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a(n) = n^2 + 4*floor(n/2)^2.
4

%I #19 Jul 08 2022 08:23:54

%S 0,1,8,13,32,41,72,85,128,145,200,221,288,313,392,421,512,545,648,685,

%T 800,841,968,1013,1152,1201,1352,1405,1568,1625,1800,1861,2048,2113,

%U 2312,2381,2592,2665,2888,2965,3200,3281,3528,3613,3872

%N a(n) = n^2 + 4*floor(n/2)^2.

%C The first bisection is A139098, the second bisection is A102083.

%H David Lovler, <a href="/A354595/b354595.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), n >= 5.

%F a(n) = A000290(n) + 4*A008794(n).

%F G.f.: x*(1 + 7*x + 3*x^2 + 5*x^3)/((1 - x)^3*(1 + x)^2).

%F E.g.f.: 2*x^2*cosh(x) + (1 + 2*x + 2*x^2)*sinh(x). - _Stefano Spezia_, Jun 07 2022

%t a[n_] := n^2 + 4 Floor[n/2]^2

%t Table[a[n], {n, 0, 90}] (* A354595 *)

%t LinearRecurrence[{1, 2, -2, -1, 1}, {0, 1, 8, 13, 32}, 60]

%o (PARI) a(n) = n^2 + 4*(n\2)^2;

%Y Cf. A000290, A008794, A102083, A139098, A247375.

%Y Cf. A213037, A327259 (diagonal), A354594, A354596.

%K nonn,easy

%O 0,3

%A _David Lovler_, Jun 01 2022