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%I #12 Feb 06 2014 03:24:52
%S 5,10,20,30,70
%N Positive numbers n such that 100*n^2/(100+n^2) are integers.
%C Rewrite 100n^2/(100+n^2) as an I : n=sqrt(100I/(100-I)), where a finite substitution for I results.
%C 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.
%e for n=20, 100*20^2/(100+20^2)=I or 80.
%t Select[Range[1000], IntegerQ[100*#^2/(100 + #^2)] &] (* _T. D. Noe_, Jan 29 2014 *)
%Y Cf. A162688.
%K nonn,fini,full
%O 1,1
%A _Larry J Zimmermann_, Jan 24 2014
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