

A080052


Value of n such that for any value of n, Pi^n is closer to its nearest integer than any value of Pi^k for 1 <= k < n.


10



1, 2, 3, 58, 81, 157, 1030, 5269, 12128, 65875, 114791, 118885, 151710
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OFFSET

1,2


COMMENTS

Robert G. Wilson v used Mathematica with a changing number of digits to accommodate 24 digits to the right of the decimal point.
At 12128 the difference from an integer is 0.000016103224605297330719...
The sequence of rounded reciprocals of the distances, b(n) = round(1/(0.5frac(Pi^a(n).5))) = round(1/abs(round(Pi^a(n))Pi^a(n))), starts { 7, 8, 159, 190, 270, 2665, 10811, 26577, 62099, 70718, ... }.  M. F. Hasler, Apr 06 2008


REFERENCES

J.M. De Koninck, Ces nombres qui nous fascinent, Entry 58, p. 21, Ellipses, Paris 2008.


LINKS



EXAMPLE

First term is 1 because this is just Pi = 3.14159....
Second term is 2 because Pi^2 = 9.869604... which is 0.13039... away from its nearest integer.
Pi^3 = 31.00627..., hence third term is 3.
Pi^58 is 0.00527... away from its nearest integer.


MAPLE

b := array(1..2000): Digits := 8000: c := 1: pos := 0: for n from 1 to 2000 do: exval := evalf(Pi^n): if (abs(exvalround(exval))<c) then c := (abs(exvalround(exval))): pos := pos+1: b[pos] := n: print(n):fi: od:
Used Maple with 8000 digits of precision and examined all n up to 2000.


MATHEMATICA

a = 1; Do[d = Abs[ Round[Pi^n]  N[Pi^n, Ceiling[ Log[10, Pi^n] + 24]]]; If[d < a, Print[n]; a = d], {n, 1, 25000}]
$MaxExtraPrecision = 10^9; a = 1; Do[d = Abs[ Round[Pi^n]  N[Pi^n, Ceiling[ Log[10, Pi^n] + 24]]]; If[d < a, Print[n]; a = d], {n, 1, 10^5}] (* Ryan Propper, Nov 13 2005 *)


PROG

(PARI) f=0; for( i=1, 99999, abs(frac(Pi^i).5)>f  next; f=abs(frac(Pi^i).5); print1(i", ")) \\ M. F. Hasler, Apr 06 2008


CROSSREFS



KEYWORD

nonn


AUTHOR

Mark Hudson (mrmarkhudson(AT)hotmail.com), Jan 22 2003


EXTENSIONS



STATUS

approved



