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A252897
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Rainbow Squares: a(n) = number of ways to pair the integers 1 to 2n so that the sum of each pair is a square.
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4
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1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 6, 18, 12, 36, 156, 295, 429, 755, 2603, 7122, 19232, 32818, 54363, 172374, 384053, 933748, 1639656, 4366714, 20557751, 83801506, 188552665, 399677820, 640628927, 2175071240, 8876685569, 32786873829, 108039828494
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OFFSET
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0,14
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COMMENTS
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The original sequence is from Henri Picciotto who asked for which n is such a pairing possible: A253472.
The name "rainbow squares" refers to the use of this problem in the elementary school classroom where children draw colored connecting "rainbows" to make the pairings.
Number of perfect matchings in the graph with vertices 1 to 2n and edges {i,j} where i+j is a square. - Robert Israel, Mar 22 2015
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LINKS
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Gordon Hamilton, Kiran S. Kedlaya, and Henri Picciotto, Square-Sum Pair Partitions, The College Mathematics Journal, Vol. 46, No. 4 (September 2015), pp. 264-269.
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EXAMPLE
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One of the solutions for n=13 consists of the following pairings of 1-26:
{1,15}, adding to 16;
{2,23}, {3,22}, {4,21}, {5,20}, {6,19}, {7,18}, {8,17}, {9,16}, {11,14}, {12, 13}, each adding to 25;
{10,26}, adding to 36;
{24,25}, adding to 49.
There are five other such pairings possible, so a(13) = 6.
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MAPLE
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F:= proc(S)
option remember;
local s, ts;
if nops(S) = 0 then return 1 fi;
s:= S[-1];
ts:= select(t -> issqr(s+t), S minus {s});
add(procname(S minus {s, t}), t = ts);
end proc:
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MATHEMATICA
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F[S_] := F[S] = Module[{s, ts}, If[Length[S] == 0, Return[1]]; s = S[[-1]]; ts = Select[S ~Complement~ {s}, IntegerQ[Sqrt[s + #]]&]; Sum[F[S ~Complement~ {s, t}], {t, ts}]];
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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