A275015 Number of neighbors of each new term in an isosceles triangle read by rows.

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, 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, 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

Offset: 1

Author

Omar E. Pol, Nov 20 2016

Keywords

• nonn
• tabl

Comments

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.

Apart from the left border and the right border, the rest of the elements are 3's.

If every "3" is replaced with a "4", we have the sequence A278290.

a(n) is also the number of new penny-penny contacts when putting pennies in a triangular arrangement.

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.

Examples

			The sequence written as an isosceles triangle begins:
.
.                     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,  3,  2;
.     1,  3,  3,  3,  3,  3,  3,  3,  2;
.   1,  3,  3,  3,  3,  3,  3,  3,  3,  2;
...
		

Crossrefs

Row sums give A008585.

Left border gives A057427.

Every diagonal that is parallel to the left border gives the elements greater than 1 of A158799.

Right border gives 0 together with A007395, also twice A057427.

Every diagonal that is parallel to the right border gives A122553.

Cf. A047931, A278290, A278317, A278354.

Programs

Mathematica

Table[Boole[n > 1] (Prepend[Reverse@ Rest@ #, First@ #] &@ Range@ n /. k_ /; k > 3 -> 3), {n, 13}] // Flatten (* or *)
Table[Boole[n > 1] (Map[Mod[#, n] &, Range@ n] /. {k_ /; k > 1 -> 3, 0 -> 2}), {n, 13}] // Flatten (* Michael De Vlieger, Nov 23 2016 *)