Almost Integer

Author

Eric W. Weisstein
November 26, 2023

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Numbers/AlmostInteger.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/AlmostInteger.html.

©2023 Wolfram Research, Inc. except for portions noted otherwise

By Category

Music

Diaschisma

(2048/2025)

N[%]

Diesis

https://en.wikipedia.org/wiki/Diesis

(128/125)

N[%]

Holdrian comma

https://en.wikipedia.org/wiki/Holdrian_comma

(2^(1/53))//N

Kleisma

(2^(-6))(3^(-5))(5^6)

N[%]

Pythagorean comma

https://en.wikipedia.org/wiki/Pythagorean_comma

(((((3/2)))^12)/(2^7))

N[%]

Schisma

((5 (3^8))/(2^15))

N[%]

Semicomma

(((3^3)(5^7))/(2^21))

N[%]

Septimal comma

(64/63)//N

Septimal diesis

(49/48)//N

Septimal kleisma

(225/224)//N

Septimal semicomma

(126/125)//N

Syntonic comma

https://en.wikipedia.org/wiki/Syntonic_comma

(81/80)//N

Pegg

notes={"A0","A#0","B0","C1","C#1","D1","D#1","E1","F1","F#1","G1","G#1","A1","A#1","B1","C2","C#2","D2","D#2","E2","F2","F#2","G2","G#2","A2","A#2","B2","C3","C#3","D3","D#3","E3","F3","F#3","G3","G#3","A3","A#3","B3","C4","C#4","D4","D#4","E4","F4","F#4","G4","G#4","A4","A#4","B4","C5","C#5","D5","D#5","E5","F5","F#5","G5","G#5","A5","A#5","B5","C6","C#6","D6","D#6","E6","F6","F#6","G6","G#6","A6","A#6","B6","C7","C#7","D7","D#7","E7","F7","F#7","G7","G#7","A7","A#7","B7","C8"};

freq=55/2 2^(Range[0,87]/12);

TextGrid[SortBy[{#[[1]],#[[2]],N[#[[2]]]}&/@Select[Transpose[{notes,freq}],0<Abs[Round[#[[2]]]-#[[2]]]<1/20&],Abs[Round[#[[3]]]-#[[3]]]&],Dividers->All,Alignment->{{Left,{Right}}}]//TraditionalForm

Units

(QuantityTF[1, "in" "ly""/(""au" "mi"")", inch light year per astronomical unit mile, (("Inches" "LightYears")/("AstronomicalUnit" "Miles"))])

UnitConvert[Quantity[1,(("Inches")/("Miles"))/(("AstronomicalUnit")/("LightYears"))]]

UnitConvert[Quantity[1,(("Inches")/("Miles"))/"AstronomicalUnit"] Quantity[365+1/4,"Days" "SpeedOfLight"]]

N[%,20]

(QuantityTF[1, (("au")^2)"s""/(""m" (("day")^2)"c"")", astronomical unit squared second per meter day squared speed of light, (((("AstronomicalUnit")^2) "Seconds")/((("Days")^2) "Meters" "SpeedOfLight"))])

Jose Garcai, pers. comm., Jun. 25, 2022

Quantity[1,(((((("AstronomicalUnit")/("Days"))))^2)/("SpeedOfLight" (("Meters")/("Seconds"))))]//UnitConvert

N[%,20]

Physics

(QuantityTF[1, "k" "e" "c""/""h", Boltzmann constant elementary charge speed of light per Planck constant, (("BoltzmannConstant" "ElementaryCharge" "SpeedOfLight")/("PlanckConstant"))])

Michael Trott, pers. comm., Apr 28, 2011

UnitConvert[Quantity[1,"BoltzmannConstant" "ElementaryCharge" "SpeedOfLight"/"PlanckConstant"]]

N[%]

((Subscript[P, C])+(7/9)) ((QuantityTF[1, (Subscript["R", ∞]^2)(Subscript["ε", 0]), Rydberg constant squared electric constant, "ElectricConstant" (("RydbergConstant")^2)]))

Michael Trott, pers. comm., Apr 28, 2011

UnitConvert[(0.1601`4+(7/9))Quantity[1, "ElectricConstant"  (("RydbergConstant")^2)]]

Simpsons FLT

(((1782^12)+(1841^12))/(1922^12))

Simpsons episode (Homer^3)

N[(((1782^12)+(1841^12))/(1922^12)),20]

{(1782^12)+(1841^12),(1922^12)}//Column

N[1-(((1782^12)+(1841^12))/(1922^12)),20]

N[((((1782^12)+(1841^12)))^(1/12)),20]

(((3987^12)+(4365^12))/(4472^12))

Simpsons episode  The Wizard of Evergreen Terrace

N[(((3987^12)+(4365^12))/(4472^12)),20]

{(3987^12)+(4365^12),(4472^12)}//Column

N[(((3987^12)+(4365^12))/(4472^12))-1,20]

N[((((3987^12)+(4365^12)))^(1/12)),20]

2nd powers

pow2=With[{n=2},DeleteCases[SortBy[Flatten[Table[{val=(((a^n)+(b^n))/(((c=Floor[((((a^n)+(b^n)))^(1/n))]))^n)),N[val,20],{a,b,c}},{a,1000},{b,1000}],1],Abs[N[#[[1]]-1]]&],{1,_,_}]];//Timing

TextGrid[Take[pow2,5],Dividers->All]

pow2=With[{n=2},DeleteCases[SortBy[Flatten[Table[{val=(((a^n)+(b^n))/(((c=Ceiling[((((a^n)+(b^n)))^(1/n))]))^n)),N[val,20],{a,b,c}},{a,1000},{b,1000}],1],Abs[N[#[[1]]-1]]&],{1,_,_}]];//Timing

TextGrid[Take[pow2,5],Dividers->All]

12th powers

pow12=With[{n=12},DeleteCases[SortBy[Flatten[Table[{val=(((a^n)+(b^n))/(((c=Floor[((((a^n)+(b^n)))^(1/n))]))^n)),N[val,20],{a,b,c}},{a,500},{b,a,500}],1],Abs[N[#[[1]]-1]]&],{1,_,_}|{_,_,{_,x_,x_}}]];//Timing

TextGrid[Take[pow12,10],Dividers->All]

pow12c=With[{n=12},DeleteCases[SortBy[Flatten[Table[{val=(((a^n)+(b^n))/(((c=Ceiling[((((a^n)+(b^n)))^(1/n))]))^n)),N[val,20],{a,b,c}},{a,500},{b,a,500}],1],Abs[N[#[[1]]-1]]&],{1,_,_}]];//Timing

TextGrid[Take[pow12c,10],Dividers->All]

j-Function

Almost Integers

Select[{#,r=Round[x=Exp[π Sqrt[#]]];N[x-r,4]}&/@Range[1000],Abs[Last[#]]<.01&]//ColumnForm

Class Number 1

Mod[h1={1,2,3,7,11,19,43,67,163},4]

{#,r=Round[x=Exp[π Sqrt[#]]],N[x-r,10]}&/@h1//ColumnForm

Class Number 2

Mod[h2={88,148,232},4]

{#,r=Round[x=Exp[π Sqrt[#/4]]],N[x-r,10]}&/@h2//ColumnForm

By Person

Barron

e

N[{(Pi Catalan)^(2/7+EulerGamma)/Catalan,E},20]

{#,N[#,20]}&@E/((Pi Catalan)^(2/7+EulerGamma)/Catalan)

EulerGamma

EulerGamma/((Pi Catalan)^(2/7+EulerGamma)/(GoldenRatio E))

N[%]

Phi

((Pi Catalan)^(2/7+EulerGamma)/Catalan^3)/2

N[{((Pi Catalan)^(2/7+EulerGamma)/Catalan^3)/2,GoldenRatio},20]

{#,N[#,20]}&@{((Pi Catalan)^(2/7+EulerGamma)/Catalan^3)/2/GoldenRatio}

Borwein

(∑_(n=1)^∞)((⌊n tanh(π)⌋)/(10^n))≈(1/81)

Tanh[π]//N

V5

Sum[((Floor[n Tanh[π]])/(10^n)),{n,∞}]

NSum[((Floor[n Tanh[π]])/(10^n)),{n,∞},WorkingPrecision->500,NSumTerms->400,NSumExtraTerms->300]-(1/81)

NSum[((Floor[n Tanh[π]])/(10^n)),{n,∞},WorkingPrecision->500,NSumTerms->400,NSumExtraTerms->300]-(1/81)

Position[Subtract@@@Partition[Table[n-Floor[n Tanh[π]],{n,10^5}],2,1],_?(#!=0&)]//Flatten

V7

Sum[((Floor[n Tanh[π]])/(10^n)),{n,∞}]

(∑_(k=-∞)^∞)(1/(10^((((k/100)))^2)))

V5

Sum[(1/(10^(((k/100))^2))),{k,-∞,∞}]//Simplify

N[{100(Sqrt[(π/(Log[10]))]),EllipticTheta[3,0,(10^(-1/10^4))]},10]

(10^(-1/10^4))//N

Block[{$MaxExtraPrecision=10^5},N[EllipticTheta[3,0,(1/(10^(1/10000)))]-100(Sqrt[(π/(Log[10]))]),10]
]

V7

Sum[(1/(10^(((k/100))^2))),{k,-∞,∞}]

Broukhis

r=x/.FindRoot[Cos[x]==x,{x,.7},WorkingPrecision->40]

r/(Pi/160)^(1/13)

1-%

Davis

N[Log[2]+Log[10,2],20]

DePompeo

CorrectDigits[{((5GoldenRatio E)/(7 Pi)),1}]

N[((5GoldenRatio E)/(7 Pi)),10]

Dubuque

1/4(Cos[1/10]+Cosh[1/10]+2 Cos[Sqrt[2]/20]Cosh[Sqrt[2]/20])

N[%4-1,50]

FullSimplify[%]

TrigExpand[%7]

TrigToExp[%7]

Simplify[%]

FullSimplify[%]

Edde

"Salim M. Edde" <salim.edde@murex.com>
Fri, 7 Sep 2007 00:20:20 -0500

N[Exp[-PolyGamma[(1/4) (2+(Sqrt[3]))]],20]

Ehlke

Thu, 07 Apr 2005 21:29:07 +0200

CorrectDigits[((π^9)/(E^8)),10]

N[((π^9)/(E^8)),20]

Galliani

1/2+1/23+1/37+1/149-1/(7*3^3*5^3*41)

N[%-EulerGamma,17]

Paolo Galliani (19 Apr 2002)

g=FullSimplify[EulerGamma^(-1/3)+1/(Pi^4((1+2EulerGamma-2/(130+Pi^2))^3))]

N[g-Zeta[3],20]

Glasser

{(BesselJ[1, 1]),(BesselJ[0, 1])}

pers. comm. 2010-02-02

N[{BesselJ[1,1],(1/94)(23+24BesselJ[0,1])},20]//Column

CorrectDigits[BesselJ[1,1],(1/94)(23+24BesselJ[0,1])]

6

vc={BesselJ[1,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^6,prec=100},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//Factor//TraditionalForm

7

vc={BesselJ[1,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^7,prec=100},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//Factor//TraditionalForm

8

vc={BesselJ[1,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^8,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//Factor//TraditionalForm

11

vc={BesselJ[1,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^11,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//TraditionalForm

{(BesselJ[2, 1]),(BesselJ[0, 1])}

4

vc={BesselJ[2,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^4,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//TraditionalForm

5

vc={BesselJ[2,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^5,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//TraditionalForm

6

vc={BesselJ[2,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^6,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//TraditionalForm

7

vc={BesselJ[2,1],BesselJ[0,1],1};
rules=Thread[Most[vc]->{a,b}];
l=With[{mult=10^7,prec=20},LatticeReduce[Transpose[Join[IdentityMatrix[Length[vc]],Round[N[{mult vc},prec]]]]]
]

expr=Most[#].vc&/@l

N[expr]

(Solve[(#/.rules)==0,a]/.(Reverse/@rules))&/@expr//TraditionalForm

Gosper

g=1-262537412640768744 Exp[-π Sqrt[163]]-196884Exp[-2π Sqrt[163]]+ 103378831900730205293632 Exp[-3π Sqrt[163]]

FullSimplify[g]

N[g,100]

Hammond

N[510Log[10,7],10]

CorrectDigits[510Log[10,7],431]

N[88Log[89],12]

CorrectDigits[88Log[89],395]

N[272Log[π,97],20]

CorrectDigits[272Log[π,97],1087]

Hickerson (sci.math)

N[163/Log[163],20]

(Subscript[h, n])

table=Table[{n,N[n!/(2Log[2]^(n+1)),20]//AccountingForm},{n,0,17}]

DisplayForm[GridBox[table,GridFrame->True,RowLines->True,ColumnLines->True,ColumnAlignments->Right]]

Contour integral

V5

a=n!/2 Sum[1/(Log[2]+2π I k)^(n+1),{k,-∞,∞}]

N[Table[a,{n,5}],20]

FullSimplify[a]

FunctionExpand[a]

FullSimplify[%]

N[Table[%,{n,5}],20]

V7

n!/2 Sum[1/(Log[2]+2π I k)^(n+1),{k,-∞,∞}]

FullSimplify[%,n∈Integers&&n>0]

Recurrence

RSolve[{f[n]==Sum[Binomial[n,k]f[n-k],{k,n}],f[1]==1},f[n],n]

f[0]:=1
f[n_]:=Sum[Binomial[n,k]f[n-k],{k,n}]

Table[f[n],{n,5}]

Exponential generating function

CoefficientList[Series[1/(2-Exp[z]),{z,0,5}],z]Table[n!,{n,0,5}]

Round

Table[Round[n!/(2Log[2]^(n+1))],{n,5}]

Hudson

Mon Oct 18, 2004

N[Log[Khinchin ] - Log[Log[Khinchin]],20]

Fri Feb 04, 2005

Sqrt[45]

N[Sqrt[45]^EulerGamma,20]

Joseph

cos((31^(1/3)))==-1

N[1+Cos[(31^(1/3))],20]

CorrectDigits[1+Cos[(31^(1/3))],0]

cos(355)==-1

N[1+Cos[355],20]

CorrectDigits[1+Cos[355],0]

erfi(erfi((1/(Sqrt[3]))))==1

N[Erfi[Erfi[1/(Sqrt[3])]],20]

CorrectDigits[Erfi[Erfi[1/(Sqrt[3])]],1]

Kobayashi

N[10((((EulerGamma^(-1/2))-1))^2),10]

CorrectDigits[{10((((EulerGamma^(-1/2))-1))^2),1}]

N[10/81(11-2Sqrt[10])-EulerGamma,20]

CorrectDigits[{10/81(11-2Sqrt[10])-EulerGamma,0}]

Le Lionnais

N[((Log[640320^3+744])/Pi)^2-163,30]

Maze and Minder

m[n_]:=Sum[Log[2]/(2^(-(k/2)) + 2^(k/2))^n, {k, -∞, ∞}]

V5

m[n]

Table[m[n],{n,3}]

N[Table[m[n],{n,5}],10]

V7

m[n]

s=Table[m[n],{n,4}]

FullSimplify[s]//TraditionalForm

N[Table[m[n],{n,5}],10]

FullSimplify[(1/2) (-4 π-Log[2]+4 I QPolyGamma[0,-((I π)/(Log[2])),(Sqrt[2])]-4 I QPolyGamma[0,((I π)/(Log[2])),(Sqrt[2])])-π]//TraditionalForm

N[%]

n==1

V5

Sum[Log[2]/(2^(-k/2) + 2^(k/2)), {k, -Infinity, Infinity}]

RealDigits[NSum[Log[2]/(2^(-k/2)+2^(k/2)),{k,-Infinity,Infinity},WorkingPrecision->30],10][[1]]

2π Sum[Sech[((2k (π^2))/(Log [2]))],{k,∞}]

2π NSum[Sech[((2k (π^2))/(Log [2]))],{k,∞},WorkingPrecision->50]

NSum[Log[2]/(2^(-k/2)+2^(k/2)),{k,-Infinity,Infinity},WorkingPrecision->100]-Pi

V7

s=Sum[Log[2]/(2^(-k/2) + 2^(k/2)), {k, -Infinity, Infinity}]-π

N[s,20]

TraditionalForm[Expand[s]/.Log[2]->1/(Subscript[l, 2])]

2π Sum[Sech[((2k (π^2))/(Log [2]))],{k,∞}]

N[%]

n==2

V5

Sum[Log[2]/(2^(-k/2) + 2^(k/2))^2, {k, -Infinity, Infinity}]

RealDigits[NSum[Log[2]/(2^(-k/2)+2^(k/2))^2,{k,-Infinity,Infinity},WorkingPrecision->30],10][[1]]

((4(π^2))/(Log[2])) Sum[k Csch[((2(π^2)k)/(Log [2]))],{k,∞}]

((4(π^2))/(Log[2]))NSum[k Csch[((2(π^2) k)/(Log [2]))],{k,∞},WorkingPrecision->50]

V7

s=Sum[Log[2]/(2^(-k/2) + 2^(k/2))^2, {k, -Infinity, Infinity}]-1

TraditionalForm[Expand[s]/.Log[2]->1/(Subscript[l, 2])]

N[%,20]

((4(π^2))/(Log[2])) Sum[k Csch[((2(π^2)k)/(Log [2]))],{k,∞}]

N[%]

Pegg

FullSimplify[(91/10)^(1/4)-33/19]

N[%,20]

Mar 04, 2002

(23/9)^5

N[%,20]

%-109

CorrectDigits[E,3-Sqrt[5/63]]

CorrectDigits[Pi,4-(105/166)^(1/3)]

CorrectDigits[EulerGamma,1/15+(35/263)^(1/3)]

Almost Triangle Dissection

label[x_]:=Module[{dd=FullSimplify@Function[r,Sqrt[r.r]][Subtract@@First[x]]},
If[IntegerQ[dd],dd,TraditionalForm[HoldForm[d]]]
]

TriangleDissection[a_,p__List]:=
Module[{ts=Triangle[a,Sequence@@#]&/@{p},
s},
s=Flatten[{Line[Last/@Vertices/@ts],Sides/@ts}];
{#,Text[label[#],Coordinates[Midpoint[#]],Background->White]}&/@s
]

{x1,y1}=Select[{x,y}/.Solve[{(x^2+y^2)==27^2,(x-30)^2+y^2==22^2},{x,y}],Positive[Last[#]]&][[1]]

{x2,y2}=Select[{x,y}/.Solve[{(x^2+y^2)==23^2,(x-30)^2+y^2==16^2},{x,y}],Positive[Last[#]]&][[1]]

d=Simplify[Sqrt[(x1-x2)^2+(y1-y2)^2]]

N[d-7,10]

Show[Graphics[TriangleDissection[30,{23,16},{27,22}]],AspectRatio->Automatic]

Lunair

pierrotlunair@yahoo.it (18 Apr 2002)

z=FullSimplify[Pi^5 (Catalan-EulerGamma)^5/(1/EulerGamma^(1/2)+1/5000)]

N[z,20]

N[Zeta[5],20]

CorrectDigits[Zeta[5],z]

Im[Zeta[5]]//N

Plouffe

FullSimplify[20 ((1/4 Sqrt[5]+1/4)Gamma[3/4]^2)/Sqrt[Pi]/Exp[5/6Pi]-1]

N[%,20]

Povolotsky

10 tanh(((28 π)/15))-((π^9)/(e^8))

Subtract@@{10Tanh[(28/15)π],((π^9)/(E^8))}//N

(((1782^12)+(1841^12))/(1922^12))

Jan. 1, 2022

But actually this appeared in Simpsons episode (Homer^3)

N[(((1782^12)+(1841^12))/(1922^12)),20]

see (Homer^3)-like above

Reira

N[22π^4,20]

%-2143

Rolli

Pascal Rolli <pascal.rolli(AT)gmx.net>
19 Feb 2004

2Nest[Cos,5,7]^2

N[2Nest[Cos,5,7]^2,20]

Sloane/Conway/Plouffe

Apparently noticed almost simultaneously around 1988 by N. J. A. Sloane, J. H. Conway, and S. Plouffe

D.~Bamberger, Nov. 26, 2023:

Also:

https://en.wikipedia.org/wiki/Talk:Mathematical_coincidence#Explaining_%7F'%22%60UNIQ--postMath-00000080-QINU%60%22'%7F

A consequence of the Jacobian theta functional identity:

Sum[(8π (k^2)-2)Exp[-π (k^2)],{k,∞}]

D.~Bamberger, Nov. 26, 2023:

Credit for the idea to differentiate the Jacobian identity at τ=i goes to Aaron Doman, a.k.a. @MathFromAlphaToOmega, who mentioned it in a comment on a video by popular math YouTuber @Mathologer (Burkard Polster). You may want to contact Aaron to ask if they are fine with being credited for it (https://www.adomanmath.com/contact). I am okay with being named if you wish to, but all I did was notice that it contradicted what's said on MathWorld, flesh out the details to confirm that it's correct, and of course get in touch with you.

Starting from the definition of the Jacobi theta function ϑ(0,τ) = Σe^(iπτn²), where the sum goes from n=-∞ to +∞, and the Jacobian identity ϑ(0,-1/τ) = √(τ/i)*ϑ(0,τ) (compare https://en.wikipedia.org/wiki/Theta_function#Jacobi_identities), differentiating on both sides gives

ϑ'(0,-1/τ) = √(-iτ)/(2τ)*ϑ(0,τ)+√(τ/i)*ϑ'(0,τ) = Σ[2πτn²-1)/(2√(-iτ)]e^(iπτn²). At the specific value τ=i, this evaluates to
Σ(iπn²-i/2)e^(-πn²).

On the other hand, one can start with just the left side, getting

ϑ'(0,-1/τ) = (iπ/τ²)Σn²e^(-iπn²/τ), which at τ=i is equal to
Σ(-iπn²)e^(-πn²).

Equating the two different ways to write ϑ'(0,-1/τ) gives Σ[(iπn²-i/2)+iπn²]e^(-πn²) = 0, which after dividing by i simplifies to
Σ(2πn²-1/2)e^(-πn²) = 0.

Split the sum into three parts (-∞ to -1, 0, and +1 to +∞), and notice that the sums from -∞ to -1 and +1 to +∞ are identical:
2Σ(2πn²-1/2)e^(-πn²) = 1/2, where the sum goes from n=1 to +∞. Multiplying by 2 gives the interesting sum
Σ(8πn²-2)e^(-πn²) = 1 (with the sum from n=1 to +∞; compare https://www.wolframalpha.com/input?i=sum+%288pi*n%C2%B2-2%29e%5E%28-pi*n%C2%B2%29+from+n%3D1+to+infinity). I haven't found anyone who has calculated it explicitly, but I am sure that sum is well known. I think the idea to combine it with the approximation π ≈ 22/7 is novel.

This converges very fast. Using just the first term (n=1) gives
e^π ≈ 8π-2 = 7π-2+π ≈ 7(22/7)-2+π = 20+π.

Funnily, the approximation π ≈ 22/7 makes the formula an order of magnitude more precise (7π-2 = 19.991..., e^π-π = 19.9991...). I guess it's by coincidence that much of the error cancels out. Even if it didn't, I think 19.991 would still be close enough to 20 to be interesting.

bug(442683): Implement Sum[(8 Pi k^2 - 2) Exp[-Pi k^2], {k, Infinity}] == 1

N[Sum[(8π (k^2)-2)Exp[-π (k^2)],{k,100}]-1,10^4]

NSum[(8π (k^2)-2)Exp[-π (k^2)],{k,2,1000}]

NSum[(8π (k^2)-2)Exp[-π (k^2)],{k,∞}]

(8π (k^2)-2)Exp[-π (k^2)]/.k->1

N[%]

N[2(4π-1)Exp[-π],20]

(E^(-π)) (-2+8 π)≈1

Exp[Pi]≈8π-2

Exp[Pi]≈8π-2=π+7π-2≈π+7(22/7)-2

π+7(22/7)-2

Stay

Date: Tue, 17 Mar 2009 11:24:45 -0700
From: Mike Stay
To: math-fun <math-fun@mailman.xmission.com>

((Exp[Pi]-Log[3])/(Log[2]))-(4/5)//FullSimplify//TraditionalForm

N[%,20]

CorrectDigits[31,-(4/5)+(((E^π)-Log[3])/(Log[2]))]

10((Exp[Pi]-Log[3])/(Log[2]))//N[#,20]&

Stoschek

N[163(Pi-E),11]

N[(Exp[Pi]-Pi),10]

N[53453/Log[53453],20]

N[((2-1)^2+(5^2-1)^2/(6^2+1))E-((2+1)^2+(5^2+1)^2/(6^2-1))^(-1),20]

Expand[((2-1)^2+(5^2-1)^2/(6^2+1))E-((2+1)^2+(5^2+1)^2/(6^2-1))^(-1)]

α=7.29735308 10^(-3);
δ=4.66920160910299067185320;

1/α

N[(28-1/δ)(1/α-137),15]

Terr

N[(GoldenRatio/(2^(Log[2]))),20]

CorrectDigits[{(GoldenRatio/(2^(Log[2]))),1}]

Trott

CorrectDigits[{Sin[11],-1}]

N[Sin[11],10]

CorrectDigits[{Sin[(2^(1/5))2017],-1}]

N[Sin[(2^(1/5))2017],20]

Weisstein

N[((7 Log[2])/64)-(131/1728),20]

N[(2411287/30240)-((100 (π^2))/9)+((1877 Log[2])/21)-((200 ((Log[2])^2))/3),20]

Expand[((80497-12040 (π^2)+105376 Log[2]-72240 ((Log[2])^2))/40320)]

N[%,20]

Equilateral Triangle Line Picking

N[(1/30) (2+4 (Sqrt[2])+(4+(Sqrt[2])) ArcSinh[1]),10]

N[Sqrt[2]-1,10]

N[(1/30) (2+4 (Sqrt[2])+(4+(Sqrt[2])) ArcSinh[1])-((Sqrt[2])-1),20]

Eight Curve

Plot[Re[EllipticPi[(1/2)*I*(I + Sqrt[7]), I*ArcSinh[Sqrt[-((2*I)/(I + Sqrt[7]))]*Tan[t]], (I + Sqrt[7])/(I - Sqrt[7])]],{t,0,π/2},PlotStyle->Red]

soln=t/.Last@NMaximize[Re[EllipticPi[(1/2)*I*(I + Sqrt[7]), I*ArcSinh[Sqrt[-((2*I)/(I + Sqrt[7]))]*Tan[t]], (I + Sqrt[7])/(I - Sqrt[7])]]- 4/10, t, WorkingPrecision -> 20]

soln-4/3

CorrectDigits[soln,4/3]

Re(Π((1/2) I (I+(Sqrt[7]));I (sinh^(-1))((Sqrt[-((2 I)/(I+(Sqrt[7])))]) tan(t))\[VerticalSeparator]((I+(Sqrt[7]))/(I-(Sqrt[7])))))

Wilson

N[Exp[6]-Pi^4-Pi^5,20]