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%I #62 Feb 16 2024 01:27:30
%S 0,12,48,108,192,300,432,588,768,972,1200,1452,1728,2028,2352,2700,
%T 3072,3468,3888,4332,4800,5292,5808,6348,6912,7500,8112,8748,9408,
%U 10092,10800,11532,12288,13068,13872,14700,15552,16428,17328,18252,19200
%N a(n) = 12*n^2.
%C Areas of perfect 4:3 rectangles (for n>0).
%C Sequence found by reading the line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Semi-axis opposite to A069190 in the same spiral. - _Omar E. Pol_, Sep 16 2011
%C (x,y,z) = (-a(n), 1 + n*a(n), 1 - n*a(n)) are solutions of the Diophantine equation x^3 + 2*y^3 + 2*z^3 = 4. - _XU Pingya_, Apr 30 2022
%H Ivan Panchenko, <a href="/A135453/b135453.txt">Table of n, a(n) for n = 0..10000</a>
%H John Elias, <a href="/A135453/a135453.png">Illustration: Even Ordered Star Perimeters</a>
%H Leo Tavares, <a href="/A135453/a135453.jpg">Illustration</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A000290(n)*12 = A001105(n)*6 = A033428(n)*4 = A016742(n)*3 = A033581(n)*2. - _Omar E. Pol_, Dec 13 2008
%F From _Amiram Eldar_, Feb 03 2021: (Start)
%F Sum_{n>=1} 1/a(n) = Pi^2/72 (A086729).
%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/144.
%F Product_{n>=1} (1 + 1/a(n)) = 2*sqrt(3)*sinh(Pi/(2*sqrt(3)))/Pi.
%F Product_{n>=1} (1 - 1/a(n)) = 2*sqrt(3)*sin(Pi/(2*sqrt(3)))/Pi. (End)
%e 192 is on the list since 16*12 is a 4:3 rectangle with integer sides and an area of 192.
%p seq(12*h^2,n=0..100); # _Muniru A Asiru_, Jan 29 2018
%t Table[12*n^2, {n, 0, 60}] (* _Stefan Steinerberger_, Dec 17 2007 *)
%t LinearRecurrence[{3,-3,1},{0,12,48},50] (* _Harvey P. Dale_, Jan 19 2020 *)
%o (PARI) a(n)=12*n^2 \\ _Charles R Greathouse IV_, Jun 17 2017
%o (GAP) List([0..100],n->12*n^2); # _Muniru A Asiru_, Jan 29 2018
%Y Cf. A000290, A001105, A016742, A033428, A033581, A086729.
%K nonn,easy
%O 0,2
%A _Ben Paul Thurston_, Dec 14 2007
%E More terms from _Stefan Steinerberger_, Dec 17 2007
%E Minor edits from _Omar E. Pol_, Dec 15 2008
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