Newton-Pepys Problem

Author

Eric W. Weisstein
October 12, 2023

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Statistics/Newton-PepysProblem.nb.

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

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

Values

Sum[Binomial[6n,x](1/(6^x))((((5/6)))^(6n-x)),{x,n,6n}]

TraditionalForm[%]

Table[(((5^(5 n)) (6^(-6 n)) (6 n)! Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)])/((5 n)!)),{n,0,10}]

N[%]

Zeilberger sum

Zeilberger, D.  "If $A_n$ Has $6n$ Dyes in a Box, With Which He lIas To Fling [at least] $n$ Sixes, Then $A_n$ Has An Easier Task Than $A_{n+1}$, at  Eaven Luck."
  Amer. Math. Monthly 103, 265, 1996.

Table[1-Sum[((2(94500(m^4)+214830(m^3)+171573(m^2)+56243m+6250)(6m)!(5^(5m+2)))/((5m+5)!m!(6^(6m+5)))),{m,0,n-1}],{n,0,10}]

N[%]

bug(440950): Implement 1 - Sum[(2 (94500 m^4 + 214830 m^3 + 171573 m^2 + 56243 m + 6250) (6 m)! 5^(5 m + 2))/((5 m + 5)! m! 6^(6 m + 5)), {m, 0, n - 1}] === (5^(5 n) 6^(-6 n) (6 n)! Hypergeometric2F1Regularized[1, -5 n, 1 + n, -(1/5)])/(5 n)!

1-2Sum[(((94500(m^4)+214830(m^3)+171573(m^2)+56243m+6250)(6m)!(5^(5m+2)))/((5m+5)!m!(6^(6m+5)))),{m,0,n-1}]//FullSimplify//Timing

sum=;

(table=Table[Evaluate@sum,{n,0,2}])//N

table//FunctionExpand

Plot

Plot

Plot[(((5^(5 n))  (6 n)! )/((6^(6 n))(5 n)!))Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],{n,0,50},PlotRange->{0,1},PlotStyle->Red,AxesLabel->{n,(Subscript[P, n])},BaseStyle->Directive[Black,FontFamily->"Times",14],Epilog->{Blue,Dashed,Line[{{0,.5},{50,.5}}]}]

Limit

Limit[(5^(5 n)) (6^(-6 n)) Binomial[6 n,n] Hypergeometric2F1[-5 n,1,1+n,-(1/5)],n->∞]

Limit[(((5^(5 n))(6 n)! )/((6^(6 n))(5 n)!))Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],n->∞]

DiscreteLimit[(5^(5 n)) (6^(-6 n)) Binomial[6 n,n] Hypergeometric2F1[-5 n,1,1+n,-(1/5)],n->∞]

DiscreteLimit[(((5^(5 n))(6 n)! )/((6^(6 n))(5 n)!))Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],n->∞]

N[(((5^(5 n)) (6^(-6 n)) (6 n)! Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)])/((5 n)!))/.n->99999]

Asymptotic (?)

FullSimplify[Series[(((5^(5 n))  (6 n)! )/((6^(6 n))(5 n)!))Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],{n,∞,0}]//Normal,n>0]

Limit[(Sqrt[(6/5)]) (E^(-n)) (n^n) Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],n->∞]

DiscreteLimit[(Sqrt[(6/5)]) (E^(-n)) (n^n) Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)],n->∞]

N[(Sqrt[(6/5)]) (E^(-n)) (n^n) Hypergeometric2F1Regularized[1,-5 n,1+n,-(1/5)]/.n->999,20]