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%I #31 Sep 08 2022 08:45:01
%S 0,1,4,16,36,64,100,144,196,256,324,400,484,576,676,784,900,1024,1156,
%T 1296,1444,1600,1764,1936,2116,2304,2500,2704,2916,3136,3364,3600,
%U 3844,4096,4356,4624,4900,5184,5476,5776,6084,6400,6724,7056,7396,7744,8100
%N a(n) and floor(a(n)/4) are both squares; i.e., squares that remain squares when written in base 4 and last digit is removed.
%C Let A(x) = (1 + k*x + (k-1)*x^2). Then the coefficients of (A(x))^2 sum to 4*k^2, where k = (n - 1). Examples: if k = 3 we have (1 + 3*x + 2*x^2)^2 = (1 + 6*x + 13x^2 + 12*x^3 + 4*x^4), and ( 1 + 6 + 13 + 12 + 4) = 36. If k = 4 we have (1 + 4*x + 3*x^2)^2 = (1 + 8*x + 22*x^2 + 24*x^3 + 9*x^4), and (1 + 8 + 22 + 24 + 9) = 64 = a(5). - _Gary W. Adamson_, Aug 02 2015
%C For n>0, a(n) are the Engel expansion of A197036. - _Benedict W. J. Irwin_, Dec 15 2016
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A004275(n)^2. - _M. F. Hasler_, Jan 16 2012
%F a(n) = 4*(-1+n)^2 for n>1; a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>4; G.f.: x*(x^3-7*x^2-x-1) / (x-1)^3. - _Colin Barker_, Sep 15 2014
%e 36 is in the sequence because 36 = 6^2 = 210 base 3 and 21 base 4 = 9 = 3^2.
%t Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {4, 16, 36}, 50]] (* _Vincenzo Librandi_, Aug 03 2015 *)
%o (PARI) concat(0, Vec(x*(x^3-7*x^2-x-1)/(x-1)^3 + O(x^100))) \\ _Colin Barker_, Sep 15 2014
%o (PARI) is_ok(n)=issquare(n) && issquare(floor(n/4));
%o first(m)=my(v=vector(m),r=0);for(i=1,m,while(!is_ok(r),r++);v[i]=r;r++;);v; /* _Anders Hellström_, Aug 08 2015 */
%o (Magma) [Floor((2*n^2)/(1 + n))^2: n in [0..60]]; // _Vincenzo Librandi_, Aug 03 2015
%Y Cf. A023110. Essentially A016742 with one addition.
%K nonn,base,easy
%O 0,3
%A _Henry Bottomley_, Jul 14 2000
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