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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.

Berman and Hanes (see link, page 81) proved in 1970 that an arrangement of 8 points on the surface of a sphere with 4 points with node degree 4 and 4 points with node degree 5 is the one with a maximum volume of their convex hull.

The polyhedron (see linked illustration) has vertices at the poles and two square rings of vertices rotated by Pi/4 against each other, with a polar angle of approx. +-62.89908285 degrees against the poles. The polyhedron is completely described by this angle and its order 16 symmetry. It would be desirable to know a closed formula representation of this angle and the volume.

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. Rising and falling diagonals are A078010 and A122552.

Apart from the left border and the right border, the rest of the elements are 6's.

For the same idea but for a right triangle see A278480; for a square array see A278545, for a square spiral see A010731; and for a hexagonal spiral see A010722.

The polyhedron (see linked illustration) with a symmetry group of order 4 has a vertex in the north pole on its axis of symmetry. The remaining 10 vertices are diametrically opposite in pairs relative to this axis of symmetry. The polar vertex has vertex degree 6. 8 vertices have vertex degree 5. 2 vertices have vertex degree 4.

This allocation seems to be the best possible approximation of a medial distribution of the vertex degrees, which is a known necessary condition for maximum volume. Of the 25 possible triangulations with vertex degree >= 4, all the others have more than 2 vertices with vertex degree 4, which leads to more pointed corners and therefore smaller volumes.

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.

Row sums give A036563(n+2), n >= 0.

From Wolfdieter Lang, Mar 23 2015: (Start)

This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).

This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.

The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.

The alternating row sums are A122553(n) = {1, repeat(3)}.

The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).

The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)

When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023