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%I #32 Dec 19 2020 03:33:17
%S 1,4,16,25,49,64,100,121,169,196,256,289,361,400,484,529,625,676,784,
%T 841,961,1024,1156,1225,1369,1444,1600,1681,1849,1936,2116,2209,2401,
%U 2500,2704,2809,3025,3136,3364,3481,3721,3844,4096,4225,4489,4624,4900
%N Squares of numbers not divisible by 3: a(n) = A001651(n)^2.
%C From Fermat's Little Theorem all these numbers are congruent to 1 mod 3.
%H Colin Barker, <a href="/A135556/b135556.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F G.f.: -x*(1+3*x+10*x^2+3*x^3+x^4) / ((1+x)^2*(x-1)^3). - _R. J. Mathar_, Feb 16 2011
%F From _Colin Barker_, Jan 26 2016: (Start)
%F a(n) = (18*n^2-6*(-1)^n*n-18*n+3*(-1)^n+5)/8.
%F a(n) = (9*n^2-12*n+4)/4 for n even.
%F a(n) = (9*n^2-6*n+1)/4 for n odd.
%F (End)
%F E.g.f.: (1/8)*( (3 + 6*x)*exp(-x) - 8 + (5 + 18*x^2)*exp(x)). - _G. C. Greubel_, Oct 19 2016
%F Sum_{n>=1} 1/a(n) = 4*Pi^2/27 (A214549). - _Amiram Eldar_, Dec 19 2020
%t LinearRecurrence[{1,2,-2,-1,1}, {1,4,16,25,49}, 25] (* or *) Table[(18*n^2-6*(-1)^n*n-18*n+3*(-1)^n+5)/8, {n,1,25}] (* _G. C. Greubel_, Oct 19 2016 *)
%t Flatten[Partition[Range[70],2,3,{1,1},{}]]^2 (* _Harvey P. Dale_, Jun 19 2018 *)
%o (PARI) isok(n) = issquare(n) && (n % 3 == 1); \\ _Michel Marcus_, Nov 02 2013
%o (PARI) Vec(-x*(1+3*x+10*x^2+3*x^3+x^4) / ( (1+x)^2*(x-1)^3 ) + O(x^100)) \\ _Colin Barker_, Jan 26 2016
%Y Cf. A001651, A001082, A214549.
%Y Partial sums of A298028.
%K nonn,easy
%O 1,2
%A _Artur Jasinski_, Nov 25 2007