Yes 'mushroomsteve' you have done it again. I thought I would slip in another number challenge incorporating one of my favourites. Another Honorary Gold Medal on it's way.
As you suggest 231 is the cumulative number of 21 and 441 is the square of 21. As with 6 c no. 21 square 36 and 8 .. c.no. 36 square 64........... I have fun with these numbers, such as the sum of the c.nos. of two adjacent numbers (e.g. 9 / 10) equals the square of the highest (100) and the difference between a c.no. and its square is equal to the c.no. of the number before. e.g. 5 c.no = 15 square 25 difference 10 which equals the c.no. of 4.
Yes, they are quite fun! The nth one is given by T(n) = n(n+1)/2, so as you say we get T(n) + T(n-1) = n(n+1)/2 + n(n-1)/2 = n^2. I remember one time trying to find on a tiny scrap of paper all of the triangular numbers that are exactly double another triangular number. They are really fun to play around with.
Top 5 medal winners so far: 'mushroomsteve' with 10, then 'PotatoBoy' with 8, followed by 'ylj' and 'overtired' with 7 and they are followed by 'colmaggio' and 'Kiwirage' with 6 each. Congratulations medal winners.