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# MRB constant

The MRB constant, named after Marvin Ray Burns, is a mathematical constant for which no closed-form expression is known. It is not known whether the MRB constant is algebraic or transcendental, nor even whether it is rational or irrational. However, it has been computed to over 6,000,000 digits of precision and terms of its simple continued fraction with no termination or obvious period of digits.

Marvin Ray Burns published his discovery of the constant in 1999. Before verifying with colleague Simon Plouffe that such a constant had not already been discovered or at least not widely published, Burns called the constant "rc" for root constant. At Plouffe's suggestion, the constant was renamed Marvin Ray Burns' Constant, and then shortened to "MRB constant" in 1999.

## Definition

The MRB constant is related to the following oscillating divergent series

$\sum _{k=1}^{\infty }(-1)^{k}~k^{1/k}=\sum _{k=1}^{\infty }(-1)^{k}~{\sqrt[{k}]{k}}.$ Its partial sums

$s_{n}=\sum _{k=1}^{n}(-1)^{k}~k^{1/k}$ are bounded within the closed interval

$\left[\liminf _{n\to \infty }\sum _{k=1}^{n}(-1)^{k}~k^{1/k},\limsup _{n\to \infty }\sum _{k=1}^{n}(-1)^{k}~k^{1/k}\right]=\left[C_{\rm {MRB}}-1,C_{\rm {MRB}}\right]$ since $\lim _{k\to \infty }k^{1/k}\,\to \,1$ , and where the MRB constant is defined as

$C_{\rm {MRB}}\equiv \limsup _{n\to \infty }\sum _{k=1}^{n}(-1)^{k}~k^{1/k}=\lim _{n\to \infty }\sum _{k=1}^{2n}(-1)^{k}~k^{1/k}=\lim _{n\to \infty }1+\sum _{k=1}^{2n+1}(-1)^{k}~k^{1/k}.$ The MRB constant can be explicitly defined by the following infinite sums

$C_{\rm {MRB}}=\lim _{n\to \infty }\sum _{k=1}^{2n}(-1)^{k}~k^{1/k}=\sum _{k=1}^{\infty }\left\{(2k)^{1/(2k)}-(2k-1)^{1/(2k-1)}\right\}=\sum _{k=1}^{\infty }\left\{{\Bigg [}(2k)^{1/(2k)}-1{\Bigg ]}-{\Bigg [}(2k-1)^{1/(2k-1)}-1{\Bigg ]}\right\}$ $=\sum _{k=1}^{\infty }(-1)^{k}~(k^{1/k}-1)$ where $\lim _{k\to \infty }k^{1/k}-1\,\to \,0$ , thus fulfilling the necessary and sufficient conditions for the convergence of an alternating series.

One should use acceleration methods when computing a numerical approximation of the MRB constant because it can be shown that one must sum a number in the order of $10^{n+1}$ iterations of $(-1)^{n}~(n^{1/n}-1)$ to get $n$ accurate digits of the MRB Constant. However, using a convergence acceleration of alternating series algorithm of Cohen-Villegas-Zagier one can compute the first 60 digits in only 100 iterations.

## CMRB

The decimal expansion of the MRB constant is

$C_{\rm {MRB}}=0.18785964246206712024851793405427323005590309490013\ldots$ (A037077)

The simple continued fraction expansion of the MRB constant is

$C_{\rm {MRB}}=0+{\cfrac {1}{5+{\cfrac {1}{3+{\cfrac {1}{10+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 5, 3, 10, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 12, 2, 17, 2, 2, 1, 1, 17, 1, 6, 4, 1, 3, 3, ...}

The first few convergents from the simple continued fraction expansion of the MRB constant are (see talk page for pattern investigation)

${\big \{}{\tfrac {0}{1}},{\tfrac {1}{5}},{\tfrac {3}{16}},{\tfrac {31}{165}},{\tfrac {34}{181}},{\tfrac {65}{346}},{\tfrac {294}{1565}},{\tfrac {359}{1911}},{\tfrac {653}{3476}},{\tfrac {1012}{5387}},{\tfrac {1665}{8863}},{\tfrac {15997}{85154}},{\tfrac {33659}{179171}},{\tfrac {421570}{2244069}},{\tfrac {876799}{4667309}},{\tfrac {15327153}{81588322}},{\tfrac {31531105}{167843953}},{\tfrac {78389363}{417276228}},{\tfrac {109920468}{585120181}},{\tfrac {188309831}{1002396409}},\ldots {\big \}}$ ### Limit (with an even number of terms) of the power tower of CMRB

$\lim _{n\to \infty }{\underset {i=1}{\overset {2n}{\rm {E}}}}c=\lim _{n\to \infty }\underbrace {c^{c^{c^{c^{.^{.^{.}}}}}}} _{2n}$ where $c$ is the MRB constant and $2n$ is the power tower height.

#### Decimal expansion of limit (with an even number of terms) of the power tower of CMRB

$\lim _{n\to \infty }{\underset {i=1}{\overset {2n}{\rm {E}}}}c=\lim _{n\to \infty }\underbrace {c^{c^{c^{c^{.^{.^{.}}}}}}} _{2n}=0.461921440164411445408588\ldots$ A052110 Decimal expansion of limit $\lim _{n\to \infty }\underbrace {c^{c^{c^{c^{.^{.^{.}}}}}}} _{2n}$ (with an even number of terms) where $c$ is the MRB constant defined in A037077.

{4, 6, 1, 9, 2, 1, 4, 4, 0, 1, 6, 4, 4, 1, 1, 4, 4, 5, 4, 0, 8, 5, 8, 8, 6, 4, 2, 6, 1, 4, 1, 9, 4, 5, 7, 8, 6, 3, 5, 0, 2, 8, 2, 8, 0, 1, 3, 6, 4, 8, 8, 2, 2, 8, 4, 4, 3, 4, 1, 6, 2, 9, 2, 7, 3, 5, 8, 9, ...}

## CMRB - 1

The decimal expansion of the MRB constant - 1 is

$C_{\rm {MRB}}-1=-0.8121403575379328797514820659\ldots$ The simple continued fraction expansion of the MRB constant - 1 is

$C_{\rm {MRB}}-1=-1+{\cfrac {1}{5+{\cfrac {1}{3+{\cfrac {1}{10+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{-1, 5, 3, 10, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 12, 2, 17, 2, 2, 1, 1, 17, 1, 6, 4, 1, 3, 3, ...}

The first few convergents from the simple continued fraction expansion of the MRB constant - 1 are

${\tfrac {-1}{1}},{\tfrac {-4}{5}},{\tfrac {-13}{16}},{\tfrac {-134}{165}},{\tfrac {-147}{181}},{\tfrac {-281}{346}},{\tfrac {-1271}{1565}},{\tfrac {-1552}{1911}},\ldots$ ## CMRB - 1/2

The Cesàro sum and the Levin u-transform sum give the midpoint of the two limit points

${\frac {(C_{\rm {MRB}})+(C_{\rm {MRB}}-1)}{2}}=C_{\rm {MRB}}-{\frac {1}{2}}$ The decimal expansion of the MRB constant - ${\frac {1}{2}}$ is

$C_{\rm {MRB}}-{\frac {1}{2}}=-0.3121403575379328797514820659\ldots$ The simple continued fraction expansion of the MRB constant - ${\frac {1}{2}}$ is

$C_{\rm {MRB}}-{\frac {1}{2}}=-1+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{10+{\cfrac {1}{20+{\cfrac {1}{2+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{-1, 1, 2, 4, 1, 10, 20, 2, 2, 2, 7, 1, 2, 2, 1, 2, 4, 9, 1, 1, 2, 4, 4, 1, ...}

## Integrated analog of the series

The integrated analog of the series is a complex-valued integral of oscillatory character

$I(2n)=\int _{1}^{2n}(-1)^{x}~{\sqrt[{x}]{x}}~dx=\int _{1}^{2n}e^{i\pi x}~x^{1/x}~dx,\quad n\in \mathbb {Z} .$ Not convergent in the continuum limit at $n\,\to \,\infty$ , the limit of the sequence of integrals with an integral difference in the upper limits $2n$ exists.

### Ultraviolet limit MI of the sequence of oscillatory integrals

The ultraviolet limit of the sequence of oscillatory integrals is defined as

$M_{I}\equiv \lim _{n\to \infty }I(2n),\quad n\in \mathbb {Z} .$ and has been evaluated by Richard J. Mathar.

#### Real and imaginary part

The decimal expansion of the real part of MI is

$\Re (M_{I})=0.0707760393115288035395\ldots$ The simple continued fraction for real part of $M_{I}$ is

$\Re (M_{I})=0+{\cfrac {1}{14+{\cfrac {1}{7+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{23+{\cfrac {1}{2+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 14, 7, 1, 2, 1, 23, 2, 1, 8, ...}

The decimal expansion of the imaginary part of MI is

$\Im (M_{I})=-0.68400038943793212918\dots$ The simple continued fraction for imaginary part of $M_{I}$ is

$\Im (M_{I})=-1+{\cfrac {1}{3+{\cfrac {1}{6+{\cfrac {1}{13+{\cfrac {1}{41+{\cfrac {1}{112+{\cfrac {1}{1+{\cfrac {1}{25+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{-1, 3, 6, 13, 41, 112, 1, 25, 1, 1, ...}

#### Absolute value of MI

The decimal expansion of absolute value $M_{I}$ is

$|M_{I}|=0.687652368927694369\ldots$ (A157852)

The simple continued fraction for absolute value of $M_{I}$ is

$|M_{I}|=0+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{4+{\cfrac {1}{1+{\cfrac {1}{24+{\cfrac {1}{1+{\cfrac {1}{4+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 1, 2, 4, 1, 24, 1, 4, 1, 2, 4, 4, 1, 8, ...}

## Series minus integrated analog of the series

By analogy with the Euler–Mascheroni constant

$\gamma \equiv \lim _{n\to \infty }\left\{\sum _{k=1}^{n}{\frac {1}{k}}-\int _{1}^{n}{\frac {dx}{x}}\right\}=\gamma _{\{1/x\}|_{1}^{n\to \infty }}$ we may define

$\gamma _{\{e^{i\pi x}~x^{1/x}\}|_{1}^{2n\to \infty }}\equiv \lim _{n\to \infty }\left\{\sum _{k=1}^{2n}(-1)^{k}~k^{1/k}-\int _{1}^{2n}I(2n)\right\},\quad n\in \mathbb {Z} ,$ $=\lim _{n\to \infty }\left\{\sum _{k=1}^{2n}e^{i\pi k}~k^{1/k}-\int _{1}^{2n}e^{i\pi x}~x^{1/x}~dx\right\},\quad n\in \mathbb {Z} .$ $=C_{\rm {MRB}}-M_{I}$ ### CMRB - MI

#### Real part of CMRB - MI

The decimal expansion of the real part of $C_{\rm {MRB}}-M_{I}$ is

$\Re (C_{\rm {MRB}}-M_{I})=0.117083603150538316709\ldots$ The simple continued fraction for real part of $C_{\rm {MRB}}-M_{I}$ is

$\Re (C_{\rm {MRB}}-M_{I})=0+{\cfrac {1}{8+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{5+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 8, 1, 1, 5, 1, 1, 1, 1, 2, 1, 5, 5, 1, ...}

#### Imaginary part of CMRB - MI

The decimal expansion of the imaginary part of $C_{\rm {MRB}}-M_{I}$ is

$\Im (C_{\rm {MRB}}-M_{I})=0.68400038943793212918\ldots$ The simple continued fraction for imaginary part of $C_{\rm {MRB}}-M_{I}$ is

$\Im (C_{\rm {MRB}}-M_{I})=0+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{6+{\cfrac {1}{13+{\cfrac {1}{41+{\cfrac {1}{112+{\cfrac {1}{1+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 1, 2, 6, 13, 41, 112, 1, 25, ...}

#### Absolute value of CMRB - MI

The decimal expansion of $|C_{\rm {MRB}}-M_{I}|$ is

$|C_{\rm {MRB}}-M_{I}|=0.693948919501972825\ldots$ The simple continued fraction for $|C_{\rm {MRB}}-M_{I}|$ is

$|C_{\rm {MRB}}-M_{I}|=0+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{5+{\cfrac {1}{\ddots }}}}}}}}}}}}}}}}$ giving the sequence

{0, 1, 2, 3, 1, 2, 1, 5, 9, 1, 8, 29, ...}

## Inquiry on Feb 28, 2013

It seems that using regularization the divergent series

$\sum _{k=1}^{\infty }(-1)^{k}~(x*k^{1/k}+y*k)=(C_{\rm {MRB}}-1/2)x-1/4y$ .

Any help here would be great!

For all complex z, the upper limit point of s(n)= $\sum _{k=1}^{n}(-1)^{k}~(k^{1/k}-z)$ is the $C_{\rm {MRB}}$ .
For all real a, the partial sums s(n)= $\sum _{k=1}^{n}(-1)^{k}~(k^{1/k}-a)$ are bounded so that their limit points form an interval [-1.+ $C_{\rm {MRB}}$ +a, $C_{\rm {MRB}}$ ] of length 1-a.

## Various forms for the MRB constant

$C_{\text{MRB}}$ $=\sum _{k}^{\infty }\left((2k)^{\frac {1}{2k}}-(2k-1)^{\frac {1}{2k-1}}\right)$ $=\sum _{n=1}^{\infty }(-1)^{n}\left(n^{\frac {1}{n}}-1\right)$ $=\sum _{n=1}^{\infty }(-1)^{n}\left(n^{\frac {x}{n}}-1\right){\text{ /. }}\,x\to 25.65665403510586285599072958839{\text{ ...}},$ $=\int _{0}^{\infty }{\text{csch}}(\pi t)\Im \left((1+it)^{\frac {1}{1+it}}\right)\,dt$ $=\int _{0}^{\text{b}}{\text{csch}}(\pi t)\Im \left((1+it)^{\frac {1}{1+it}}\right)\,dt{\text{ /. }}\,{\text{b}}\to 1.32379817612535131470834{\text{...,}}$ $=\int _{0}^{\infty }{\text{csch}}(\pi t)\Im \left((1+it)^{\frac {x}{1+it}}\right)\,dt{\text{ /. }}\,x\to 25.65665403510586285599072958839{\text{ ...}},$ $={\underset {k=1}{\overset {\infty }{-\sum }}}{\frac {(-1)^{k}}{k!}}\eta ^{(k)}k$ ${\text{where }}\eta ^{(k)}k{\text{ denotes the }}k{\text{th derivative of the Dirichlet eta function at }}k.$ $={\underset {k=1}{\overset {\infty }{-\sum }}}{\frac {c_{k}}{k!}}\eta ^{(k)}0{\text{/.}}c_{k}=\sum _{j=1}^{\infty }(-1)^{j}j^{k-j}\left({\begin{array}{c}k\\j\\\end{array}}\right)=\{-1,-1,2,9,4,-95,-414,49,10088,55521{\text{...}}\}{\text{where }}\eta ^{(k)}0{\text{ denotes the }}k{\text{th derivative of the Dirichlet eta function at }}0.$ The proof is found here. 

$f(x)=(-1)^{x}\left(x^{1/x}-1\right);{\text{CMRB}}=\Im \left(\int _{0}^{\infty }{\frac {2f(1-{\text{it}})}{e^{2\pi t}-1}}\,dt\right).$ $g(x)=x^{1/x};{\text{CMRB}}=i\int _{0}^{\infty }{\frac {g(1-it)-g(1+it)}{\exp(\pi t)-\exp(-\pi t)}}dt.$ ## Many integral forms for the MRB constant

Let

{{x = 25.656654035105862855990729 ... and
{u = -3.20528124009334715662802858},
{u = -1.975955817063408761652299},
{u = -1.028853359952178482391753},
{u = 0.0233205964164237996087020},
{u = 1.0288510656792879404912390},
{u = 1.9759300365560440110320579},
{u = 3.3776887945654916860102506},
{u = 4.2186640662797203304551583} or

$u=\infty$ }


or

let
{{x = 1 and
{u = 2.451894470180356539050514},
{u = 1.333754341654332447320456} or

$u=\infty$ }


then

$\int _{0}^{u}{\text{csch}}(\pi t)\Im \left((1+it)^{\frac {x}{1+it}}\right)\,dt=C_{\rm {MRB}}.$ ## Many proper integral forms for the MRB constant

$gx=x^{1/x};{\text{CMRB}}=\sum _{k=1}^{\infty }(-1)^{k}(gk-1)$ $w={\frac {i(t-b)}{t-a}};$ $p=g(w+1)\csc(\pi w)$ ${\text{CMRB}}=\int _{a}^{b}{\frac {(b-a)\Re (p)}{(t-a)^{2}}}\,dt,a,b\in \mathbb {C} .$ ## Non-trivial relationship between The MRB constant (CMRB) and its integrated analog (CMKB)

Let g(x) = x^{1/x}.

$\mathrm {CMRB} =\lim \limits _{N\to \infty }\sum \limits _{n=1}^{2N}(-1)^{n}g(n)=\lim \limits _{N\to \infty }\int \limits _{0}^{2N}{\frac {\Im g(1+it)}{\sinh(\pi t)}}dt.$ The proof is found here. 

$\mathrm {CMKB=M1} =\lim \limits _{N\to \infty }\int \limits _{1}^{2N}(-1)^{x}g(x)dx=(-i)\lim \limits _{N\to \infty }\int \limits _{0}^{2N}{\frac {g(1+it)}{\exp(\pi t)}}dt$ $=\lim \limits _{N\to \infty }\int \limits _{1}^{2N}e^{i\pi t}g(t)dt=(-i)\lim \limits _{N\to \infty }\int \limits _{0}^{2N}e^{i^{2}\pi t}g(1+it)dt.$ The proof is found here. .

Here are the results from Mathematica in checking those formulas:

 Clear[g]; g[x_] = x^(1/x); CMRB = N[NSum[(-1)^n (g[n] - 1), {n, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 57], 30]

(* 0.187859642462067120248517934054*)

g[x_] = x^(1/x); CMRB -  NIntegrate[Im[g[1 + I t]/(Sinh[Pi t])], {t, 0, Infinity},
WorkingPrecision -> 30]

(* 0.*10^-31*)

g[x_] = x^(1/x); Timing[ MKB = N[NIntegrate[Exp[I Pi t] (g[t]), {t, 1, Infinity},
WorkingPrecision -> 57], 30] - I/Pi]

(*{0.03125,
0.070776039311528803539528021830 - 0.684000389437932129182744459993 I}*)

g[x_] = x^(1/x); Timing[MKB - (N[NIntegrate[(-1)^t (g[t]), {t, 1, Infinity},
WorkingPrecision -> 57, MaxRecursion -> 3500], 30] - I/Pi)] // Quiet

(* {30.8438, 6.3*10^-29 - 6.00*10^-28 I}*)

g[x_] = x^(1/x); Timing[ MKB -
(-N[ I NIntegrate[(g[(1 + t I)])/( Exp[Pi t]), {t, 0, Infinity},  WorkingPrecision -> 57] + I/Pi, 30])]

(* {0.046875, 0.*10^-31 + 0.*10^-31 I}*)

g[x_] = x^(1/x); Timing[ MKB - (-N[
I NIntegrate[Exp[I^2 Pi t] (g[1 + t I]), {t, 0, Infinity},  WorkingPrecision -> 57] + I/Pi, 30])]

(* {0.03125, 0.*10^-31 + 0.*10^-31 I}*)


There is another connection; it follows.

${\text{In the following equations Capital}}\Im {\text{ means the imaginary part , and lower case i is the imaginary unit i.e. sqrt(-1)}}.$ $g(x)=x^{1/x}{\text{ and }}f(x)=(-1)^{x}(g(x)-1)$ ${\text{CMRB}}=\sum _{n=1}^{\infty }fn=i\int _{0}^{\infty }{\frac {g(1-it)-g(1+it)}{e^{\pi t}-e^{-\pi t}}}\,dt=\int _{0}^{\infty }{\frac {\Im (g(1+it))}{\sinh(\pi t)}}\,dt=\int _{0}^{\infty }{\frac {2f(1-it)}{e^{2\pi t}-1}}\,dt.$ ${\text{Re(MKB)}}==i{\underset {N\to \infty }{\int }}_{0}^{2N}{\frac {g(1-it)-g(1+it)}{2e^{\pi t}}}=\Im \int _{0}^{\infty }{\frac {f(1-it)-f(1+it)}{e^{2\pi t}+1}}\,dt=\Im \int _{0}^{\infty }{\frac {f(1+it)+f(1-it)}{e^{2\pi t}-1}}\,dt$ ${\text{NoMKB}}=\Im \int _{0}^{\infty }{\frac {f(1-it)-f(1+it)}{e^{2\pi t}-1}}dt.$ ${\text{NoMKB+Re(MKB)}}=CMRB.$ ${\text{Im(MKB)}}=-i{\underset {N\to \infty }{\text{lim}}}\int _{0}^{2N}{\frac {g(it+1)+g(1-it)}{2e^{\pi t}}}\,dt.$ $f(x)=e^{i\pi x}\left(1-(x+1)^{\frac {1}{x+1}}\right),$ ${\text{CMRB}}=\sum _{n=0}^{\infty }fn,$ ${\text{M2}}=\int _{0}^{\infty }ft\,dt,=\int _{0}^{i\infty }ft\,dt,$ ${\text{part}}=\int _{0}^{i\infty }{\frac {{\text{Im}}(2(f(-t)))+(f(-t)-ft)}{-1+e^{-2i\pi t}}}\,dt,$ $(1-i){\text{CMRB}}={\text{M2}}-{\text{part}}.$ 