Let T_n = 1 + 2 + 3 + ... + n = n(n+1)/2. Then T_n is the
n'th triangular number. The n'th tetrahedral number is
T_1 + T_2 + T_3 + ... + T_n = n(n+1)(n+2)/6 [1]. We can
repeat this process to form the sum of the first n tetrahedral
numbers to get the n'th 4-tetrahedral number n(n+1)(n+2)(n+3)/24,
and so on for the higher order versions [2].
[1] http://mathworld.wolfram.com/TetrahedralNumber.html
http://en.wikipedia.org/wiki/Tetrahedral_number
http://mathforum.org/workshops/usi/pascal/pascal_tetrahedral.html
[2] http://www.math.toronto.edu/mathnet/questionCorner/tetnumbers.html
Apparently, this is all fairly well known. However, I wonder if
the analogous situation for multiplication (or exponentiation)
replacing addition has been studied.
For example, let's call 1*2*3*...*n = n! the 1'st order factorial
of n, 1!*2!*3!*...*n! the 2'nd order factorial of n, and so on.
Are there any interesting mathematical issues going on with these
higher order factorial numbers?
Dave L. Renfro
Well, obviously 1!*2!*3!*...*n! = 1^n*2^{n-1}*3^{n-2}*...*(n-1)^2*n^1,
so it looks like such operations can be reduced to
multiplication/exponentiation.
For exponentiation however, the situation seems trivially uninteresting, since
if we interpret power towers the usual way (from top to bottom) all such
expressions will equal 1:
1^2^3^...^n = 1
1^(1^2)^(1^2^3)^...^(1^2^3^...^n) = 1
1^(1^(1^2))^(1^(1^2)^(1^2^3))^...^(1^(1^2)^(1^2^3)^...^(1^2^3^...^n)) = 1
If we exclude 1 from the towers above, sufficient bounds can be gotten both
from left and right, but I seriously doubt there are any closed form
expressions for these towers, since such expressions don't even exist for
non-trivial power towers with same exponents.
> Dave L. Renfro
--
Ioannis