[math-fun] Oblong conjecture
Oblong conjecture: any n × (n+1) oblong can be divided into round(n^(1/3)) + 7 or fewer squares. Most oblongs seem to require exactly round(n^(1/3)) + 6 squares. True until at least 388. I've updated https://math.stackexchange.com/questions/2057290/oblongs-into-minimal-square... with corrected data. --Ed Pegg Jr
That’s very pretty! What are the best known bounds? Is there a heuristic reason to expect n^(1/3) behavior? Jim Propp On Tuesday, October 23, 2018, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Oblong conjecture: any n × (n+1) oblong can be divided into round(n^(1/3)) + 7 or fewer squares. Most oblongs seem to require exactly round(n^(1/3)) + 6 squares.
True until at least 388. I've updated https://math.stackexchange.com/questions/2057290/oblongs- into-minimal-squares with corrected data.
--Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
No known bounds that I know of. I have no idea why the cube root fits within 1 for cases 19 to 388. So far as I know the first oblong escaping the bound by more than 1 will be 969, which can be done with 14 squares. My bound suggests 16 squares. On Tue, Oct 23, 2018 at 12:14 PM James Propp <jamespropp@gmail.com> wrote:
That’s very pretty! What are the best known bounds? Is there a heuristic reason to expect n^(1/3) behavior?
Jim Propp
On Tuesday, October 23, 2018, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Oblong conjecture: any n × (n+1) oblong can be divided into round(n^(1/3)) + 7 or fewer squares. Most oblongs seem to require exactly round(n^(1/3)) + 6 squares.
True until at least 388. I've updated https://math.stackexchange.com/questions/2057290/oblongs- into-minimal-squares with corrected data.
--Ed Pegg Jr _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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