A222655 a(n) = 16*n^4 + 4.

4, 20, 260, 1300, 4100, 10004, 20740, 38420, 65540, 104980, 160004, 234260, 331780, 456980, 614660, 810004, 1048580, 1336340, 1679620, 2085140, 2560004, 3111700, 3748100, 4477460, 5308420, 6250004, 7311620, 8503060, 9834500, 11316500, 12960004, 14776340

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

Cf. A211412.

Sequence in context: A032081 A034235 A352412 * A000847 A188810 A120443

Adjacent sequences: A222652 A222653 A222654 * A222656 A222657 A222658

Keyword

nonn,easy

Author

Larry J Zimmermann, Mar 10 2013

Status

approved