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%I #18 Feb 16 2016 06:07:38
%S 4,17,72,304,1283,5414,22845,96397,406757,1716352,7242319,30559689,
%T 128949662,544115986,2295951781,9687997993,40879475731,172495033261,
%U 727860031657,3071278144467,12959565068034,54684179957837,230745362360740,973653116715681,4108426630946045
%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(4,17).
%C The empirical g.f. / recurrence agrees with the original definition for at least 2000 terms (and a(2000) ~ 10^1250). - _M. F. Hasler_, Feb 11 2016
%H Colin Barker, <a href="/A022031/b022031.txt">Table of n, a(n) for n = 0..1000</a>
%F Empirical g.f.: -(x^6+x^5+x^4+x^3-x-4) / ((x-1)*(x^6+2*x^5+3*x^4+4*x^3+4*x^2+3*x-1)). - _Colin Barker_, Sep 18 2015
%F a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. a(n+1)/a(n) ~ 4.219599938... as n -> oo. - _M. F. Hasler_, Feb 11 2016
%o (PARI) a=[4,17];for(n=2,2000,a=concat(a,ceil(a[n]^2/a[n-1])-1));A022031(n)=a[n+1] \\ _M. F. Hasler_, Feb 11 2016
%Y Cf. A022018 - A022025, A022026 - A022032.
%K nonn
%O 0,1
%A _R. K. Guy_
%E Edited by _M. F. Hasler_, Feb 11 2016
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