|
%I #26 Mar 04 2024 16:13:37
%S 5,26,135,700,3629,18813,97527,505582,2620947,13587040,70435478,
%T 365138879,1892887004,9812762803,50869551972,263708740319,
%U 1367071205166,7086923541985,36738748574433,190454382472052,987319198674433,5118281802804775,26533271760636405,137548993480193164
%N 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(5,26).
%C The empirical g.f. / recurrence agrees with the original definition for at least 2000 terms (and a(2000) ~ 10^1430). - _M. F. Hasler_, Feb 11 2016
%H Colin Barker, <a href="/A022032/b022032.txt">Table of n, a(n) for n = 0..1000</a>
%F Empirical g.f.: -(x^6+x^5+x^4+x^3-x-5) / (x^7+x^6+x^5+x^4-x^2-5*x+1). - _Colin Barker_, Sep 18 2015
%F a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. - _M. F. Hasler_, Feb 11 2016
%t (* This empirical recurrence should not be used to extend the data. *) LinearRecurrence[{5, 1, 0, -1, -1, -1, -1}, {5, 26, 135, 700, 3629, 18813, 97527}, 24] (* _Jean-François Alcover_, Dec 12 2016 *)
%o (PARI) a=[5,26];for(n=2,2000, a=concat(a, ceil(a[n]^2/a[n-1])-1));A022032(n)=a[n+1] \\ _M. F. Hasler_, Feb 11 2016
%Y Cf. A022018 - A022025, A022026 - A022031.
%K nonn
%O 0,1
%A _R. K. Guy_
%E Edited by _M. F. Hasler_, Feb 11 2016
|
