|
| |
|
|
A100009
|
|
Iterated pentatope numbers, starting with Ptop(2) = 5. The pentatope number of the pentatope number of the pentatope number ... of 2.
|
|
10
| | |
|
|
|
OFFSET
| 0,1
|
|
|
COMMENTS
| This can been seen as a 4-dimensional parallel to the 3-dimensional A099179 (Iterated tetrahedral numbers) and 2-dimensional A099053 Iterated triangular numbers. This need not start at 2. For example, starting at a(0) = 17, which is not a pentatope number, we have a(1) = Ptop(17) = 17*(17+1)*(17+2)*(17+3)/24 = 4845 = 3 * 5 * 17 * 19 {which happens to be the product of two twin-prime pairs); a(2) = Ptop(Ptop(17)) = Ptop(4845) = 4845*(4845+1)*(4845+2)*(4845+3)/24 = 22988020743780.
|
|
|
REFERENCES
| J. V. Post, "Iterated Triangular Numbers", preprint.
J. V. Post, "Iterated Polygonal Numbers", preprint.
|
|
|
LINKS
| Hyun Kwang Kim, On Regular Polytope Numbers.
J. V. Post, Table of Polytope Numbers, Sorted, Through 1,000,000.
|
|
|
FORMULA
| a(1) = 2; for n>1, pentatope number Ptop(n) = n*(n+1)*(n+2)*(n+3)/4!; for k>1, a(k+1) = Ptop(a(k)) = (a(k))*(a(k)+1))*(a(k)+2))*(a(k)+3))/24.
|
|
|
EXAMPLE
| a(3) = 1088430 because a(0) = 2 is the seed for this instance of the more general recurrence, a(1) = Ptop(2) = 2*(2+1)*(2+2)*(2+3)/24 = 5; a(2) = Ptop(Ptop(2)) = Ptop(5)= 5*(5+1)*(5+2)*(5+3)/24 = 70; a(3) = Ptop(Ptop(Ptop(2))) = Ptop(70)= 70*(70+1)*(70+2)*(70+3)/24 = 1088430.
|
|
|
CROSSREFS
| Cf. A099053, A099179, A000332.
Sequence in context: A175169 A066933 A132496 * A167218 A013045 A007506
Adjacent sequences: A100006 A100007 A100008 * A100010 A100011 A100012
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 16 2004
|
| |
|
|
