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%I #25 Sep 08 2022 08:45:23
%S 0,2,7,15,27,42,60,82,107,135,167,202,240,282,327,375,427,482,540,602,
%T 667,735,807,882,960,1042,1127,1215,1307,1402,1500,1602,1707,1815,
%U 1927,2042,2160,2282,2407,2535,2667,2802,2940,3082,3227,3375,3527,3682,3840
%N a(n) = floor((5*n^2+1)/3).
%C If n < a(m) then all integer solutions (i,j,k) of 2*i^2 + 2*j^2 + 3*k^2 - 2*i*k = n satisfy max(|i|,|j|,|k|) < m.
%H G. C. Greubel, <a href="/A113422/b113422.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).
%F G.f.: (2*x + 3*x^2 + 3*x^3 + 2*x^4)/((1-x)^2(1-x^3)).
%F a(-n) = a(n).
%F a(3n) = 15n^2, a(3n+1) = 15n^2 + 10n + 2.
%F a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5), a(0)=0, a(1)=2, a(2)=7, a(3)=15, a(4)=27. - _Harvey P. Dale_, Nov 28 2013
%F a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+3)/3). - _Wesley Ivan Hurt_, Mar 12 2015
%p A113422:=n->floor((5*n^2+1)/3): seq(A113422(n), n=0..100); # _Wesley Ivan Hurt_, Mar 12 2015
%t Floor[(5*Range[0,50]^2+1)/3] (* or *) LinearRecurrence[{2,-1,1,-2,1},{0,2,7,15,27},50] (* _Harvey P. Dale_, Nov 28 2013 *)
%o (PARI) a(n)=(5*n^2+1)\3
%o (Magma) [Floor((5*n^2+1)/3) : n in [0..100]]; // _Wesley Ivan Hurt_, Mar 12 2015
%K nonn,easy
%O 0,2
%A _Michael Somos_, Oct 31 2005
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