cp's OEIS Frontend

A022023 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,30).

%I A022023 #25 Feb 16 2016 08:35:54
%S A022023 6,30,151,761,3836,19337,97477,491378,2477019,12486565,62944332,
%T A022023 317300149,1599498817,8063016906,40645382751,204891935393,
%U A022023 1032852992092,5206575364849,26246162074765,132305973770306,666949729466899,3362069972805741,16948075698414380
%N A022023 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(6,30).
%C A022023 This coincides with the linearly recurrent sequence defined by the expansion of (6 - 5*x^2)/(1 - 5*x - x^2 + 4*x^3) only up to n <= 69. - _Bruno Berselli_, Feb 11 2016
%H A022023 Colin Barker, <a href="/A022023/b022023.txt">Table of n, a(n) for n = 0..1000</a>
%F A022023 a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - _M. F. Hasler_, Feb 10 2016
%p A022023 A022023 := proc(n)
%p A022023     option remember;
%p A022023     if n <= 1 then
%p A022023         op(n+1,[6,30]) ;
%p A022023     else
%p A022023         a := procname(n-1)^2/procname(n-2) ;
%p A022023         if type(a,'integer') then
%p A022023             a+1 ;
%p A022023         else
%p A022023             ceil(a) ;
%p A022023         fi;
%p A022023     end if;
%p A022023 end proc: # _R. J. Mathar_, Feb 10 2016
%o A022023 (PARI) a=[6,30];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ _M. F. Hasler_, Feb 10 2016
%Y A022023 Cf. A022018 - A022025, A022026 - A022032.
%K A022023 nonn
%O A022023 0,1
%A A022023 _R. K. Guy_
%E A022023 Double-checked and extended to 3 lines of data by _M. F. Hasler_, Feb 10 2016