cp's OEIS Frontend

A022027 Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(2,16).

%I A022027 #31 Jul 13 2023 09:48:09
%S A022027 2,16,127,1008,8000,63492,503904,3999232,31739888,251903488,
%T A022027 1999230976,15866888256,125927492096,999423012864,7931916549888,
%U A022027 62951622430720,499615287394304,3965194632954880,31469750573916160,249759543441752064,1982215569002196992,15731845549721911296
%N A022027 Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(2,16).
%C A022027 Not to be confused with the Pisot T(2,16) sequence, which is A013730. - _R. J. Mathar_, Feb 13 2016
%H A022027 Colin Barker, <a href="/A022027/b022027.txt">Table of n, a(n) for n = 0..1000</a>
%H A022027 <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>
%F A022027 Conjectures: a(n) = 8*a(n-1)-4*a(n-3). G.f.: -(x^2-2) / (4*x^3-8*x+1). - _Colin Barker_, Sep 18 2015
%F A022027 a(n+1) = ceiling(a(n)^2/a(n-1))-1 for n>0. - _M. F. Hasler_, Feb 11 2016
%t A022027 RecurrenceTable[{a[1] == 2, a[2] == 16, a[n] == Ceiling[a[n-1]^2 / a[n-2] - 1]}, a, {n, 30}] (* _Vincenzo Librandi_, Feb 12 2016 *)
%o A022027 (PARI) a=[2, 16]; for(n=2, 1000, a=concat(a, ceil(a[n]^2/a[n-1])-1)); A022027(n)=a[n+1] \\ _M. F. Hasler_, Feb 11 2016
%o A022027 (Magma) I:=[2,16]; [n le 2 select I[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..30]]; // _Vincenzo Librandi_, Feb 12 2016
%Y A022027 Cf. A022018 - A022025, A022026 - A022032.
%K A022027 nonn
%O A022027 0,1
%A A022027 _R. K. Guy_
%E A022027 Double-checked (original definition agrees with g.f. / recurrence for n=0..1000), extended and edited by _M. F. Hasler_, Feb 11 2016