%I #12 Nov 19 2018 07:19:38
%S 1,2,4,8,17,19,23,35,221,424,3846,16708,19142,19937,55188,87368
%N Numbers k such that Pi^k  1/phi is closer to its nearest integer than any value of Pi^j  1/phi for 1 <= j < k.
%C phi is the Golden ratio (1 + sqrt(5))/2.
%C At k = 3846 the discrepancy is 0.0000887984081945...
%C From _Ryan Propper_, Jul 27 2005: (Start)
%C At n = 16708 the discrepancy from an integer is 0.00006159...
%C At n = 19142 the discrepancy from an integer is 0.00003501...
%C At n = 19937 the discrepancy from an integer is 0.00001498...
%C At n = 55188 the discrepancy from an integer is 0.00001048...
%C At n = 87368 the discrepancy from an integer is 0.00000693...
%C (End)
%C As 1/phi = phi  1, the sequence is equivalent to "Numbers k such that Pi^k  phi is closer to its nearest integer than any value of Pi^j  phi for 1 <= j < k."  _David A. Corneth_, Nov 19 2018
%e The first term is 1 because this is just Pi  1/phi = 2.52355...
%e The second term is 2 because Pi^2  1/phi = 9.25157...
%e The next term is 4 because Pi^4  1/phi is closer to an integer than Pi^3  1/phi.
%t $MaxExtraPrecision = 10^6; p = 2/(1+Sqrt[5]); b = 1; Do[a = Abs[N[Round[Pi^n  p]  (Pi^n  p), 30]]; If[a < b, Print[n]; b = a], {n, 1, 10^5}] (* _Ryan Propper_, Jul 27 2005 *)
%o (PARI) upto(n) = my(c = 2, phi = (1 + sqrt(5)) / 2, res = List, r = 2); Pik = 1; for(i = 1, n, Pik *= Pi; c = frac(Pik  phi); c = min(c, 1c); if(c < r, listput(res, i); r = c)); res \\ _David A. Corneth_, Nov 19 2018
%Y Cf. A079490, A080052, A080279, A080280.
%K more,nonn
%O 1,2
%A Mark Hudson (mrmarkhudson(AT)hotmail.com), Feb 13 2003
%E a(12)a(16) from _Ryan Propper_, Jul 27 2005
