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.

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

Offset: 1

Author

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

Keywords

• nonn

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/(0.5-frac(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

Examples

			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.
		

References

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

Crossrefs

Cf. A079490, A137994, A137995.

Programs

Maple

b := array(1..2000): Digits := 8000: c := 1: pos := 0: for n from 1 to 2000 do: exval := evalf(Pi^n): if (abs(exval-round(exval))
				

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 *)

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

Extensions

More terms from Carlos Alves and Robert G. Wilson v, Jan 23 2003

One more term from Ryan Propper, Nov 13 2005

a(11)-a(13) from Jeremy Elson, Nov 13 2011