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%I #32 Mar 04 2023 05:08:06
%S 0,0,0,0,2,2,2,3,6,6,6,8,12,12,12,15,20,20,20,24,30,30,30,35,42,42,42,
%T 48,56,56,56,63,72,72,72,80,90,90,90,99,110,110,110,120,132,132,132,
%U 143,156,156,156,168,182,182,182,195,210,210,210,224,240
%N a(n) = floor(n/4)*ceiling((n+2)/4).
%C The three fixed points of this sequence are 0, 12 and 15. - _Bernard Schott_, Feb 27 2023
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,0,-2,2,-2,1).
%F a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-5) + 2*a(n-6) - 2*a(n-7) + a(n-8). - _R. J. Mathar_, Mar 11 2012
%F a(n) = A002265(n)*A002265(n+5). - _Michel Marcus_, Feb 27 2023
%F From _Amiram Eldar_, Mar 04 2023: (Start)
%F Sum_{n>=4} 1/a(n) = 15/4.
%F Sum_{n>=4} (-1)^n/a(n) = 1/4. (End)
%p Sequence = seq(floor(n/4)*ceiling((n+2)/4), n=0..60); # _Bernard Schott_, Mar 01 2023
%t LinearRecurrence[{2,-2,2,0,-2,2,-2,1},{0,0,0,0,2,2,2,3},80] (* _Harvey P. Dale_, Nov 05 2014 *)
%o (Python)
%o def A038715(n): return (n>>2)*(1+(n+1>>2)) # _Chai Wah Wu_, Feb 02 2023
%o (PARI) a(n) = (n\4)*((n+5)\4); \\ _Michel Marcus_, Feb 27 2023
%Y Cf. A002265.
%K nonn
%O 0,5
%A _N. J. A. Sloane_, May 02 2000
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