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Least number k such that abs(A011541(n+1) - A011541(n)*k^3) is a minimum.
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%I #17 Jun 13 2016 09:05:07

%S 9,37,43,19,79

%N Least number k such that abs(A011541(n+1) - A011541(n)*k^3) is a minimum.

%C If b is the sum of two positive cubes in exactly n ways, then b*c^3 is the sum of two positive cubes in at least n ways for all c > 0. So A011541(i)*c^3 is a candidate for unknown Taxi-cab numbers. a(6) = 79 is an interesting example that is related to this case. Additionally, the benefit of this simple fact is the determination of upper bounds for unknown Taxi-cab numbers in relatively easy way.

%C The inequalities that are given in the comment section of A011541 are:

%C A011541(7) <= 101^3*A011541(6),

%C A011541(8) <= 127^3*A011541(7),

%C A011541(9) <= 139^3*A011541(8).

%C So a(6) <= 101, a(7) <= 127, a(8) <= 139.

%e a(5) = 79 because abs(A011541(6) - A011541(5)*79^3) = abs(24153319581254312065344 - 48988659276962496*79^3) = 0

%Y Cf. A011541.

%K nonn,hard,more

%O 1,1

%A _Altug Alkan_, May 26 2016