Something for sci.math's amateur mathematicians? (Part 2)

For Part 1, see:

http://groups.google.com/group/sci.math/msg/722ec5cae2c3f48a

Ioannis Galidakis showed that the problem I raised there,
which I had obviously not given much thought to, seems
to be either fairly uninteresting or virtually intractable
(depending on how it's interpreted). I think the problem
this time will be more interesting. The problem -- really
two problems -- is at the end of this post.

The n'th triangular number is

T(n) = 1 + 2 + 3 + ... + n = n(n+1)/2.

The n'th tetrahedral (or pyramidal) number is

T(1) + T(2) + T(3) + ... + T(n) = n(n+1)(n+2)/6.

You can get this last result by replacing each
T(k), for k = 1, 2, ..., n, with (1/2)k + (1/2)k^2.
Then regroup terms so that all the (1/2)k terms are
together (and use the triangular number formula
to get a closed form expression for their sum)
and all the (1/2)k^2 terms are together (and use
the sum of the first n squares formula to get
a closed form expression for their sum):

[ (1/2)(1) + (1/2)(1^2) ]

+ [ (1/2)(2) + (1/2)(2^2) ]

+ [ (1/2)(3) + (1/2)(3^2) ]

....

+ [ (1/2)(n) + (1/2)(n^2) ]

= (1/2)*(1 + 2 + 3 + ... + n)

+ (1/2)*(1^2 + 2^2 + 3^2 + ... + n^2)

= (1/2)*[ n(n+1)/2 ] + (1/2)*[ n(n+1)(2n+1) ]

= (1/12)n(n+1)*[ 3 + (2n+1) ]

= (1/12)n(n+1)(2n+4) = (1/6)n(n+1)(n+2)

Let's denote the n'th triangular number by T(n,2)
and the n'th tetrahedral number by T(n,3). Then
the n'th hypertetrahedral number is

T(n,4) = T(1,3) + T(2,3) + T(3,3) + ... + T(n,3),

and expanding things like I did above, you'll get

T(n,4) = n(n+1)(n+2)(n+3)/24.

This can be continued, and one can show that
(follow the URL below)

T(n,k) = C(n+k-1, k), where C(i,j) is the

usual binomial coefficient number "i choose j".

http://www.math.toronto.edu/mathnet/questionCorner/tetnumbers.html

I seem to recall that Fermat proved this identity
for T(n,k).

Instead of the triangular numbers, if we begin with
the sum of the first n p'th powers, there are two natural
ways we could continue:

(1) Take the sum of the first n of these sum-of-p'th-powers
numbers, and continue the process k times like above.

(2) Take the sum of the p'th powers of these sum-of-p'th-powers
numbers, and continue the process k times like above.

I would imagine that various relations analogous to
T(n,k) = C(n+k-1, k) are known in each case, but I'm
not aware of them. (This doesn't mean much, because I
know very little about the field of combinatorics.)
If such relations are not known, it seems to me that
working them out might be something useful to work on
and it ought to be publishable in some expository or
recreational math journal.

Dave L. Renfro

.

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