

A272259


Irregular triangle read by rows: Row n>=32 gives the smallest square loop, i.e., lexicographically earliest circular permutation of length n such that any two adjacent numbers sum to a perfect square.


2



1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15, 1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 20, 29, 7, 18, 31, 33, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15, 1, 3, 13, 12, 4, 32, 17, 8, 28, 21, 15, 34, 30, 19, 6, 10, 26, 23, 2, 14, 22, 27, 9, 16, 33, 31, 18, 7, 29, 20, 5, 11, 25, 24
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OFFSET

32,2


COMMENTS

T(n) gives the smallest Hamiltonian cycle in the corresponding undirected unweighted graph with n vertices and edges satisfying the square sum condition, so this is also a solution to the Traveling Salesman Problem.
There are no circular solutions for n < 32.
T(32) = A112663(k1), 1 <= k <= 32.
Row n has length n, and we start with row n = 32.


LINKS

Martin Renner, Table of n, a(n) for n = 32..663


EXAMPLE

Table starts with
n = 32: 1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15.
n = 33: 1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 20, 29, 7, 18, 31, 33, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15.


MAPLE

with(GraphTheory):
n:=32; # Vertices from 1 to n
E:={}: # Edges
for a from 1 to n do
for b from a+1 to n do
if type(sqrt(a+b), integer) then E:={op(E), {a, b}}: fi:
od:
od:
G:=Graph(E);
Tn:=TravelingSalesman(G)[2, 1..n];


CROSSREFS

Cf. A071984, A112663.
Sequence in context: A088029 A126815 A053619 * A112663 A220288 A201101
Adjacent sequences: A272256 A272257 A272258 * A272260 A272261 A272262


KEYWORD

nonn,tabf


AUTHOR

Martin Renner, Apr 23 2016


STATUS

approved



