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Could it be possible that the mass of the proton can be calculated by a series of integer sequences? Or is this just a curiosity?

$$\sum_{m=1}^{\infty } \frac{1}{10^{26}(m^2+1)_{2m}}=$$

NSum[1/Pochhammer[m^2+1,2m], {m,1,\[Infinity]}, WorkingPrecision -> 50] /10^26 

First seven digits match the proton's mass in kilograms.

$1.6726218229590580987863882056891582636342622102204\times10^{-27}$
$1.672621\times10^{-27}$ - from OEIS revised 11/15/12
$1.672621777\times10^{-27}$ - from Wikipedia

What's to say that sometime in the future, the proton's mass won't be made more accurate by adding $4.5\times10^{-35}$ to the current number?

Edit to explain motivation

Whenever I get a result I don't recognize, I look it up on OEIS. I found this number.

I posted on Mathematica.SE with the intention of asking for advice on how to prove that it converges. That would make this number a constant.

If this is a "fluke" or the result of "small numbers," it's still worth exploring.

Edit: It does converge.

Final Thoughts

$f_{p}=0.16726218229590580987863882056891582636342622102204$

is the 0-dimensional value of a fractal know as the Hilbert Curve.

To get the minimal 3-dimensional value: $f \times 10^{((dimension+1)!)}$ where $0\le dimension \le 3$.

This results in the value for a $1\times 1\times 1$ cube (coincidentally, the definition of the gram.)

To get kilograms: $f \times 10^{((dimension+1)!+3)}$.

I posit that the fractalness is the stabilizing influence on the proton.

Coda

I agree with everyone that I have been wrong-headed about the importance of this constant. I have posted the constant on OEIS A219733. Thanks for your patience.

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Nice! but the problem here is that mass is a dimensionful quantity so the same integer representation may not hold in other units. It could perhaps be more useful if one can try to represent (say) the ratio of proton and electron mass as some such expression in terms of integers. –  user10001 Nov 12 '12 at 11:36
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Minor point: a proton isn't 167 grams. –  Mark Hurd Nov 12 '12 at 14:29
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Everyone--or at least the PDG--knows the proper mass of a proton is 938.272 (in $\mathrm{MeV}/c^2$, of course). What Mark, Arnold and I are getting at is that you are treating the the unit "kiligram" as if it has some fundamental importance when it is quite arbitrary. –  dmckee Nov 12 '12 at 21:28
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No. You're missing the crucial physics of the issue. Please read the comments and the answers in detail and think about the physics of what we're saying instead of concentrating on the numbers. –  Emilio Pisanty Nov 13 '12 at 15:48
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This is completely wrong. The correct proton mass is 1.14610984 $\times 10^{-28}$ slug. What is this kilogram junk science? –  Jerry Schirmer Nov 14 '12 at 22:52
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3 Answers

up vote 40 down vote accepted

To formalize dushya's comment as an answer:

Since the kilogram is an arbitrary, man-made unit, the actual numerical value of the proton mass in kilograms is meaningless (i.e. it's as good as its value in pounds, ounces, stones, solar masses, $\textrm{MT}/c^2$, etc.). The true fundamental constants of nature are dimensionless: they have the same value in every unit system. Thus dimensional constants like $c$, $\hbar$, $G$, and indeed $m_p$ and $m_e$, are not very meaningful and can be set to $1$ with a judicious choice of units (which is done quite often).

True fundamental constants are often ratios of dimensional quantities such as the fine structure constant, $$\alpha=\frac{e^2/4\pi\epsilon_0}{\hbar c},$$ which quantifies how strong, on a quantum scale, the electromagnetic interaction is. In terms of mass, the constants you'd like to predict are things like the ratio $m_p/m_e\approx 1800$, and so on.

Given that, the formula you have found is just a fluke: a consequence of the fact that we chose as our basic unit of mass the mass of a cube of water whose sides measure one hundred-millionth of a quarter of a meridian.


EDIT, given the long comment thread:

@Fred, let me try and rephrase this a bit to see if I can bring out the arbitrariness we're talking about well up to the surface. The real number you have discovered is the inverse of the one you posted: $$\frac{10^{26}}{\sum_{m=1}^\infty \frac{1}{(m^2+1)_{2m}}}\approx 5.978638 \times 10^{26},$$ which appears to approximate within experimental error the number of protons and neutrons that will fit - at sea level and at "room" temperature - a cubical box about yea big in side containing that particular common chemical that you find in drinking fountains, kitchen sinks, lakes, and even falling out of the sky (on Earth) rather often. 

Since the proton really is quite fundamental, any stabilizing influence of the fractalness needs to account for the size of the Earth, its predominant climate a hundred years ago, the abundance of water in it, and the detailed chemical state of the brains of a number of mainly French gentlemen that sat down a while ago to try and make unit systems (which are always arbitrary) at least simple to work with.

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This does allow you to make a prediction--- the significance of the match tells you approximately the number of attempts you have made to get such a coincidence. The ratio of the mass of a proton to a mass of a steel cylinder in Paris was determined by the psychology of some French revolutionaries.

But from the accuracy you get, one can be 99.99% sure that you automatically went through an automated search of hundreds of thousands of integer sequences, probably by going through the database, and compared to a list of about 100 physical constants. When one is doing such a search, one should say the number of attempts.

A historical case where something like this was significant was when people in the early part of the 19th century noticed that ${1\over \sqrt{\epsilon_0 \mu_0}}= c$. This came as a surprise back then (the system of units didn't make it a definition as it does today, nor was it obvious from the relativity postulate). There aren't many more cases of numerology like this being significant.

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I'm not so sure if a formal search by one person was performed; if it was a search, a search over what? The end values of m; and it stopped at 3? It might be more likely that a bunch of mathematicians are doing a bunch of calculations independently; the result of one looks familiar; and that person posts the question here. –  Clayton Stanley Nov 13 '12 at 2:36
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@claytontstanley: I believe the fellow did a systematic search over the database of integer sequences. I considered the possibility of a distributed accident, it is possible of course, but I think it is unlikely in this case. To get 7 digits of matching you need 10,000 people searching through a hundred different sequences each, I doubt this. It is also a use of mathematica and the integer sequence database, so perhaps it was a publicity stunt by a Wolfram employee. Who knows. –  Ron Maimon Nov 13 '12 at 2:59
    
It's nothing more than the sum of the reciprocals of the products of the numbers between consecutive squares. –  Fred Kline Nov 13 '12 at 6:24
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@FredDanielKline: And I'm sure that it's one of the million or so integer sequences in the database. –  Ron Maimon Nov 13 '12 at 6:28
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Tanquam ex ungue leonem. I knew after the first two lines, without seeing the signature, that I was reading Ron (+1) –  Eduardo Guerras Valera Nov 22 '12 at 22:33
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no, I am afraid you have not discovered a physically relevant relation until you prove what is the relation between the series and the terms defining the theory in the UV (such as $g_{3}$ at some scale $\mu$, $m_{u,d}$, electromagnetic corrections...). I would be much more impressed if you could get at the same time a similar formula for the neutron mass which is 'almost' like a proton beside the small up- and donw-quark masses and the electric charge. As I said, these things should enter in your formula in order to give you $m_{proton}\neq m_{neutron}$.

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Well yes, but you could argue that this tells you soemthing about the strong force or the proton or whatever, up to the point where you have to consider the breaking of isospin symmetry. The reason why this is a flue isn't proton/neutron asymmetry--it's that pure math can't tell you anything about quantities measured in kilograms. –  Jerry Schirmer Nov 12 '12 at 17:38
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maybe I haven't written properly what I wanted to stress: even though you come up with an improved formula that gives $m_{proton}$ in terms of e.g. $m_{W}$ (rather than a random unit of mass such as kg), it still means nothing to me until you show how the terms in the series arise from QCD. People use to forget that we know very well the theory behind the strong interactions (at least up to energies of few TeV) which is QCD that is defined by physical quantities such as $\Lambda_{QCD}$, $m_{u,d}$, $\theta_{QCD}$. –  argopulos Nov 12 '12 at 19:57
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one more comment: it's just damn hard to calculate $m_{proton}$ from first principles, nothing more. If you come with a formula and say nothing more than showing that it matches the measured value of $m_{proton}$ then it's pretty much like re-stating what the proton mass is (in a very contrived way), rather than explain it. –  argopulos Nov 12 '12 at 20:02
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sure, but there certainly have been physical discoveries made by finding ''accidents'' in data (usually physical quantities not depending on some supposed independent variable), and then coming up with theoretical explanations for those accidents. If (for example--I'm making things up), you had some relation like $m_{p} = (1-e^{-\pi})m_{n}$ or whatever, it could potentially be a reason to look for the answer in QCD, or modifications thereof. –  Jerry Schirmer Nov 12 '12 at 20:21
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