The triangle is generated from the product of matrix A and matrix B, i.e., A * B where A = the infinite lower triangular matrix:
1 0 0 0 0 ...
1 1 0 0 0 ...
1 1 1 0 0 ...
1 1 1 1 0 ...
1 1 1 1 1 ...
... and B = the infinite lower triangular matrix:
1 0 0 0 0 ...
1 3 0 0 0 ...
1 3 5 0 0 ...
1 3 5 7 0 ...
1 3 5 7 9 ...
...
Row sums give the square pyramidal numbers
A000330.
T(n+0,0)=1*n=
A000027(n+1); T(n+1,1)=3*n=
A008585(n); T(n+2,2)=5*n=
A008587(n); T(n+3,3)=7*n=
A008589(n); etc. So T(n,0)*T(n,1)=3*n*(n+1)=
A028896(n) (6 times triangular numbers). T(n,1)*T(n,2)/10=3*n*(n+1)/2=
A045943(n) for n>0 T(n,2)*T(n,3)/10=7/2*n*(n+1)=
A024966(n) for n>1 (7 times triangular numbers), etc.
Consider the following array, signed as shown:
...
1, 3, 5, 7, 9, 11, ...
2, -6, 10, -14, 18, -22, ...
3, 9, 15, 21, 27, 33, ...
4, -12, 20, -28, 36, -44, ...
5, 15, 25, 35, 45, 55, ...
6, -18, 30, -42, 54, -66, ...
7, 21, 35, 49, 63, 77, ...
...
Let each term (+, -)k = (+, -) phi^(-k).
Consider the inverse terms of the Lucas series (1/1, 1/3, 1/4, 1/7, ...).
By way of example, let q = phi = 1.6180339...; then
...
1/1 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + q^(-9) + ...
1/3 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
1/4 = q^(-3) + q^(-9) + q^(-15) + q^(-21) + q^(-27) +...
1/7 = q^(-4) - q^(-12) + q^(-20) - q^(-28) + q^(-36) + ...
1/11 = q^(-5) + q^(-15) + q^(-25) + q^(-35) + q^(-45) + ...
...
Relating to the Pell series, the corresponding "Lucas"-like series is (2, 6, 14, 34, 82, 198, ...) such that herein, q = 2.414213... = (1 + sqrt(2)).
Then analogous to the previous set,
...
1/2 = q^(-1) + q^(-3) + q^(-5) + q^(-7) + ...
1/6 = q^(-2) - q^(-6) + q^(-10) - q^(-14) + q^(-18) + ...
... (End)
