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User: Larry J Zimmermann

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A236391 Positive numbers n such that 100*n^2/(100+n^2) are integers.

Original entry on oeis.org

5, 10, 20, 30, 70

Offset: 1

Larry J Zimmermann, Jan 24 2014

Rewrite 100n^2/(100+n^2) as an I : n=sqrt(100I/(100-I)), where a finite substitution for I results.

Instead of 100, we could use 72 and get n=8,24,48,64; we could use 162 and get n=54,108,144,150. Many other values are possible.

			for n=20, 100*20^2/(100+20^2)=I or 80.

Cf. A162688.

Select[Range[1000], IntegerQ[100*#^2/(100 + #^2)] &] (* T. D. Noe, Jan 29 2014 *)

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

Original entry on oeis.org

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

Larry J Zimmermann, Mar 10 2013

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.

			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.

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).

Cf. A211412.

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 *)

a(n)=16*n^4+4 \\ Charles R Greathouse IV, Mar 21 2013

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)

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User: Larry J Zimmermann

A236391 Positive numbers n such that 100*n^2/(100+n^2) are integers.

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