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For even n, a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n); for odd n, the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n); a(0) = 4, a(1) = 16.
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%I #22 Jul 13 2023 09:56:05

%S 4,16,63,249,984,3889,15370,60745,240075,948819,3749901,14820274,

%T 58572352,231488326,914882931,3615779646,14290202610,56477415835,

%U 223208766625,882160643536,3486455360919,13779090092886,54457408494633,215225339261149,850608722312629,3361756570848769

%N For even n, a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n); for odd n, the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n); a(0) = 4, a(1) = 16.

%C Original definition: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n).

%C This original definition would lead to sequence 4, 16, 63, 248, 976, 3841, ... which agrees to over 2000 terms with the conjectured g.f. = (4 - x^2)/(1 - 4*x + x^3). - _M. F. Hasler_, Feb 11 2016

%H <a href="/index/Ph#Pisot">Index entries for Pisot sequences</a>

%F Conjecture: a(n) = 4*a(n-1)-a(n-3)+a(n-4). G.f. = (4-x^2+x^3)/(1-4*x+x^3-x^4). - _Colin Barker_, Feb 16 2012

%F a(n) = ceiling(a(n-1)^2/a(n-2))-1 for even n > 0, a(n) = floor(a(n-1)^2/a(n-2))+1 for even n > 0. - _M. F. Hasler_, Feb 11 2016

%o (PARI) a=[4,16];for(n=2,2000,a=concat(a,if(bittest(n,0),a[n]^2\a[n-1]+1,ceil(a[n]^2/a[n-1])-1)));A022030(n)=a[n+1] \\ _M. F. Hasler_, Feb 11 2016

%Y Cf. A022026 - A022032, A022018 - A022025.

%K nonn

%O 0,1

%A _R. K. Guy_

%E Edited (definition changed to fit data, extended to 3 lines) by _M. F. Hasler_, Feb 11 2016