Resistance across opposite vertices of n-dimensional cube with each edge at one ohm resistance is $$R_n=\sum_{k=0}^{n-1}\frac1{(n-k){n\choose k}}=\frac1n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}.$$ The proof is simple: we can identify nodes with equal voltages, and there are ${n\choose k}$ nodes with a distance $k$ to a given point.

From another hand $$R_n=\frac1{2^n}\sum_{k=1}^{n}\frac{2^k}{k},$$ because both sums satisfy the same recurrence relation $$R_n=\frac1n+\frac12R_{n-1}.$$

We can consider the sum $\sum_{k=1}^{n}\frac{2^k}{k}$ as a partial sum of $p$-adic logarithm $$\log_p(1+x)=-\sum_{k=1}^{\infty}\frac{(-x)^k}{k}$$ at the point $x=2$ (it is well defined for $p=2$). Using two formulas for $R_n$ we can get a simple application. We can find the value of $2$-adic logarithm at the point $-2$: $$-\log_2(-2)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2^k}{k}= \lim_{n\to\infty}\frac{2^n}n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}=0.$$ It is not a surprise (see § 4.4.11 from Cohen (2007), Number theory, Volume I: Tools and Diophantine equations).

First question: why does $2$-adic logarithm arise in combinatorial problem?

Second question: do you know any more connections between combinatorial and $p$-adic objects?

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    $\begingroup$ That the series vanishes at $-2$ is unsurprising because, unlike the complex log, the $p$-adic log is a homomorphism, to a torsion-free group, so that it has to vanish at torsion points. You’ll also get zero when you plug in $x=i-1$. $\endgroup$ – Lubin Oct 29 '13 at 13:57
  • $\begingroup$ Yes, but does it have a combinatorial sencs? $\endgroup$ – Alexey Ustinov Oct 29 '13 at 14:16
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    $\begingroup$ Do you have access to MathSciNet? I typed in Anywhere: p-adic and Primary Classification: 05, and got 37 hits. $\endgroup$ – Gerry Myerson Oct 29 '13 at 22:45
  • $\begingroup$ Thank you, Gerry. I've found an example concerning p-adic properties of alternating sign matrices (arrays of 0, 1 and −1, such that the entries of each row and column add up to 1 and the non-zero entries of a given row/column alternate) $\endgroup$ – Alexey Ustinov Oct 30 '13 at 2:06

The following article https://arxiv.org/abs/0904.1757 (The Hypercube of Resistors, Asymptotic Expansions, and Preferential Arrangements, by Nicholas Pippenger. Published in Mathematics Magazine 83(N5) (2010), 331-346) might be relevant for this question (as well as for Asymptotic rate for $\sum\binom{n}k^{-1}$).


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