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%I #52 Sep 08 2022 08:44:35
%S 1,4,9,1,4,9,16,4,9,16,25,9,16,25,36,16,25,36,49,25,36,49,64,36,49,64,
%T 81,49,64,81,100,64,81,100,121,81,100,121,144,100,121,144,169,121,144,
%U 169,196,144,169,196,225,169,196,225,256,196,225,256,289,225
%N A Kutz sequence.
%C The pattern is obvious: after the initial three terms, we have four successive squares.
%C Another description of the same sequence: array read by rows, with four columns, in which row n lists n^2, (n+1)^2, (n+2)^2, n^2. - _Omar E. Pol_, Sep 28 2011
%H Vincenzo Librandi, <a href="/A007892/b007892.txt">Table of n, a(n) for n = 1..10000</a>
%H R. E. Kutz, <a href="http://www.jstor.org/stable/3027304">Two unusual sequences</a>, Two-Year College Mathematics Journal, 12 (1981), 316-319.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,2,-2,0,0,-1,1).
%F a(n) = (floor((-1)^n+(n+5)/2)-3*floor((n+6)/4))^2. [_Arkadiusz Wesolowski_, Sep 27 2011]
%F a(n) = (n-3*floor(n/4))^2. [_Arkadiusz Wesolowski_, Sep 28 2011]
%F G.f.: x*(1+3*x+5*x^2-8*x^3+x^4-x^5-3*x^6+4*x^7)/((1-x)^3*(1+x+x^2+x^3)^2). a(n) = (A110657(n-1)+1)^2 = ((2*n-6*(-1)^((n-1)*n/2)-3*(-1)^n+9)/8)^2. [_Bruno Berselli_, Sep 28 2011]
%t Table[(n - 3*Floor[n/4])^2, {n, 60}] (* _Arkadiusz Wesolowski_, Sep 29 2011 *)
%t Rest[Flatten[Table[Range[n,n+3]^2,{n,0,20}]]] (* _Harvey P. Dale_, Oct 24 2015 *)
%o (Magma) [(n-3*Floor(n/4))^2: n in [1..60]]; // _Vincenzo Librandi_, Sep 28 2011
%o (PARI) a(n)=(floor((-1)^n+(n+5)/2)-3*floor((n+6)/4))^2 \\ _Charles R Greathouse IV_, Sep 28 2011
%o (Maxima) makelist(((2*n-6*(-1)^((n-1)*n/2)-3*(-1)^n+9)/8)^2,n,1,60); \\ _Bruno Berselli_, Sep 28 2011
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_.
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