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Sum of terms in slice [n, n] = 3^(2n); example : 1+2+1+2+8+15+8+1+2+8+8+2+1+2+1 = 3^4 = 81 for the slice [2, 2].

Sum of terms in 2D slice [n, 1] is 3^(n+1).

Sum of terms in 2D slice [1, k] is 3^(1+k).

Sum of terms in 2D slice [2, k] is 3^(2+k).

Sum of terms in 2D slice [n, 2] is 3^(n+2).

Let j be a given slice of the hexagonal pyramid. For j = 0, there are no elements.

For j > 0, let a[x] to a[x+y-1] represent the elements of the slice, where x is the (j-1)th element of A019298 and y is the j-th element of A077043. Each slice j consists of j rows of varying column length, numbered 0 to j-1.

The length of the first row of slice j is given by floor((j+1)/2) and the last row by floor(j/2)+1, where by convention the last row is always greater or equal in length to the first row.

The floor(j/2)th row is j columns in length and any row before it is given by the formula floor((j+1)/2) + row#. For rows after the floor(j/2)th row, the length is given by floor(j/2) + j - row#.

The elements a[x] to a[x+y-1] are thus layed out as a concatenated series of rows of varying column lengths as specified above.

Thus for a given slice j, the element at row row# and column col# is represented by a[x + floor((j+1)/2) * row# + row# * (row# - 1) / 2 + col# ] when row# <= floor(j/2) and by a[x + y - (floor(j/2) + 1) * (j - row#) - (j - row#) * (j - row# + 1) / 2 + col#] otherwise, where x and y are defined above and row# and col# start counting from 0.

The elements of a for a given slice j, row# and col#, represented by the coordinate pair (row#, col#), is given by the following recursive relation:

For j = 1, there is 1 element whose value is 1 at (0, 0). Call this Slice[1] whose first and only element forms a0 = 1.

For j > 1, each element (row#, col#) is given by the sum of the 3 elements above it in the pyramid. If the preceding slice does not contain one of the cells specified because the coordinates are invalid for that slice, the value is assumed to be 0.

The cells above can be found using the following formula for a given cell Slice[j](row#, col#):

If j is odd:

If row# > floor(j/2):

Sum:

Slice[j-1](row#, col#-1)

Slice[j-1](row#-1,col#)

Slice[j-1](row#-1,col#-1)

Otherwise:

Sum:

Slice[j-1](row#, col#)

Slice[j-1](row#-1,col#)

Slice[j-1](row#-1,col#-1)

Otherwise:

If row# > floor(j/2):

Sum:

Slice[j-1](row#, col#)

Slice[j-1](row#-1,col#)

Slice[j-1](row#,col#-1)

Otherwise:

Sum:

Slice[j-1](row#, col#)

Slice[j-1](row#,col#-1)

Slice[j-1](row#-1,col#-1)

Each slice is also a solution to the Prouhet-Tarry-Escott problem for a given n and k. The slices[n,k] in sequence A109672 map to the slices here by the relation k + n = j - 1, where k = n (j odd) or k = n + 1 (j even).

When j is even, k = n - 1 would also be a solution to the Pascal hexagonal pyramid, however here the k = n + 1 solution is chosen. For j even, the slices are also given by A109673.

Only 3 of the 6 hexagonal vertices have corresponding cells in the slice below them. Only every other vertex has a cell below it and all vertices with cells below them are always separated by 2 edges.

By convention, when constructing Slice[j] for j odd, the uppermost vertices of Slice[j-1] are chosen to have cells below them and for j even the 2 vertices adjacent to the uppermost vertices of Slice[j-1] are chosen.