The mass of the proton times its charge radius is very close to 4ħ/c. Is this a coincidence? The claim is $r_p m_p = 4 L_0 M_0 = 4\hbar/c$, where $r_p$ and $m_p$ are the proton's charge radius and mass, and $L_0$ and $M_0$ are the Planck length and mass.  
Using the muon measurement $r_p=0.84087 \times 10^{-15}$ m,  $m_p=1.6726219 \times 10^{-27}$ kg, $r_p m_p = 1.40646 \times 10^{−42}$ kg m.
$4\hbar/c = 1.40706915 \times 10^{-42}$ kg m, agreement to $0.04\%$.   That seems pretty incredible.
Another way to put it is as a "prediction" of $r_p = \dfrac{4\hbar}{m_p c} = .84124$ fm.
With the error bars, the muon measurement is $.84087 \pm .00039$ fm, so call it min $.84048$ fm, max $.84126$ fm. The "prediction" is very close to the max.
The CODATA value for the proton's charge radius (electron measurement) is $.8751 \pm .0061 $ fm, close but not so incredibly close as the muon measurement.
Coincidence?
 A: In one sense, this should not be a coincidence, because the mass and the charge radius are actually determined by more fundamental quantities (quark masses, strong force coupling constant). So given that those quantities have their particular values, in theory it inexorably determines what "mass times charge radius" is going to be. 
What should be a coincidence is the appearance of the number 4, when "mass times charge radius" is expressed in natural units. However, since charge radius depends on the distribution of charge inside the proton, it's not inconceivable that, e.g., the distribution of "partons" (quarks and gluons), the distribution of momentum among them, etc., follows some principle which really does imply that the answer is "4 + small corrections". At least, I can't think of a simple reason this should not be so. 
For me the deepest fact is that fundamental theory should predict something for this quantity. That prediction might be 4, or it might be e^1.37, but it must be something close to 4. 
Update: A comment in chat by @Rococo allows @dandb's observation to be expressed in a very crisp way:
The charge radius of the proton (in muonic hydrogen) is almost exactly four times the reduced Compton wavelength of the proton.
Update 2017: Via P.R. Silva (eqn 6), I have run across a heuristic model of the nucleon in which M = 4/R (in natural units). Here R is the radius of the bag in the "bag model". See Xiangdong Ji, "Mass of the hadron", slide 20. I have not found where this argument originates, but a remark in a 1994 paper by Ji (see paragraph beginning "In the chiral limit...", on the final page) hints at it. 
A: Another attractive feature of this conjecture is that it is similar to another conjecture related to hadrons that is known to be true: that the spin of a hadron is equal to the sum of the spins of the quarks in the hadron (which come in discrete half integer increments), even though non-quark partons in the hadron have non-zero spins that "magically" cancel out in the total for reasons that are not well understood (i.e. the "proton spin crisis")). Until we understand why this is the case for spin in hadrons, we can't rule out that this conjecture is exactly true for related reasons.
I'd also note that the in lots of other areas of the Standard Model (e.g. some of the more obscure relationships between electroweak constants in the Standard Model), there are lots of known exact relationships between Standard Model, so it wouldn't be a priori unreasonable to wonder if there was such a relationship here, particularly given that charge radius is an electroweak phenomenon.  I don't think that there is any known Standard Model constant relationship that can explain why this conjecture should be true, but the precision of the the conjectured relationship is sufficiently great that it isn't unreasonable to entertain the possibility that it is exactly or exactly subject to small corrections, true.
