[seqfan] Re: Percentages
israel at math.ubc.ca
israel at math.ubc.ca
Mon Dec 7 22:19:33 CET 2015
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.
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