[seqfan] Re: Percentages

[israel at math.ubc.ca]

Equivalently, b = 2a+1 is a solution of b^2 == 1 (mod 400).  There are 8 
possibilities for b (mod 400): 1, 49, 151, 199, 201, 249, 351, 399.
The corresponding solutions are 
  a =  200 m 
      24 + 200 m
      75 + 200 m
      99 + 200 m
      100 + 200 m
      124 + 200 m
      175 + 200 m
      199 + 200 m
for nonnegative integers m.

Cheers,
Robert        


On Dec 7 2015, Eric Angelini wrote:

>
>Hello SeqFans,
>this rather dull seq might be of interest...
>
>S=24,75,99,100,124,175,199,200,224,
>275,299,300,324,375,399,400,...
>
>The idea is that a(n) percent of [a(n)+1] is an integer.