d-1)^{2^{m-1-d}}\pmod{2^B}.$$
\end{theorem}
The advantage is that now $h(m)$ requires $2^{d-1}+m-2-d$ multiplications (always reducing mod $2^B$) compared with $2^{m-2}-1$ multiplications if using (\ref{hdef}).
We compare computing $\od(2^{49}!)$ mod $2^{50}$ via Proposition \ref{hprop} and computing it using Theorem \ref{thm2}. In either case, we compute $h(m)$ for $m<19$ using (\ref{hdef}). In Table \ref{50} we list, for even values of $m$ between 20 and 44, the approximate number of multiplications required to compute $h(m)$ using (\ref{hdef}) (the {\bf Old} way) and using Theorem \ref{thm2} (the {\bf New} way).  Note that $h(m)$ must then be raised to the $50-m$ power mod $2^{50}$.


\renewcommand{\arraystretch}{1.2}
\begin{table}[h]
\centering
\begin{tabular}{c|ccccccccccccc}
$m$&$20$&$22$&$24$&$26$&$28$&$30$&$32$&$34$&$36$&$38$&$40$&$42$&$44$\\
$d$&$17$&$16$&$15$&$14$&$13$&$12$&$11$&$10$&$9$&$8$&$7$&$6$&$5$\\
\hline
Old&$2^{18}$&$2^{20}$&$2^{22}$&$2^{24}$&$2^{26}$&$2^{28}$&$2^{30}$&$2^{32}$&$2^{34}$&$2^{36}$&$2^{38}$&$2^{40}$&$2^{42}$\\
New&$2^{16}$&$2^{15}$&$2^{14}$&$2^{13}$&$2^{12}$&$2^{11}$&$2^{10}$&$2^9$&$2^8$&$2^7$&$2^6$&$2^5$&$2^4$
\end{tabular}
\caption{Two ways of computing $\od(2^{49}!)$ mod $2^{50}$.}
\label{50}
\end{table}


\bigskip


In preliminary discussion, Schoenfield \seqnum{A359349} wrote ``If the conjecture can be proved, then I think I could compute the first 105 terms (of $z$) with a program that would take maybe a few days to run.''



\section{A formula for the 2-adic integer $K$ in Theorem \ref{thm1}}\label{sec2}
In this section, we prove Theorem \ref{thm1} and give a formula for the 2-adic integer $K$ that occurs in it.
We begin by reviewing the proof  \cite{MAA} of existence of the 2-adic integer $z$, as some of the ingredients will be useful later.

\begin{lemma} \label{bij} Let $I_e=\{i:2^{e-1}<i\le 2^e\}$ and $S_e=\{j: j\text{ odd and }1\le j<2^e\}$. Then $\od:I_e\to S_e$ is bijective.\end{lemma}
\begin{proof} The inverse function $\phi$ is defined by $\phi(u)=2^tu$ where $t=\max\{k:2^ku\le 2^e\}$.\end{proof}
\begin{lemma} \label{Gauss} If $e\ge3$, the product of all odd positive integers less than $2^e$ is $\equiv2^e+1\pmod{2^{e+1}}$.\end{lemma}
\begin{proof} We begin with the proof  \cite[Lemma 1]{Gran} that the product is $\equiv1$ (mod $2^e$). Pair each element with its inverse in $\Z/2^e$. Only $\pm1$ and $2^{e-1}\pm1$ equal their own inverse, and their product is 1, while all other pairs yield 1.

Let $P$ be the set of pairs $(a,b)\in S_e\times S_e$ with $a<b$  and $ab\equiv 2^e+1$ (mod $2^{e+1}$). If $(a,b)\in P$, then so is $(2^e-b,2^e-a)$ since $a+b$ is even. Moreover, $(a,b)\ne (2^e-b,2^e-a)$ since, if so, then $a(2^e-a)\equiv 2^e+1$ (mod $2^{e+1}$), which cannot occur since $a^2\equiv 1$ (mod 8). Thus the cardinality of $P$ is even, and hence the product of all $ab$ with $(a,b)\in P$ is $\equiv1$ (mod $2^{e+1}$). Other pairs $(c,d)$ with $cd\equiv1$ (mod $2^e$) have $cd\equiv1$ (mod $2^{e+1}$). Finally we have $1$, $2^e-1$, and $2^{e-1}\pm1$, whose product is $\equiv 2^e+1$ (mod $2^{e+1}$).
\end{proof}
\begin{corollary} If $e\ge3$, then $\od(2^{e-1}!)\equiv\od(2^e!)\pmod{2^e}$.\end{corollary}
\begin{proof} By Lemmas \ref{bij} and \ref{Gauss}, we have
$$\frac{\od(2^e!)}{\od(2^{e-1}!)}=\prod_{i\in I_e}\od(i)=\prod_{j\in S_e}j\equiv1\pmod{2^e}.$$\end{proof}
\begin{corollary}\label{zdef} There is a $2$-adic integer $z$ which equals $\od(2^{e-1}!)$ $(\bmod\  2^e)$ for all $e$.\end{corollary}

The stronger (mod $2^{e+1}$) part of Lemma \ref{Gauss} is not needed here, but will be used shortly.
A stronger version of the next  result will be proved in Theorem \ref{hard}.
\begin{proposition} \label{weak1} If $e\ge2$ and $S_e$ is as above, then
$$\prod_{i\in S_e}i\equiv\prod_{i\in S_e}(2^e+i)\pmod{2^{2e}}.$$\end{proposition}
\begin{proof} If $S$ is a set of cardinality $n$, let $\sihat_i(S)=\sigma_{n-i}(S)$, where $\sigma$ is the usual elementary symmetric polynomial. Then $\sihat_1(S_e)$ is divisible by $2^e$ since, for odd $j\le 2^{e-1}-1$, the expression
$$\prod_{\substack{i\in S_e\\i\ne j}}i+\prod_{\substack{i\in S_e\\i\ne 2^e-j}}i$$ is divisible by $2^e$.
We have
$$\prod_{i\in S_e}(2^e+i)-\prod_{i\in S_e}i=\sum_{j>0}2^{je}\sihat_j(S_e)\equiv 2^{e}\sihat_1(S_e)\equiv0\pmod{2^{2e}}.$$
\end{proof}

Let $(2^e-1)!!=\odprod(1,2^e-1)$.
\begin{corollary}\label{wcor} If $e\ge2$, then $\dfrac{(2^e-1)!!-1}{2^e}\equiv\dfrac{(2^{e+1}-1)!!-1}{2^{e+1}} \pmod{2^{e-1}}$.\end{corollary}
\begin{proof} By Lemma \ref{Gauss}, the two expressions are odd integers. We show that their ratio is $\equiv1$ (mod $2^{e-1}$).
Let $A=(2^e-1)!!-1$. By Lemma \ref{Gauss}, $A=2^eu$ with $u$ odd. By Proposition \ref{weak1}, $\odprod(2^e+1,2^{e+1}-1)=A+1+k2^{2e}$ for some integer $k$. The 
desired ratio is
\begin{align*}
\frac{(A+1)(A+1+k2^{2e})-1}{2A}
& = \frac{A^2+2A+(A+1)k2^{2e}}{2A} \\
& =2^{e-1}u+1+\frac{(A+1)k2^{2e}}{2^{e+1}u}\equiv1\pmod{2^{e-1}}.
\end{align*}
\end{proof}

\begin{definition} We define $w$ to be the 2-adic integer which is congruent to $\dfrac{(2^e-1)!!-1}{2^e} \pmod {2^{e-1}}$ for all $e$.\label{wdef}\end{definition}

 This is well-defined by Corollary \ref{wcor}. One can compute that the binary expansion of $w$ ends $\cdots 1001110011001$.

We have introduced two 2-adic integers, $z$ and $w$. The next result shows that their product equals the difference between the unstable parts of consecutive rows of Table \ref{table} in a metastable range.
\begin{theorem}\label{prod}If $e\ge3$, then
$$\frac{\od(2^e!)-\od(2^{e-1}!)}{2^e}\equiv zw\pmod{2^{e-1}}.$$\end{theorem}
\begin{proof} By Lemma \ref{bij}, we have $(2^e-1)!!=\dfrac{\od(2^e!)}{\od(2^{e-1}!)}$.
Thus
$$\frac{(2^e-1)!!-1}{2^e}=\frac{\od(2^e!)-\od(2^{e-1}!)}{2^e\od(2^{e-1}!)},$$
so $$\frac{(2^e-1)!!-1}{2^e}\cdot\od(2^{e-1}!)=\frac{\od(2^e!)-\od(2^{e-1}!)}{2^e}.$$
The result follows now from Corollary \ref{zdef} and Definition \ref{wdef}.
\end{proof}
\begin{example} \label{expl2} The binary expansion of $zw$ ends $\cdots011000010011$, obtained by multiplying the binary numbers in (\ref{z12}) and just after Definition \ref{wdef}. The numbers $\od(2^7!)$ and $\od(2^6!)$ agree mod $2^7$. From Table \ref{table}, beginning in the $2^7$ position, the binary expansion of $\od(2^7!)$ ends $\cdots 1010011$, while that of $\od(2^6!)$ ends with eight 0's. The difference agrees with $zw$ mod $2^6$.\end{example}

We now state the main theorem of this section.
\begin{theorem} \label{Kthm} The $2$-adic integer $K$ of Theorem \ref{thm1} equals $\ {-zw}$.
\end{theorem}
\begin{proof}[Proof of Theorems \ref{thm1} and \ref{Kthm}.] The difference
$\stab(e+1,d)-\uns(e,d)$, as described in the paragraph preceding Theorem \ref{thm1}, equals $\dfrac{\od(2^{e+d}!)-\od(2^e!)}{2^{e+1}}$ (mod $2^d$). We have
\begin{align*}
\frac{\od(2^{e+d}!)-\od(2^e!)}{2^{e+1}}
& = \sum_{i=1}^d\frac{\od(2^{e+i}!)-\od(2^{e+i-1}!)}{2^{e+1}} \\
& = \sum_{i=1}^d2^{i-1}\frac{\od(2^{e+i}!)-\od(2^{e+i-1}!)}{2^{e+i}}\\
\end{align*}
\begin{align*}
& \equiv\sum_{i=1}^d2^{i-1}zw\pmod{2^e}\\
& \equiv\sum_{i=1}^\infty2^{i-1}zw\pmod{2^d}\\
& = -zw.\qedhere\end{align*}
\end{proof}

\begin{example} \label{expl3} The binary expansion of $-zw$ ends $\cdots010111101101$, since adding this to the $zw$ in Example \ref{expl2} equals 0. Add that to the binary number obtained by reversing the order of the first $12$ bits after the space on line $14$ of Table \ref{table}, and you obtain the binary number obtained by reversing the order of the last $12$ bits before the space on line $26$. This illustrates Theorem \ref{thm1} with $d=12$ and $e=14$.\end{example}

\section{Proof of Theorem \ref{thm2}}\label{sec3}
In this section, we prove Theorem \ref{thm2} and some mild generalizations. The bulk of our work is the following strengthening of Proposition \ref{weak1}, the proof of which appears later.
\begin{theorem} \label{hard} With $S_e$ as defined in Lemma \ref{bij}, and $A$ any integer, if $e\ge2$, then
$$\prod_{i\in S_e}(A2^e+i)\equiv\prod_{i\in S_e}i\pmod{2^{3e-1}}.$$
\end{theorem}
\begin{corollary} \label{ABcor} For any integers $A$, $B$, $j$, and $e\ge2$, we have
$$\prod_{i\in S_e}(A2^e+i)^{2^j}\equiv\prod_{i\in S_e}(B2^e+i)^{2^j}\pmod{2^{3e-1+j}}.$$
\end{corollary}
\begin{proof} It is elementary that if $\a\equiv\b$ (mod $2t$), then $\a^2\equiv\b^2$ (mod
$4t$). We apply this iteratively to Theorem \ref{hard}, and then both expressions in the corollary are congruent to $\prod i^{2^j}$.\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}.]
We write the conjectured congruences in succession, beginning
\begin{align*}
\odprod(2^{m-1}+1,2^m-1)
& \equiv \odprod(2^{m-1}+1,2^{m-1}+2^{m-2}-1)^2\pmod {2^{3m-7}},\end{align*}
then
\begin{align*}
\odprod(2^{m-1}+1,2^{m-1}+2^{m-2}-1)^2 &\equiv \odprod(2^{m-1}+1,2^{m-1}+2^{m-3}-1)^{2^2}\pmod{2^{3m-9}},\end{align*}
with arbitrary entry
\begin{align*}
\odprod(2^{m-1}+1,2^{m-1}+2^{d+1}-1)^{2^{m-2-d}}
\equiv&\odprod(2^{m-1}+1,2^{m-1}+2^d-1)^{2^{m-1-d}}\pmod{2^{2d+m-3}}.
\end{align*}
After canceling $\odprod(2^{m-1}+1,2^{m-1}+2^d-1)^{2^{m-2-d}}$, this becomes
\begin{align*}
& \odprod(2^{m-1}+2^d+1,2^{m-1}+2^{d+1}-1)^{2^{m-2-d}} \\
& \quad\quad \equiv \odprod(2^{m-1}+1,2^{m-1}+2^d-1)^{2^{m-2-d}}\pmod{2^{2d+m-3}}.
\end{align*}
We can restate this as
$$\prod_{i\in S_d}(2^{m-1}+2^d+i)^{2^{m-2-d}}\equiv\prod_{i\in S_d}(2^{m-1}+i)^{2^{m-2-d}}\pmod{2^{2d+m-3}},$$
and this is a consequence of Corollary \ref{ABcor}.
\end{proof}
We prove the following two lemmas, from which Theorem \ref{hard} follows easily.
\begin{lemma}\label{lem1} If $e\ge2$, then $\sihat_1(S_e)\equiv 2^{2e-2}\pmod{2^{2e-1}}$.\end{lemma}
\begin{lemma}\label{lem2} If $e\ge2$, then $\sihat_2(S_e)\equiv2^{e-2}\pmod{2^{e-1}}$.\end{lemma}
\begin{proof}[Proof of Theorem \ref{hard}.] We have
$$\prod_{i\in S_e}(A2^e+i)-\prod_{i\in S_e}i=\sum_{j>0}(A2^e)^j\sihat_j(S_e)\equiv0\pmod{2^{3e-1}}$$
by Lemmas \ref{lem1} and \ref{lem2}, with the argument slightly different for the two parities of $A$.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem1}.] If $e\ge2$, then
$$\sihat_1(S_e)=\sum_{i=0}^{2^{e-2}-1}\biggl(\frac{(2^e-1)!!}{2i+1}+\frac{(2^e-1)!!}{2^e-1-2i}\biggr)=2^e\sum_{i=0}^{2^{e-2}-1}\frac{(2^e-1)!!}{(2i+1)(2^e-1-2i)}.$$
Let $H_e=\displaystyle \sum_{i=0}^{2^{e-2}-1}\frac{(2^e-1)!!}{(2i+1)(2^e-1-2i)}$. We prove by induction that $H_e\equiv2^{e-2}$ mod $2^{e-1}$, which implies the lemma.

The claim is true for $e=2$. Assume it true for $e-1$. We have
$$H_e\equiv\sum_{i=0}^{2^{e-2}-1}\frac{((2^{e-1}-1)!!)^2}{(2i+1)(2^{e-1}-2i-1)}\pmod{2^{e-1}}.$$
The summands for $i$ and $2^{e-2}-1-i$ are equal. Thus $H_e\equiv 2(2^{e-1}-1)!!H_{e-1}$ (mod $2^{e-1}$). By the induction hypothesis, we obtain $H_e\equiv2^{e-2}$ (mod $2^{e-1}$), as desired.
\end{proof}



The following results will be used in the proof of Lemma \ref{lem2}.
\begin{lemma}\label{8} If $e\ge3$, then, of the $2^{e-1}$ numbers $i^2$ $\bmod$ $2^e$, $i\in S_e$, there are exactly four having each of the $2^{e-3}$ values less than $2^e$ and $\equiv1\pmod 8$.\end{lemma}
\begin{proof} Each of the  $2^{e-3}$ numbers, being $\equiv1$ (mod 8), is a quadratic residue, and so must occur as $i^2$ for some $i\in S_e$. Each occurs in at least four ways since
for odd $i<2^{e-1}$, the numbers $i$, $2^{e-1}-i$, $i+2^{e-1}$, and $2^e-i$ are distinct numbers with the same square mod $2^e$. By the Pigeonhole Principle, the claimed partitioning must hold.
\end{proof}
\begin{lemma}\label{9} For $e\ge3$, we have
$$\sihat_1(1,9,\ldots,2^e-7,1,9,\ldots,2^e-7,1,9,\ldots,2^e-7,1,9,\ldots,2^e-7)\equiv 2^{e-1}\pmod{2^e}.$$
\end{lemma}
\begin{proof}The proof is by induction. The claim is true for $e=3$ and $4$;
we have $\sihat_1(1,1,1,1)=4$ and $\sihat_1(1,9,1,9,1,9,1,9)=4\cdot9^4+4\cdot9^3=4\cdot9^3\cdot10$.
For arbitrary $e$, our expression equals $4\cdot9^3\cdots(2^e-7)^3\cdot\sihat_1(1,9,\ldots2^e-7)$. Because of the 4,
we can consider $\sihat_1(1,9,\ldots,2^e-7)$ mod $2^{e-2}$, so we obtain an odd multiple of
$4\cdot\Si$ with
$$\Si=\sihat_1(1,9,\ldots,2^{e-2}-7,1,9,\ldots,2^{e-2}-7,1,9,\ldots,2^{e-2}-7,1,9,\ldots,2^{e-2}-7).$$
By the induction hypothesis, $\Si\equiv2^{e-3}$ (mod $2^{e-2}$), and so our desired expression is $\equiv2^{e-1}$ (mod $2^e$).\end{proof}


\begin{proposition}\label{sqprop} If $e\ge1$, then $\displaystyle\sum_{i\in S_e}\dfrac{((2^e-1)!!)^2}{i^2}\equiv 2^{e-1}\pmod{2^e}$.\end{proposition}
\begin{proof} By Lemma \ref{8},  it equals the expression in Lemma \ref{9}.\end{proof}

\begin{proof}[Proof of Lemma \ref{lem2}.]

 Let $D_e=\{(a,b)\in S_e\times S_e:\ a<b\}$. Note that $$\sihat_2(S_e)=\displaystyle\sum_{(a,b)\in D_e}\dfrac{(2^e-1)!!}{a\cdot b},$$ denoted  by $T_e$. Write $T_e=T_{1,e}+T_{2,e}$, where
 $$T_{1,e}=\sum_{\substack{(a,b)\in D_e\\ a\not\equiv b\ (2^{e-1})}}\frac{(2^e-1)!!}{a\cdot b}\quad\text{and}\quad
 T_{2,e}=\sum_{\substack{(a,b)\in D_e\\ a\equiv b\ (2^{e-1})}}\frac{(2^e-1)!!}{a\cdot b}.$$
Each summand of $T_{2,e}$ corresponds to a unique element of $S_{e-1}$, and so
$$T_{2,e}\equiv\sum_{a\in S_{e-1}}\frac{((2^{e-1}-1)!!)^2}{a^2}\equiv2^{e-2}\pmod{2^{e-1}}$$
by Proposition \ref{sqprop}.

We prove $T_{1,e}\equiv0 \pmod {2^{e-1}}$ by induction. It is true when $e=3$ as we obtain four summands, each with denominator 3. Assume validity for $e-1$.
Every element of $D_{e-1}$ corresponds to four  summands of $T_{1,e}$ which are equal mod $2^{e-1}$.
 We obtain
 $$T_{1,e}\equiv 4\sum_{(a,b)\in D_{e-1}}\frac{((2^{e-1}-1)!!)^2}{a\cdot b}=4(2^{e-1}-1)!!(T_{1,e-1}+T_{2,e-1})\equiv0\pmod{2^{e-1}},$$
 using the induction hypothesis for $4T_{1,e-1}$ and the already-proved result for $4T_{2,e-1}$.
\end{proof}

\section{Acknowledgments}
 We thank Andrew Granville for providing an alternate proof of Lemma \ref{lem1}, involving properties of Bernoulli numbers. We thank the referee for several comments which improved the exposition.

\begin{thebibliography}{9}

\bibitem{MAA} D.\ M.\ Davis, Binomial coefficients involving infinite powers of primes, {\it Amer.\ Math.\ Monthly} {\bf 121} (2014) 734--737.

\bibitem{Gran} A.\ Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers. In {\it Organic Mathematics},
CMS Conf.\  Proc., Vol.\ 20, Amer. Math. Soc., 1997, pp.~253--276.
\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B50; Secondary 11E95, 11A15.

\noindent \emph{Keywords:}  2-adic integer, 2-power, factorial, integer sequence.

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\noindent (Concerned with sequence
\seqnum{A359349}.)

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\vspace*{+.1in}
\noindent
Received October 30 2025;
revised version received May 7 2026; May 10 2026.
Published in {\it Journal of Integer Sequences}, May 15 2026.

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Return to \href{https://cs.uwaterloo.ca/journals/JIS/}{Journal of Integer Sequences home page}.
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