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A100009 Iterated pentatope numbers, starting with Ptop(2) = 5. The pentatope number of the pentatope number of the pentatope number ... of 2. +0
11
2, 5, 70, 1088430, 58478088289964410654020, 487260744950926456755434821807825520379852064518083517864587673941517278739124700409362355 (list; graph; listen)
OFFSET

0,1

COMMENT

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.

Adjacent sequences: A100006 A100007 A100008 this_sequence A100010 A100011 A100012

Sequence in context: A133004 A066933 A132496 this_sequence A013045 A007506 A042693

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post (jvospost2(AT)yahoo.com), Nov 16 2004

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Last modified October 13 09:05 EDT 2008. Contains 145008 sequences.


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