OFFSET
0,1
COMMENTS
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.
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.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
2^2 + (4*n^2)^2 = (4*n^2 + 4*n + 2)*(4*n^2 - 4*n + 2).
a(n) = 4 * A211412(n).
From Amiram Eldar, May 18 2023: (Start)
Sum_{n>=0} 1/a(n) = tanh(Pi/2)*Pi/16 + 1/8.
Sum_{n>=0} (-1)^n/a(n) = 1/8 + sech(Pi/2)*Pi/16. (End)
EXAMPLE
For n=1, the two squares are 4 and 16; the rectangle is 10 X 2.
For n=2, the two squares are 4 and 256; the rectangle is 26 X 10.
MATHEMATICA
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 *)
PROG
(PARI) a(n)=16*n^4+4 \\ Charles R Greathouse IV, Mar 21 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Larry J Zimmermann, Mar 10 2013
STATUS
approved