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Floor(r*a(n-1)) - floor(r*a(n-2)), where r = 3/2, a(0) = 1, a(1) = 1.
4

%I #14 Jan 07 2020 12:58:59

%S 1,1,0,-1,-2,-1,1,3,3,0,-4,-6,-3,4,11,10,-1,-17,-24,-10,21,46,38,-12,

%T -75,-95,-30,98,192,141,-77,-327,-375,-72,455,790,503,-431,-1401,

%U -1455,-81,2061,3213,1728,-2227,-5933,-5559,561,9180,12929,5623,-10959

%N Floor(r*a(n-1)) - floor(r*a(n-2)), where r = 3/2, a(0) = 1, a(1) = 1.

%H Clark Kimberling, <a href="/A275865/b275865.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = floor(3*a(n-1)/2) - floor(3*a(n-2)/2).

%t r = 3/2; a[0] = 1; a[1] = 1; z = 100;

%t a[n_] := a[n] = Floor[r*a[n - 1]] - Floor[r*a[n - 2]];

%t t = Table[a[n], {n, 0, z}]

%t RecurrenceTable[{a[0]==a[1]==1,a[n]==Floor[(3a[n-1])/2]-Floor[ (3a[n-2])/ 2]},a,{n,60}] (* _Harvey P. Dale_, Jan 07 2020 *)

%Y Cf. A275862, A275863, A275864.

%K easy,sign

%O 0,5

%A _Clark Kimberling_, Aug 12 2016