For all n when b = 243*n^6 and {a and c} vary accordingly to form the AP. So when n = 1, b = 243 and a can be any number from 1 to 242 and c is adjusted accordingly. If {a, b, c} are in AP, the sum of a, b, and c will always be 3b.
The first few values of b are :-
{243, 15552, 177147, 995328, 3796875, 11337408, 28588707, 63700992,129140163, 243000000}.
The 6th root of 3 times those numbers are {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
So there will be an infinite set of them where the total number in each set is 1 less than 243*n^6 for each n in the set.
Lets assume the numbers {a, b, c} > 0
For all n when b = 243*n^6 and {a and c} vary accordingly to form the AP. So when n = 1, b = 243 and a can be any number from 1 to 242 and c is adjusted accordingly. If {a, b, c} are in AP, the sum of a, b, and c will always be 3b.
The first few values of b are :-
{243, 15552, 177147, 995328, 3796875, 11337408, 28588707, 63700992,129140163, 243000000}.
The 6th root of 3 times those numbers are {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
So there will be an infinite set of them where the total number in each set is 1 less than 243*n^6 for each n in the set.
Paul.