cp's OEIS Frontend

A022021 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(5,20).

%I A022021 #24 Feb 16 2016 08:53:01
%S A022021 5,20,81,329,1337,5434,22086,89767,364852,1482917,6027219,24497237,
%T A022021 99567416,404685244,1644816681,6685249720,27171759829,110437838993,
%U A022021 448867366641,1824392026070,7415121953942,30138277741915,122495056843392,497873139253657,2023572780632275
%N A022021 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(5,20).
%C A022021 This coincides with the linearly recurrent sequence defined by the expansion of (5 - 4*x^2)/(1 - 4*x - x^2 + 3*x^3) only up to n <= 39. - _Bruno Berselli_, Feb 11 2016
%H A022021 Colin Barker, <a href="/A022021/b022021.txt">Table of n, a(n) for n = 0..1000</a>
%F A022021 a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - _M. F. Hasler_, Feb 10 2016
%p A022021 A022021 := proc(n)
%p A022021     option remember;
%p A022021     if n <= 1 then
%p A022021         op(n+1,[5,20]) ;
%p A022021     else
%p A022021         a := procname(n-1)^2/procname(n-2) ;
%p A022021         if type(a,'integer') then
%p A022021             a+1 ;
%p A022021         else
%p A022021             ceil(a) ;
%p A022021         fi;
%p A022021     end if;
%p A022021 end proc: # _R. J. Mathar_, Feb 10 2016
%o A022021 (PARI) a=[5,20];for(n=2,30,a=concat(a,a[n]^2\a[n-1]+1));a \\ _M. F. Hasler_, Feb 10 2016
%Y A022021 Cf. A022018 - A022025, A022026 - A022032.
%K A022021 nonn
%O A022021 0,1
%A A022021 _R. K. Guy_
%E A022021 Double-checked and extended to 3 lines of data by _M. F. Hasler_, Feb 10 2016