

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

The next term has 90 digits.
This can been seen as a 4dimensional parallel to the 3dimensional A099179 (Iterated tetrahedral numbers) and 2dimensional A007501 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 twinprime 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

Table of n, a(n) for n=0..4.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 6575.
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. A007501, A099179, A000332.
Sequence in context: A175169 A066933 A132496 * A167218 A013045 A290864
Adjacent sequences: A100006 A100007 A100008 * A100010 A100011 A100012


KEYWORD

easy,nonn


AUTHOR

Jonathan Vos Post, Nov 16 2004


STATUS

approved



