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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).
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%I #25 Feb 16 2016 08:35:54

%S 6,30,151,761,3836,19337,97477,491378,2477019,12486565,62944332,

%T 317300149,1599498817,8063016906,40645382751,204891935393,

%U 1032852992092,5206575364849,26246162074765,132305973770306,666949729466899,3362069972805741,16948075698414380

%N 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 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 Colin Barker, <a href="/A022023/b022023.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n+1) = floor(a(n)^2/a(n-1))+1 for all n > 0. - _M. F. Hasler_, Feb 10 2016

%p A022023 := proc(n)

%p option remember;

%p if n <= 1 then

%p op(n+1,[6,30]) ;

%p else

%p a := procname(n-1)^2/procname(n-2) ;

%p if type(a,'integer') then

%p a+1 ;

%p else

%p ceil(a) ;

%p fi;

%p end if;

%p end proc: # _R. J. Mathar_, Feb 10 2016

%o (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 Cf. A022018 - A022025, A022026 - A022032.

%K nonn

%O 0,1

%A _R. K. Guy_

%E Double-checked and extended to 3 lines of data by _M. F. Hasler_, Feb 10 2016