d-pyramidal number?
So, the sum of the first n natural numbers is the nth triangular number. And the sum of the first n triangular numbers is the nth tetrahedral number.
Is there a term that generalizes this concept? Something along the lines of a "d-pyramidal number", where if d = 1 you get the natural numbers, for d = 2 you get triangular numbers, and for d = 3 you get tetrahedral numbers, and so on?
I was trying to sleep last night with the Twelve Days of Christmas going through my head and wondered how many gifts were given at the end of the 12 days. (It's 364.) As I was showering this morning I went through several strategies to derive this answer (along the way, lamenting that if I wasn't dealing with integers then I could just do a quick integral, and that my visualization skills in three dimensions is not very good), and on the bus I ended up coming up with (n + d - 1) choose d as the general answer for the nth d-pyramidal number.
(Actually I just had an uglier product--the pithy version of the formula didn't occur to me until I went online searching for the right terminology and found a site that shows the answer in a Pascal's triangle, and made me go, "duh, what I have can be expressed more concisely!")
Is there a term that generalizes this concept? Something along the lines of a "d-pyramidal number", where if d = 1 you get the natural numbers, for d = 2 you get triangular numbers, and for d = 3 you get tetrahedral numbers, and so on?
I was trying to sleep last night with the Twelve Days of Christmas going through my head and wondered how many gifts were given at the end of the 12 days. (It's 364.) As I was showering this morning I went through several strategies to derive this answer (along the way, lamenting that if I wasn't dealing with integers then I could just do a quick integral, and that my visualization skills in three dimensions is not very good), and on the bus I ended up coming up with (n + d - 1) choose d as the general answer for the nth d-pyramidal number.
(Actually I just had an uglier product--the pithy version of the formula didn't occur to me until I went online searching for the right terminology and found a site that shows the answer in a Pascal's triangle, and made me go, "duh, what I have can be expressed more concisely!")