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a(n) = 6 + round(n^3) - (minimal number of squares in a dissection of an (n) X (n+1) oblong into squares).
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%I #20 May 17 2019 17:52:47

%S 5,4,3,3,3,3,1,1,2,2,1,1,1,1,1,1,1,2,0,1,1,1,1,1,1,1,1,0,0,1,0,0,0,1,

%T 0,0,0,0,0,0,0,0,1,0,1,1,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,

%U -1,0,0,0,0,0,0,0,-1,-1,0,0,0,0,0,0,-1,0,-1,0,-1,0,-1,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0

%N a(n) = 6 + round(n^3) - (minimal number of squares in a dissection of an (n) X (n+1) oblong into squares).

%C After a(18)=2, all terms through a(387) are in (-1,0,1). The first known term outside of this range is a(969) >= 2.

%H B. Felgenhauer, <a href="http://int-e.eu/~bf3/squares/">Filling Rectangles with Integer-Sided Squares</a>

%H Ed Pegg Jr, <a href="http://demonstrations.wolfram.com/MinimallySquaredRectangles/">Minimally Squared Rectangles</a>

%H Ed Pegg Jr on StackExchange, <a href="http://math.stackexchange.com/questions/2057290/oblongs-into-minimal-squares">Oblongs into minimal squares</a>, Dec 13 2016.

%F a(n) = 6 + A105209(n) - A279317(n).

%Y Cf. A105209, A279317.

%K hard,sign

%O 1,1

%A _Ed Pegg Jr_, Oct 26 2018