A275015 - OEIS

%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