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# The enumeration of Eulerian numbers.

I very strongly advise against the enumeration of the Eulerian numbers as given on the page. Everyone working with Eulerian numbers will observe that the enumeration of these numbers has not been standardized. Donald Knuth himself changed his enumeration in the course of time. So extreme care is advisable when reading the literature.

Also Wikipedia does not get things straight in their article, some statements refer to the enumeration given here and some other statements to another enumeration. The situation in the OEIS comments on the sequence is also far from being consistent.

However things changed. In the recent years the convention seems to stabilize (for good reasons). I advise to follow the enumeration below:

 n\k 0 1 2 3 4 5 6 0 1 1 1 2 1 1 3 1 4 1 4 1 11 11 1 5 1 26 66 26 1 6 1 57 302 302 57 1 7 1 120 1191 2416 1191 120 1

This is the enumeration of Graham, Knuth and Patashnik, Concrete Mathematics and it is the enumeration of the Digital Library of Mathematical Functions by the American National Institute of Standards and Technology (which is the successor of the famous Handbook) -- compare Table 26.14.1 --. — Peter Luschny 16:48, 23 July 2010 (UTC)

Thanks for that recommendation!

I felt too that k should start at 0 (the recurrence formula begs for it,) I'll gladly make the changes! I didn't dare change the convention of Euler's Number Triangle on MathWorld (it chose $A_{n,1}=A_{n,n}=1\,$ ).

From the recursion rule, with $k\in \{1,...,d\}\,$ :

$E(1,1)=1,\,$ $E(d,k)=[d-(k-1)]\ E(d-1,k-1)+k\ E(d-1,k),\ d\geq 2,\ k\in \{1,...,d\}\,$ so obviously begs for (although less so with the k+1 on the right...) $k\in \{0,...,d-1\}\,$ $E(0,0)=E(1,0)=1,\,$ $E(d,k)=(d-k)\ E(d-1,k-1)+(k+1)\ E(d-1,k),\ d\geq 2,k\in \{0,...,d-1\}\,$ I'll also use the number triangles standard (n,k) instead of (d,k) where d refered to the dimension in nd (for which the Eulerian polynomials where a core part of the generating function.)

And the extratriangular 0! = 1 for degree zero too!

I'm updating everything right now!

Thanks

Daniel Forgues, July 23, 2010

Also, concerning the Eulerian polynomials, with $k\in \{1,...,d\}\,$ :

$E_{0}(x)=1,\,$ $E_{d}(x)=\sum _{k=1}^{d}E(d,k)\ x^{k-1},\ d\geq 1\,$ is the d th Eulerian polynomial  whose Eulerian numbers:

$E(d,k)|_{k=1}^{d}=\{1,(2^{d}-d-1),...,(2^{d}-d-1),1\},\ d\geq 1\,$ can be recursively generated with the triangle of Eulerian numbers.

very much begs for $k\in \{0,...,d-1\}\,$ :

$E_{0}(x)=1,\,$ $E_{d}(x)=\sum _{k=0}^{d-1}E(d,k)\ x^{k},\ d\geq 1\,$ is the d th Eulerian polynomial  whose Eulerian numbers:

$E(d,k)|_{k=0}^{d-1}=\{1,(2^{d}-d-1),...,(2^{d}-d-1),1\},\ d\geq 1\,$ can be recursively generated with the triangle of Eulerian numbers.

Thanks

Daniel Forgues, July 23, 2010