Editing Pentatope number (section)

== Properties ==
Two of every three pentatope numbers are also [[pentagonal number]]s. To be precise, the {{math|(3''k'' − 2)}}th pentatope number is always the <math>\left(\tfrac{3k^2 - k}{2}\right)</math>th pentagonal number and the {{math|(3''k'' − 1)}}th pentatope number is always the <math>\left(\tfrac{3k^2 + k}{2}\right)</math>th pentagonal number. The {{math|(3''k'')}}th pentatope number is the [[pentagonal number|generalized pentagonal number]] obtained by taking the negative index <math>-\tfrac{3k^2 + k}{2}</math> in the formula for pentagonal numbers. (These expressions always give [[integer]]s).<ref name="oeis">{{Cite OEIS|A000332}}</ref>

The [[infinite sum]] of the [[Multiplicative inverse|reciprocals]] of all pentatope numbers is {{sfrac|4|3}}.<ref>{{citation|title=Sums of the inverses of binomial coefficients|journal=Fibonacci Quarterly|year=1981|url=http://www.fq.math.ca/Scanned/19-5/rockett.pdf|first=Andrew M.|last=Rockett|volume=19|issue=5|pages=433–437|doi=10.1080/00150517.1981.12430049 }}. Theorem 2, p.&nbsp;435.</ref> This can be derived using [[telescoping series]].

:<math>\sum_{n=1}^\infty \frac{4!}{n(n+1)(n+2)(n+3)} = \frac{4}{3}.</math>

Pentatope numbers can be represented as the sum of the first {{mvar|n}} [[tetrahedral number]]s:<ref name="oeis"/>

:<math>P_n = \sum_{ k =1}^n \mathrm{Te}_k,</math>

and are also related to tetrahedral numbers themselves:

:<math>P_n = \tfrac{1}{4}(n+3) \mathrm{Te}_n.</math>

No [[prime number]] is the predecessor of a pentatope number (it needs to check only −1 and {{nowrap|1=4 = 2<sup>2</sup>}}), and the largest [[semiprime]] which is the predecessor of a pentatope number is 1819.

Similarly, the only primes preceding a [[Figurate_number#Triangular_numbers_and_their_analogs_in_higher_dimensions|6-simplex number]] are [[83 (number)|83]] and 461.