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a(n) = 16*n^4 + 4.
1

%I #34 May 18 2023 10:40:53

%S 4,20,260,1300,4100,10004,20740,38420,65540,104980,160004,234260,

%T 331780,456980,614660,810004,1048580,1336340,1679620,2085140,2560004,

%U 3111700,3748100,4477460,5308420,6250004,7311620,8503060,9834500,11316500,12960004,14776340

%N a(n) = 16*n^4 + 4.

%C Derivation. Given: x^2 + y^2 = (x + y + sqrt(2*x*y))*(x + y - sqrt(2*x*y)). Let y = 2*x*n^2 to eliminate the radicals. Now we get x^2 + (2*x*n^2)^2 = (x + 2*x*n^2 + 2*x*n)*(x + 2*x*n^2 - 2*x*n). For this sequence, x=2 and n = 1,2,3,... Now we have 2^2 + (2*2*n^2)^2 = (2 + 4*n^2 + 4*n)*(2 + 4*n^2 - 4*n) or 4 + 16*n^4 = 4 + 16*n^4. Therefore this formula generates 2 squares equal to a rectangle of the same area.

%C Application. Tie a string to the middle of a rod with 2 squares on one end (e.g., 2^2 and 4^2) and a rectangle on the other end (e.g., 10 X 2); then hold it up and it balances (same area on both ends) which is what a Calder Mobile is.

%H Harvey P. Dale, <a href="/A222655/b222655.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 (5,-10,10,-5,1).

%F 2^2 + (4*n^2)^2 = (4*n^2 + 4*n + 2)*(4*n^2 - 4*n + 2).

%F a(n) = 4 * A211412(n).

%F From _Amiram Eldar_, May 18 2023: (Start)

%F Sum_{n>=0} 1/a(n) = tanh(Pi/2)*Pi/16 + 1/8.

%F Sum_{n>=0} (-1)^n/a(n) = 1/8 + sech(Pi/2)*Pi/16. (End)

%e For n=1, the two squares are 4 and 16; the rectangle is 10 X 2.

%e For n=2, the two squares are 4 and 256; the rectangle is 26 X 10.

%t 16*Range[0,40]^4+4 (* or *) LinearRecurrence[{5,-10,10,-5,1},{4,20,260,1300,4100},40] (* _Harvey P. Dale_, Nov 28 2014 *)

%o (PARI) a(n)=16*n^4+4 \\ _Charles R Greathouse IV_, Mar 21 2013

%Y Cf. A211412.

%K nonn,easy

%O 0,1

%A _Larry J Zimmermann_, Mar 10 2013