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a(n) = (2^n + 1)^2.
7

%I #31 May 14 2021 17:22:16

%S 4,9,25,81,289,1089,4225,16641,66049,263169,1050625,4198401,16785409,

%T 67125249,268468225,1073807361,4295098369,17180131329,68720001025,

%U 274878955521,1099513724929,4398050705409,17592194433025

%N a(n) = (2^n + 1)^2.

%H H. Bottomley, <a href="/A060919/a060919.gif">Illustration of initial terms</a>

%H I. Strazdins, <a href="https://doi.org/10.1023/A:1005769927571">Universal affine classification of Boolean functions</a>, Acta Applic. Math. 46 (1997), 147-167.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-14,8).

%F a(n) = A000051(n)^2. - _R. J. Mathar_, Nov 27 2015

%F G.f.: ( -4+19*x-18*x^2 ) / ( (x-1)*(2*x-1)*(4*x-1) ). - _R. J. Mathar_, Nov 27 2015

%o (PARI) a(n)=(2^n + 1)^2

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_.