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%I #3 Oct 02 2013 15:47:21
%S 4,122,221,320,1030,1120,1210,1300,67199,68189,69179,77198,78188,
%T 79178,87197,88187,89177,97196,98186,99176,1000300,1010200,1020100,
%U 1030000,100409738,100508738,100607738,100706738,100805738,100904738,101409638
%N Numbers which on adding their reversal gives perfect cubes.
%C For entries not ending in zero, their reversal is also present.
%e a(2)=122 because 122+221=343 which is 7^3.
%t Do[ If[ IntegerQ[(n + FromDigits[ Reverse[ IntegerDigits[n]]]) ^(1/3)], Print[n]], {n, 1, 101508638}]
%K base,nonn
%O 1,1
%A _Shyam Sunder Gupta_, Mar 16 2003
%E Edited and extended by _Robert G. Wilson v_, Mar 16 2003
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