When originally toying with this idea I generated a sequence [while examining the expression (3n^2+9*n +8)/2 ] which appeared to be in slight variance with Sloane A082840; this proved to be a matter of computer programming precision.

Charlie notes:
The formula given in Sloane for the ordinal of A (i.e., the nth triangular number), is:

With a=2+sqrt(3), b=2-sqrt(3)
a(n)=-3/2+(1/12)(a-2b+5)a^n+(1/12)(b-2a+5)b^n.

Charlie's computer listing substitutes the above values for "a" and "b" in this line:
ans = -3 / 2 + (1 / 12) * (a - 2 * b + 5) * a ^ n + (1 / 12) * (b - 2 * a + 5) * b ^ n

Doing that, and then replacing "n" with "A1", you may copy this expression into cell B1 of a spreadsheet:
= -3 / 2 + (1 / 12) * ((2+SQRT(3)) - 2 * (2-SQRT(3)) + 5) * (2+SQRT(3)) ^ A1 + (1 / 12) * ((2-SQRT(3)) - 2 * (2+SQRT(3)) + 5) * (2-SQRT(3)) ^A1

Copy down Column A the incremental values for "n" beginning with 1 in A1. Copy the formula down Column B. The values obtained are Sloane's A082840. I note Charlie further addresses a related sequence at Sloane, A133161, in his discussion.

Substituting a Column B value into the set:
{ N * (N + 1)/2, (N + 1)*(N + 2)/2, (N +2)*(N + 3)/2 } yields three consecutive triangular numbers whose sum is also triangular.

Brian Smith offered me this expression:
=(3/8+9/(4*sqrt[48])) * (7+sqrt[48])^n + (3/8-9/(4*sqrt[48])) * (7-sqrt[48])^n + 1/4 but I was to later realise that it yielded the sums of three consecutive triangles, this forms Sloane's A129803.

As I noted with Charlie's expression, you may do similarly with Brian Smith's. Copy incremental values for n into Column A beginning at A1, copy the expression:
=(3/8+9/(4*SQRT(48))) * (7+SQRT(48))^A1 + (3/8-9/(4*SQRT(48))) * (7-SQRT(48))^A1 + 1/4
into B1, and then copy that formula downwards in Column B. This gives the consecutive triangular sums of three consecutive triangular numbers.

Those triangles may be found by evaluating the equation:
S = (3N^2 + 9N + 8)/2 where S is the triangular sum (as per Brian Smith's expression).
The three triangles are:
{ N * (N + 1)/2, (N + 1)*(N + 2)/2, (N +2)*(N + 3)/2 }

We have two means at our disposal to find any required set.
1. For any required set we may enter a value (for A1) into the expansion I've made of Charlie's derivation, build the set of three triangles and then calculate the triangular sum.
2. For any required set we may enter a value (for A1)into the Brian Smith expression, evaluate the Sloane value which is given by A082840 and build the triangles.

I trust that Charlie and Brian's contributions have allowed for me to present a complete solution.

Comments: (
You must be logged in to post comments.)