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%I #49 Apr 24 2018 05:52:24
%S 0,1,2,1,3,2,1,3,3,2,1,3,3,3,2,1,3,3,3,3,2,1,3,3,3,3,3,2,1,3,3,3,3,3,
%T 3,2,1,3,3,3,3,3,3,3,2,1,3,3,3,3,3,3,3,3,2,1,3,3,3,3,3,3,3,3,3,2,1,3,
%U 3,3,3,3,3,3,3,3,3,2,1,3,3,3,3,3,3,3,3,3,3,3,2,1,3,3,3,3,3,3,3,3,3,3,3,3,2
%N Number of neighbors of each new term in an isosceles triangle read by rows.
%C To evaluate a(n) consider only the neighbors of a(n) that are present in the isosceles triangle when a(n) should be a new term in the triangle.
%C Apart from the left border and the right border, the rest of the elements are 3's.
%C If every "3" is replaced with a "4", we have the sequence A278290.
%C a(n) is also the number of new penny-penny contacts when putting pennies in a triangular arrangement.
%C For the same idea but for a right triangle see A278317; for a square array see A278290, for a square spiral see A278354; and for a hexagonal spiral see A047931.
%e The sequence written as an isosceles triangle begins:
%e .
%e . 0;
%e . 1, 2;
%e . 1, 3, 2;
%e . 1, 3, 3, 2;
%e . 1, 3, 3, 3, 2;
%e . 1, 3, 3, 3, 3, 2;
%e . 1, 3, 3, 3, 3, 3, 2;
%e . 1, 3, 3, 3, 3, 3, 3, 2;
%e . 1, 3, 3, 3, 3, 3, 3, 3, 2;
%e . 1, 3, 3, 3, 3, 3, 3, 3, 3, 2;
%e ...
%t Table[Boole[n > 1] (Prepend[Reverse@ Rest@ #, First@ #] &@ Range@ n /. k_ /; k > 3 -> 3), {n, 13}] // Flatten (* or *)
%t Table[Boole[n > 1] (Map[Mod[#, n] &, Range@ n] /. {k_ /; k > 1 -> 3, 0 -> 2}), {n, 13}] // Flatten (* _Michael De Vlieger_, Nov 23 2016 *)
%Y Row sums give A008585.
%Y Left border gives A057427.
%Y Every diagonal that is parallel to the left border gives the elements greater than 1 of A158799.
%Y Right border gives 0 together with A007395, also twice A057427.
%Y Every diagonal that is parallel to the right border gives A122553.
%Y Cf. A047931, A278290, A278317, A278354.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Nov 20 2016
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