Re: [math-fun] Equal sums of powers
<< Prouhet's Thue-Morse based solution turns out to be ... Solution to what? —Dan
See https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem (via Hans Havermann) << Prouhet used the Thue–Morse sequence to construct a solution with m = 2^l for any l . Namely, partition the numbers from 0 to m-1 into evil and odious numbers; then the sets of the partition give a solution to. For instance, for m = 8 and l = 4 , Prouhet's solution is: 0^1 + 3^1 + 5^1 + 6^1 + 9^1 + 10^1 + 12^1 + 15^1 = 1^1 + 2^1 + 4^1 + 7^1 + 8^1 + 11^1 + 13^1 + 14^1 ; 0^2 + 3^2 + 5^2 + 6^2 + 9^2 + 10^2 + 12^2 + 15^2 = 1^2 + 2^2 + 4^2 + 7^2 + 8^2 + 11^2 + 13^2 + 14^2 ; 0^3 + 3^3 + 5^3 + 6^3 + 9^3 + 10^3 + 12^3 + 15^3 = 1^3 + 2^3 + 4^3 + 7^3 + 8^3 + 11^3 + 13^3 + 14^3 ;
On 7/2/17, Dan Asimov <dasimov@earthlink.net> wrote:
<< Prouhet's Thue-Morse based solution turns out to be ...
Solution to what?
—Dan
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