likewise for cycle numbers (for fun, not useful here),
-
![{\displaystyle {\begin{aligned}&\Delta ^{n}\left(\lambda j.\left[{k+j \atop j}\right]\right)(0)\\=&[x^{k}]\left(\sum _{j=0}^{n}\left((-1)^{n-j}*{n \choose j}*\prod _{i=0}^{k+j-1}(1+i*x)\right)\right)\\=&(-1)^{n}*[x^{k}]\left(\prod _{i=0}^{k-1}\left(1+i*x\right)*\sum _{j=0}^{n}\left((-n)^{\overline {j}}*\left({\frac {1}{x}}+k\right)^{\overline {j}}*{\frac {x^{j}}{j!}}\right)\right)\\=&(-1)^{n}*[x^{0}]\left(\left({\frac {1}{x}}\right)^{\overline {k}}*{_{2}F_{0}}\left({-n,{\frac {1}{x}}+k \atop ~};x\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0fb105c64be436989d3c96d06857ebd8dd2e4208)
![{\displaystyle \Delta ^{n}\left(\lambda j.\left[{k+j \atop j}\right]\right)(0)=(-1)^{n}*\left[{\frac {x^{k}}{k!}}\right]{_{2}F_{1}}\left({-n,k+1 \atop 1};{\frac {\log \left({\frac {1}{1-x}}\right)}{x}}\right)=\sum _{\pi \in \mathrm {partitions(\{1..n\})} }\left(\sum _{j=|\pi |}^{n}\left({n \choose j}*(-1)^{n-j}*{\frac {(k+j)!}{(j-|\pi |)!}}\right)*\prod _{b\in \pi }\left({\frac {b!}{b+1}}\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/7d152db069e2a1ca263f376dc340e94639e30018)
- (seemingly first given in [3])
- and
- note that since
, we have an explicit form for the generalised harmonic numbers 
(so 
), and an o.g.f. 
, where the polylogarithm is defined 
and 
Interesting part
Define ![{\displaystyle a(n,k)=\mathrm {conv} \left({\lambda i.\left[{n+1 \atop i}\right] \atop \lambda i.(-1)^{i}*\Delta ^{n}\left(\lambda j.\left\{{k-i+j \atop j}\right\}\right)(0)}\right)(k)=\sum _{i=0}^{k}\left(\left[{n+1 \atop n+1-i}\right]*(-1)^{i}*\sum _{j=0}^{n}\left((-1)^{n-j}*{n \choose j}*\left\{{k-i+j \atop j}\right\}\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/914d8b0ab50d1e204edd33ae64d77dd590734d72)
The above formulae give the g.f. forms ![{\displaystyle a(n,k)=[x^{n-k}]\left(\prod _{i=1}^{n}(i-x)*{_{1}F_{1}}\left({-n \atop 1-x};-x\right)\right)=\sum _{j=0}^{n}\left({n \choose j}*[x^{n-j-k}]\left(\prod _{i=j+1}^{n}(i-x)\right)\right)=\sum _{j=0}^{n}\left({n \choose j}*[x^{k}]\left(\prod _{i=j+1}^{n}(i*x-1)\right)\right)=\sum _{j=0}^{n}\left({n \choose j}*[x^{n-k-j}]\left(\left(j+1-x\right)^{\overline {n-j}}\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ed5bb9c5c0df7d406ed8ab2b429ae38e6a2bcc26)
, which give the combinatorial interpretation 
, explicitly 
(upper limits on outer sums imposed by cardinality)
A serieswise interpretation of 
is that ![{\displaystyle [x^{k}]\left({\frac {\sum _{k=0}^{\infty }(a(n,k)*x^{k})}{\prod _{i=1}^{n}(1-i*x)}}\right)=\mathrm {conv} \left({\lambda k.a(n,k) \atop \lambda k.\left\{{n+k \atop n}\right\}}\right)(k)=\Delta ^{n}\left(\lambda x.\left\{{k+x \atop x}\right\}\right)(0)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/35d14b61151d9a85f219e9cdb1b8c2b2003ac213)
, so one can perform its definition backwards, 
diagonals 
where we define 
, so giving its g.f. as ![{\displaystyle f_{C}(n,j)=([x^{n}]-[x^{j}])(F_{C}(x))}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b055bb5142e91323882cb035d08ad5cd17d168e5)
, we have 
then 
and ![{\displaystyle f_{0}(n,j)=[n>0]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/55e0d2a4bd2e80e6e51a09d5701b7781fd5dbd49)
, 
and 
, 
and 
, 
so we have
![{\displaystyle a(n,n-0)=n!=\left[{n+1 \atop 1}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8fa264a24715ccf4a15851dbca4ae714e9da03b4)
![{\displaystyle a(n,n-1)=n!*(n-H_{n})=n*\left[{n+1 \atop 1}\right]-\left[{n+1 \atop 2}\right]=A067318(n)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/66e9655212a9090e9b85aa8f3ba4445ebd37547e)
[n 2]![{\displaystyle a(n,n-2)=n!*\left({\frac {n*(n-1)}{4}}-n*(H_{n}-1)+{\frac {H_{n}^{2}-H_{n}^{(2)}}{2}}\right)={\frac {n \choose 2}{2}}*\left[{n+1 \atop 1}\right]-n*\left(\left[{n+1 \atop 2}\right]-\left[{1+1 \atop 2}\right]\right)+\left[{n+1 \atop 3}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/144729d085144cffd7f78964f9ca2747d08314d7)
proof: we can sum the generating function for constant 
over all 
(or rather, all 
, due to its reversedness) by setting 
and evaluating it, reducing it to 
, the sum over Lah numbers by which it's defined.
Another definition is
![{\displaystyle A000262(n)=\sum _{k=0}^{n}\left(A355260(n,k):=\left[{n \atop k}\right]*\sum _{i=0}^{k}\left\{{k \atop i}\right\}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ecbe1450bdce19657c33e99e875e5a46a9d3a8ae)
, where
A355260(n,k)=|
A055924(n,k)|, the latter having an explanation from David Callan as counting the cardinality of the subset where the (
) lists contain a total of
cycles
Note that per Vaclav Kotesovec's comment (ignoring the Laurent series), 
From [5], we have 
.
Proposition: This corresponds with 
.
Again using the generating function,
![{\displaystyle [x^{k}]\left(\sum _{n=k}^{2*k}\left(\sum _{j=0}^{n}\left({n \choose j}*\prod _{i=j+1}^{n}(i*x-1)\right)\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/c684a1d250beeb238de73195b992455d901d947f)
The Lagrange inversion theorem gives
![{\displaystyle {\begin{aligned}A058349(k+1)&=\left[{\frac {x^{k}}{k!}}\right]\left(\left({\frac {x}{x+2*(1-\cosh(x))}}\right)^{k+1}\right)\\&=\left[{\frac {x^{k}}{k!}}\right]\left({\frac {1}{\left(1-{\frac {2*(1-\cosh(x))}{x}}\right)^{k+1}}}\right)\\&=\left[{\frac {x^{k}}{k!}}\right]\left({\frac {1}{\left(1-2*\sum _{n=1}^{\infty }\left({\frac {x^{2*n-1}}{(2*n)!}}\right)\right)^{k+1}}}\right)\\&=\left[{\frac {x^{k}}{k!}}\right]\left(\mathrm {gf} \left(\lambda k.\sum _{P\in (\mathbb {N} \setminus 2\mathbb {N} )^{\infty } \atop \sum (P)=k}\left(\prod _{b\in P}((b+1)!)\right)\right)^{k+1}\right)\\&=k!*\sum _{P\in \mathbb {N} ^{k+1} \atop \sum (P)=k}\left(\prod _{b\in P}\left(\sum _{P\in (\mathbb {N} \setminus 2\mathbb {N} )^{\infty } \atop \sum (P)=b}\left(\prod _{b\in P}((b+1)!)\right)\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d8206caee41a85e30af45ddcdd4eb5b6815816cd)
Series-parallel posets can be constructed through two operations that group a subset of nodes into a metanode that behaves as a single one, connecting them in parallel and in series respectively. ([6] explains a little bit more context.) These can be considered more abstractly as groupings into sets and lists; each type may only contain nodes and the other type (since subgroupings of the same type would be merged together), and the 'connected' criterion makes the tree rooted at a list.
Whereas thus far we have been concerned mostly with o.g.f.'s, in this part we use e.g.f.'s, since they make nested combinatorial structures isomorphic with arithmetic using Faà di Bruno (the opposite way around from above; getting g.f.'s from interpretations)
Given the e.g.f. 
of a sequence enumerating nonempty [structure]s on nodes, the e.g.f. of the number of sets of [structure]s with nodes between them is 
, whereas the number of lists of [structure]s is 
So we can partition the nodes into a set of lists with 
(A000262), and permit nodes to lie in the set by adding 
to the 
's input for 
(A052844[n 3])
The other way around, we can partition nodes into a list of sets with 
(A000670), then permit nodes to lie in the list by adding to the 
's input for 
(A006155[n 4]).
so the e.g.f. enumerating 'stripey trees' rooted at lists (which has 
by convention) is 
, where 
enumerates them rooted at sets (we subtract the 
th and 
th powers, since a list cannot consist of a single set and vice versa).
can be rewritten 
so 
, giving the functional inverse that R. C. Read obtained!
working backwards, 
, ie. 
, gives that for 
, 
however, we can also go the other way around, as 
. ))={\begin{cases}0&n=0\\1&n=1\\-2&n\ \mathrm {even} \\0&\mathrm {else} \end{cases}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/84ccbec7350fefadf3ec71e8c42dab02c9dab850)
defining the aerated array 
to be the number of partitions of labelled objects into a set of even-cardinality nonempty subsets, then ![{\displaystyle \left\{\left\{{n \atop k}\right\}\right\}=[x^{n}]\left(\prod _{j=1}^{k}\left({\frac {(2*j-1)*x^{2}}{1-(j*x)^{2}}}\right)\right)=[x^{n}]\left({\frac {1}{k!*2^{k}}}*\sum _{j=1}^{k}\left({\frac {(-1)^{k-j}*{2*k \choose k-j}}{1-(j*x)^{2}}}\right)\right)={\frac {1+(-1)^{n}}{k!*2^{k}}}*\sum _{j=1}^{k}\left((-1)^{k-j}*{2*k \choose k-j}*j^{n}\right)=\left[{\frac {x^{n}}{n!}}\right]\left({\frac {(\cosh(x)-1)^{k}}{k!}}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/486c06a1e69a712e63f846adb676cc5e815e82b9)
(x+2*(1-\cosh(x))))&={\begin{cases}1&n=1\\0&\mathrm {else} \end{cases}}\\&=\sum _{k=0}^{n}\left(A058349(k)*\sum _{P\in \mathrm {partitions} (\{0..n-1\}) \atop |P|=k}\left(\prod _{b\in P}\left({\begin{cases}0&|b|=0\\1&|b|=1\\-2&|b|\ \mathrm {even} \\0&\mathrm {else} \end{cases}}\right)\right)\right)\\&=\sum _{k=0}^{n}\left(A058349(k)*\sum _{m=0 \atop n-m\ \mathrm {even} }^{k}\left(\sum _{P\in \mathrm {partitions} \left(\left\{0..n-m-1\right\}\right) \atop {|P|=k-m \atop \forall b\in P\colon |b|\ \mathrm {even} }}\left((-2)^{|P|}\right)\right)\right)\\&=\sum _{k=0}^{n}\left(A058349(k)*\sum _{m=0 \atop n-m\ \mathrm {even} }^{k}\left((-2)^{k-m}*\left|\left\{P\in \mathrm {partitions} \left(\left\{0..n-m-1\right\}\right)\colon |P|=k-m,\forall b\in P\colon |b|\ \mathrm {even} \right\}\right|\right)\right)\\&=\sum _{k=0}^{n}\left(A058349(k)*\sum _{m=0 \atop n-m\ \mathrm {even} }^{k}\left((-2)^{k-m}*\left\{\left\{{n-m \atop k-m}\right\}\right\}\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/7af8ba71d75e1714719b562eb7468217dbc1363c)
so we need only prove that for 
,
again per Kotesovec, 
Notes
- ↑ replacing the Bernoulli numbers with their upper-exclusively-defined variant
produces the array ![{\displaystyle (-1)^{n-k}*\left[{n \atop k}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/586131fdfc59712c33e8e65193a06e69e37a4b9a)
, which is the matrix inverse (rather than backwards recurrence-relation/diagonal-polynomial continuation) of the array 
and is used (usually notated 
) in some places like the DLMF
- ↑ Philippe Deléham's formula in A067318 mentions that it is also
for a triangle defined by the DELTA operator he first defined in A084938 (see also Luschny's sequence transformation entry and R. J. Mathar's write-up), and that triangle (A088996) curiously has a leading diagonal coinciding with this page's 
as well, however besides these it is quite different (ie. its left edge is orthogonal, not knightwise)
- ↑ counts over the same structure as A000262 except count each partitioning into a set of lists
with multiplicity equal to 
, since each one can be either nested into its own set or freestanding in the list
- ↑ this sequence does not currently contain this combinatorial interpretation! representation of the first few terms:
a(0)=1=|{[]}|
a(1)=2=|{[0],[{0}]}|
a(2)=9=|{[0,1],[1,0],[{0},1],[0,{1}],[{1},0],[1,{0}],[{0},{1}],[{1},{0}],[{0,1}]}|
, meaning (from Kotesovec's asymptotics) 
References
- ↑ Donald Knuth, 1992. Two notes on notation
- ↑ Victor Adamchik, 1997. On Stirling numbers and Euler sums. Journal of Computational and Applied Mathematics, Volume 79, Issue 1. (DoI, author's webpage)
- ↑ J. H. Conway and R. K. Guy (1996). The Book of Numbers, New York: Springer-Verlag, page 258
note that they use 
, differing from most literature (and this) which use 
for the zeta partial sums
it seems they didn't know about the more elegant version in terms of upper-exclusive sums
- ↑ it was not well-known to me; Sriram V. Pemmaraju, Cycle Structure of Permutations (lecture note, autumn 2001) explains it as follows
write your permutation as a sequence of cycles, each one with its minima first, and sort cycles (keyed by first element) from largest to smallest, and concatenate them, then each cycle becomes a left-to-right minimum.
Likewise, one can send them back by placing partitions before indexes of left-to-right minima, treat parts as cycles and convert from cycle to ordinary notation.
- ↑ Ronald C. Read, September 26, 1991. Graphical enumeration by cycle-index sums: first steps toward a unified treatment (Research Report CORR 91-19, University of Waterloo)
- ↑ Steven R. Finch, July 7, 2003. Series-Parallel Networks
)=\mathrm {res} _{x}\left({\frac {f(x)}{x^{n+1}}}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/155afd6ed448a8f5c6606e3c4aca30f70ba9eb71)
![{\displaystyle \left\{{n \atop k}\right\}=\left[{-k \atop -n}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/acfe09dc1b5ff9e2769a4fc1dc89ad16ab36aeff)
![{\displaystyle {\begin{aligned}\left[{n \atop k}\right]&=[x^{k}]\left(\prod _{i=0}^{n-1}(i+x)\right)=\sum _{S\subseteq \{0..n-1\} \atop |S|=n-k}\left(\prod _{i\in S}(i)\right)\\&=\left[{\frac {x^{n}}{n!}}\right]\left({\frac {\log \left({\frac {1}{1-x}}\right)^{k}}{k!}}\right)={\frac {n!}{k!}}*\sum _{P\in \mathbb {N} ^{k} \atop \sum (P)=n}\left({\frac {1}{\prod _{b\in P}(b)}}\right)={\frac {n!}{k!}}*\sum _{b=0}^{n}\left({\frac {{\frac {(k-1)!}{(n-b)!}}*\left[{n-b \atop k-1}\right]}{b}}\right)=\sum _{b=0}^{n}\left({\frac {n^{\underline {b}}}{b*k}}*\left[{n-b \atop k-1}\right]\right)\\&=\left[{\frac {x^{n-k}}{k^{\overline {n-k}}}}\right]\left(\left({\frac {x}{1-e^{-x}}}\right)^{n}\right)=k^{\overline {n-k}}*\sum _{P\in \mathbb {N} ^{n} \atop \sum (P)=n-k}\left(\prod _{b\in P}\left({\frac {B_{b}^{+}}{b!}}\right)\right)=k^{\overline {n-k}}*\sum _{b=0}^{n-k}\left({\frac {B_{b}^{+}}{b!}}*{\frac {\left[{n-1 \atop k+b-1}\right]}{(k+b-1)^{\overline {n-k-b}}}}\right)={\frac {n-1}{k-1}}*\sum _{b=0}^{n-k}\left(B_{b}^{+}*{k+b-2 \choose b}*\left[{n-1 \atop k+b-1}\right]\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/c8dbb448938adcbc8860e608ef78cb5c6ac3bc88)
![{\displaystyle {\begin{aligned}\left\{{n \atop k}\right\}&=[x^{n-k}]\left(\prod _{i=1}^{k}\left({\frac {1}{1-i*x}}\right)\right)=\sum _{P\in \mathbb {N} ^{k} \atop \sum (P)=n-k}\left(\prod _{i=1}^{k}(i^{P_{i}})\right)=\sum _{b=0}^{n-k}\left(k^{b}*\left\{{n-1-b \atop k-1}\right\}\right)\\&=\left[{\frac {x^{n}}{n!}}\right]\left({\frac {(e^{x}-1)^{k}}{k!}}\right)={\frac {n!}{k!}}*\sum _{P\in \mathbb {N} ^{k} \atop \sum (P)=n}\left({\frac {1}{\prod _{b\in P}(b!)}}\right)={\frac {n!}{k!}}*\sum _{b=0}^{n}\left({\frac {{\frac {(k-1)!}{(n-b)!}}*\left\{{n-b \atop k-1}\right\}}{b!}}\right)=\sum _{b=0}^{n}\left({\frac {n \choose b}{k}}*\left\{{n-b \atop k-1}\right\}\right)\\&=\left[{\frac {x^{n-k}}{k^{\overline {n-k}}}}\right]\left(\left({\frac {x}{\log \left(1+x\right)}}\right)^{n}\right)=k^{\overline {n-k}}*\sum _{P\in \mathbb {N} ^{n} \atop \sum (P)=n-k}\left(\prod _{b\in P}(G_{b})\right)=k^{\overline {n-k}}*\sum _{b=0}^{n-k}\left(G_{b}*{\frac {\left\{{n-1 \atop k+b-1}\right\}}{(k+b-1)^{\overline {n-k-b}}}}\right)=(n-1)*\sum _{b=0}^{n-k}\left(G_{b}*k^{\overline {b-1}}*\left\{{n-1 \atop k+b-1}\right\}\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/032a15fb3695cd475461df133f1698fda7942136)
![{\displaystyle \left[{n \atop k}\right]={\frac {n}{k}}*\sum _{b=0}^{n-k}\left({\frac {B_{b}^{+}}{b!}}*k^{\overline {b}}*\left[{n-1 \atop k+b-1}\right]\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8bd36d78176d7a0fe8441642e3033b145fe8d6e5)
![{\displaystyle \left\{{k+j \atop j}\right\}=[x^{k}]\left(\prod _{i=1}^{j}\left({\frac {1}{1-i*x}}\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/8af8c2ba03a1f506d5b82453aaa85f334ca710ee)
![{\displaystyle [x^{k}]\left(\prod _{i=1}^{j}(1+i*x)\right)=\sum _{i=0}^{j}\left(\left[{j+1 \atop j+1-i}\right]*x^{i}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/69455ee93bc78d7830c41ad19e5cd3f1696e28d6)
![{\displaystyle \left\{{n \atop k}\right\}=\sum _{i=0}^{k-1}\left(\left[{k+1 \atop k-i}\right]*(-1)^{i}*\left\{{n-1-i \atop k}\right\}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b7e2c125ac3187298bbf750bcf382d1616b48044)
![{\displaystyle \left[{n+1 \atop k}\right]=n*\left[{n \atop k}\right]+\left[{n \atop k-1}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/78b855f76b4e18983dd9cefc870fb2cc2611264b)
![{\displaystyle s(n,k)={\frac {\left[{n+1 \atop k}\right]}{n!}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3df754a416166314328efb14b7a984d8e8049e1c)
![{\displaystyle {\begin{aligned}\left[{n \atop k+1}\right]&={\frac {(n-1)!}{k!}}*\sum _{i=0}^{k-1}\left((1-k)^{\overline {i}}*H_{n-1}^{(i+1)}*{\frac {(k-1-i)!}{(n-1)!}}*\left[{n \atop k-i}\right]\right)\\&={\frac {(n-1)!}{k!}}*\sum _{i=0}^{k-1}\left((-1)^{i}*(k-1)^{\underline {i}}*H_{n-1}^{(i+1)}*{\frac {(k-1-i)!}{(n-1)!}}*\left[{n \atop k-i}\right]\right)\\&={\frac {(n-1)!}{k!}}*\sum _{i=0}^{k-1}\left((-1)^{i}*{\frac {(k-1)!}{(k-1-i)!}}*H_{n-1}^{(i+1)}*{\frac {(k-1-i)!}{(n-1)!}}*\left[{n \atop k-i}\right]\right)\\&={\frac {1}{k}}*\sum _{i=0}^{k-1}\left((-1)^{i}*H_{n-1}^{(i+1)}*\left[{n \atop k-i}\right]\right)\\&={\frac {1}{k}}*\sum _{i=1}^{k}\left((-1)^{i-1}*H_{n-1}^{(i)}*\left[{n \atop k+1-i}\right]\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dabe59faadc7721fdb7bee671da878dd05777cd4)
![{\displaystyle \left[{n \atop k}\right]={\frac {1}{k-1}}*\sum _{i=0}^{k-2}\left(H_{n-1}^{(i+1)}*(-1)^{i}*\left[{n \atop k-1-i}\right]\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3dc0aa7fe877d7c387ba61898145ee6c85aba031)
![{\displaystyle {\begin{aligned}\left[{n+1 \atop k+1}\right]&=n!*\sum _{P\in \mathrm {orderedPartitions} (k)}\left(\prod _{i=0}^{\#(P)-1}\left({\frac {(-1)^{P_{i}-1}*H_{n}^{(P_{i})}}{k-\sum _{j=0}^{i-1}(P_{j})}}\right)\right)\\&=n!*\sum _{P\in \mathrm {orderedPartitions} (k)}\left((-1)^{k-\#(P)}*\prod _{i=0}^{\#(P)-1}\left({\frac {H_{n}^{(P_{i})}}{\sum _{j=0}^{i}(P_{j})}}\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d3103ef8cf33ae490d76900f4522dbd6f19b531d)
![{\displaystyle \left[{n+1 \atop k+1}\right]=n!*(-1)^{k}*\sum _{P\in \mathbb {N} ^{\infty } \atop \sum iP_{i}=k}\left(\prod _{i=0}^{\infty }\left({\frac {\left({\frac {-H_{n}^{(i)}}{i}}\right)^{P_{i}}}{P_{i}!}}\right)\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ee8061e87280d00822bd0be0e8caa8bcc18965cf)
![{\displaystyle \left\{{k+j \atop j}\right\}=[x^{j}]\left({\frac {\sum _{i=0}^{k}\left(\left\langle \left\langle {k \atop i}\right\rangle \right\rangle *x^{i}\right)}{(1-x)^{2*k+1}}}\right)}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/087efa1800f69f02126c2eeb6cabceec4f220357)
![{\displaystyle {\begin{aligned}&\Delta ^{n}\left(\lambda j.\left\{{k+j \atop j}\right\}\right)(0)\\=&[x^{k}]\left(\sum _{j=0}^{n}\left((-1)^{n-j}*{n \choose j}*\prod _{i=1}^{j}\left({\frac {1}{1-i*x}}\right)\right)\right)\\=&[x^{k}]\left(\sum _{j=0}^{n}\left((-1)^{n-j}*{n \choose j}*{\frac {\left({\frac {-1}{x}}\right)^{j}}{(1-{\frac {1}{x}})^{\overline {j}}}}\right)\right)\\=&(-1)^{n}*[x^{k}]\left(\sum _{j=0}^{n}\left({\frac {(-n)^{\overline {j}}}{(1-{\frac {1}{x}})^{\overline {j}}}}*{\frac {\left({\frac {-1}{x}}\right)^{j}}{j!}}\right)\right)\\=&(-1)^{n}*[x^{k}]\left({_{1}F_{1}}\left({-n \atop 1-{\frac {1}{x}}};{\frac {-1}{x}}\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6881a3f8f3138082c9dd535a61908f832f0e15b6)
![{\displaystyle {\begin{aligned}=&\mathrm {res} _{x}\left({\frac {(-1)^{n}*k!}{x^{k+1}}}*\sum _{j=0}^{n}\left({\frac {(-n)^{\overline {j}}*(k+1)^{\overline {j}}}{1^{\overline {j}}}}*{\frac {\left({\frac {e^{x}-1}{x}}\right)^{j}}{j!}}\right)\right)\\=&(-1)^{n}*\left[{\frac {x^{k}}{k!}}\right]\left({_{2}F_{1}}\left({-n,k+1 \atop 1};{\frac {e^{x}-1}{x}}\right)\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/1bd18cfca52bceee1fe2c106127788a625e68e18)
)\right)\right)\\=&(-1)^{n}*k!*\sum _{\pi \in \mathrm {partitions(\{1..n\})} }\left(\sum _{j=|\pi |}^{\infty }\left(\left[{\frac {x^{j}}{j!}}\right](f(x))*{\frac {1^{j-|\pi |}}{(j-|\pi |)!}}\right)*\prod _{b\in \pi }\left({\frac {1}{b+1}}\right)\right)\\=&(-1)^{n}*k!*\sum _{\pi \in \mathrm {partitions(\{1..n\})} }\left({\frac {\sum _{j=|\pi |}^{\infty }\left({\frac {(-1)^{j}*n!*(k+j)!}{k!*j!*(n-j)!*(j-|\pi |)!}}\right)}{\prod _{b\in \pi }(b+1)}}\right)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dbd9d1ba14dce76f26a0ead1c7a1a5714beac3d4)
![{\displaystyle \left[{n \atop k}\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/53e244be41a40ffd185e9edea632ba557f6ad945)
