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

5, 10, 20, 30, 70

Offset

1,1

Comments

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.

Links

Table of n, a(n) for n=1..5.

Example

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

Mathematica

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

Crossrefs

Cf. A162688.

Sequence in context: A027884 A370532 A331141 * A026357 A117518 A107486

Adjacent sequences: A236388 A236389 A236390 * A236392 A236393 A236394

Keyword

nonn,fini,full

Author

Larry J Zimmermann, Jan 24 2014

Status

approved