A058878 -id:A058878 - cp's OEIS frontend

A340030 Triangle read by rows: T(n,k) is the number of hypergraphs on n labeled vertices with k edges and all vertices having even degree, 0 <= k < 2^n.

1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 7, 7, 0, 0, 1, 1, 0, 0, 35, 105, 168, 280, 435, 435, 280, 168, 105, 35, 0, 0, 1, 1, 0, 0, 155, 1085, 5208, 22568, 82615, 247845, 628680, 1383096, 2648919, 4414865, 6440560, 8280720, 9398115, 9398115, 8280720, 6440560, 4414865, 2648919, 1383096, 628680, 247845, 82615, 22568, 5208, 1085, 155, 0, 0, 1

Offset: 0

Andrew Howroyd, Jan 09 2021

Hypergraphs are graphs in which an edge is connected to a nonempty subset of vertices rather than exactly two of them. An edge is a nonempty subset of vertices.
Equivalently, T(n,k) is the number of subsets of {1..2^n-1} with k elements such that the bitwise-xor of the elements is zero.
Also the coefficients of polynomials p_{n}(x) which have the representation
  p_{n}(x) = (x + 1)^(2*(n - 1) - 1)*q_{n - 1}(x), where q_{n}(x) are the polynomials defined in A340263, and n >= 2. - Peter Luschny, Jan 10 2021

			Triangle begins:
[0]  1;
[1]  1, 0;
[2]  1, 0, 0,  1;
[3]  1, 0, 0,  7,   7,   0,   0,   1;
[4]  1, 0, 0, 35, 105, 168, 280, 435, 435, 280, 168, 105, 35, 0, 0, 1;

Andrew Howroyd, Table of n, a(n) for n = 0..2046 (rows 0..10)
Wikipedia, Hypergraph.

Rows 3..8 are A002394, A010085, A010086, A010087, A010088, A010089.
Row sums are A016031(n+1).
Column k=3 gives A006095.
Cf. A058878, A281123, A340312, A340263.

T(n,k) = {(binomial(2^n-1, k) + (-1)^((k+1)\2)*(2^n-1)*binomial(2^(n-1)-1, k\2))/2^n}
{ for(n=0, 5, print(vector(2^n, k, T(n,k-1)))) }

T(n,k) = (binomial(2^n-1, k) + (-1)^ceiling(k/2)*(2^n-1)*binomial(2^(n-1)-1, floor(k/2)))/2^n.
T(n,2*k) + T(n,2*k+1) = binomial(2^n-1, k)/2^n = A281123(n,k).
T(n, k) = T(n, 2^n-1-k) for n >= 2.

A033678 Number of labeled Eulerian graphs with n nodes.

Original entry on oeis.org

1, 0, 1, 3, 38, 720, 26614, 1858122, 250586792, 66121926720, 34442540326456, 35611003057733928, 73321307277341501168, 301201690357187097528960, 2471354321681605983102370864, 40525241311304939167532163726672

Offset: 1

N. J. A. Sloane and Geoffrey Mess (mess(AT)math.ucla.edu)

From Petros Hadjicostas, Feb 20 2021: (Start)
See the comments for A058878 about the different (and sometimes confusing) terminology regarding even and (connected or not) Euler graphs.
Cao (2002) uses the term "connected labeled Eulerian graphs" in the title of his Section 4.3, where this sequence appears, and the term "labeled Eulerian graph" in some of the discussion of that section. The author does cite the definition of Harary and Palmer (1973) for an Euler or Eulerian graph (as a connected even graph).
Note that all graphs counted by this sequence, by A058878, and by the triangular arrays A228550 and A341743 are assumed to be simple (with no loops and no multiple edges). Read (1962), however, indicates how to solve similar counting problems in the case of graphs with loops and/or multiple edges. (End)

F. Harary and E. Palmer, Graphical Enumeration, (1973), p. 12, Eq. (1.4.6).
E. M. Palmer in L. W. Beineke and R. J. Wilson, Selected Topics in Graph Theory, Academic Press, NY, 1978, p. 385ff.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=1..50
Huantian Cao, AutoGF: An Automated System to Calculate Coefficients of Generating Functions, thesis, 2002.
Huantian Cao, AutoGF: An Automated System to Calculate Coefficients of Generating Functions, thesis, 2002 [Local copy, with permission]
math.stackexchange.com, Is it possible disconnected graph has euler circuit? [sic], August 30, 2015.
Ronald C. Read, Euler graphs on labelled nodes, Canadian Journal of Mathematics, 14 (1962), 482-486; see the discussion in Section 4, following Eq. (8) on p. 486.

Cf. A228550 (with multiple components)

Row sums of A341743.

A033678 := proc(n) option remember; local k; if n=1 then 1 else 2^binomial(n-1,2)-(1/n)*add(k*binomial(n,k)*2^binomial(n-k-1,2)*A033678(k),k=1..n-1); fi; end;

n = 16; (Series[ Log[ 1 + Sum[ 2^( (p-1)(p-2)/2 )x^p/(p!), {p, 1, n} ] ], {x, 0, n} ] // CoefficientList[#, x]& // Rest) * Range[n]! (* truncated exponential generating function *)
(* Second program: *)
a[n_] := a[n] = If[n == 1, 1, 2^Binomial[n-1, 2]-(1/n)*Sum[k*Binomial[n, k]*2^Binomial[n-k-1, 2]*a[k], {k, 1, n-1}]]; Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Feb 11 2014, after Maple *)

@cached_function
def A033678(n):
    if n == 1: return 1
    return 2^binomial(n-1, 2)-sum(k*2^((k-n+1)*(k-n+2)/2)*binomial(n,k)*A033678(k) for k in (1..n-1))/n
[A033678(n) for n in (1..16)] # Peter Luschny, Jan 17 2016

A340312 Triangle read by rows: T(n,k) is the number of subsets of {0..2^n-1} with k elements such that the bitwise-xor of all the subset members gives zero, 0 <= k <= 2^n.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 7, 14, 7, 0, 1, 1, 1, 1, 0, 35, 140, 273, 448, 715, 870, 715, 448, 273, 140, 35, 0, 1, 1, 1, 1, 0, 155, 1240, 6293, 27776, 105183, 330460, 876525, 2011776, 4032015, 7063784, 10855425, 14721280, 17678835, 18796230, 17678835, 14721280, 10855425, 7063784, 4032015, 2011776, 876525, 330460, 105183, 27776, 6293, 1240, 155, 0, 1, 1

Offset: 0

Jianing Song, Jan 04 2021

Sum_{k=0..2^n} T(n, k) gives the total number of subsets with bitwise-xor of all the subset members zero. There are in total 2^(2^n - n) such subsets of {0, 1, ..., 2^n-1}, see A300361 and the Mathematics Stack Exchange link below.
Equivalently, T(n, k) is the number of subsets of the vector space (F_2)^n such that the sum of elements in the subset is the zero vector.
T(n, k) is symmetric, that is, T(n, k) = T(n, 2^n-k) for k = 0..2^n, since if the bitwise-xor of the members in S is zero, then the complement of S in {0, 1, ..., 2^n-1} also has this property.

			Triangle begins:
[0]  1, 1;
[1]  1, 1, 0;
[2]  1, 1, 0, 1, 1;
[3]  1, 1, 0, 7, 14, 7, 0, 1, 1;
[4]  1, 1, 0, 35, 140, 273, 448, 715, 870, 715, 448, 273, 140, 35, 0, 1, 1;
[5]  1, 1, 0, 155, 1240, 6293, 27776, 105183, 330460, 876525, 2011776, 4032015, 7063784, 10855425, 14721280, 17678835, 18796230, 17678835, 14721280, 10855425, 7063784, 4032015, 2011776, 876525, 330460, 105183, 27776, 6293, 1240, 155, 0, 1, 1;
T(n,0) = 1 since the bitwise-xor of all the elements in the empty set is the identity of bitwise-xor (0), hence the empty set meets the requirement.
T(n,1) = 1 since the only such subset is {0}.
T(n,2) = 0 since no distinct a, b have a ^ b = 0.
T(n,3) = A006095(n): if distinct a, b, c have a ^ b ^ c = 0, then c = a ^ b, and a, b must both be nonzero since a = 0 implies b = c. On the other hand, if a, b are nonequal and are both nonzero, then c = a ^ b has c != a and c != b since c = a implies b = 0. So the total number of triples (a, b, c) is (2^n-1)*(2^n-2), and the total number of subsets {a, b, c} is (2^n-1)*(2^n-2)/3! = A006095(n).
T(n,4) = A016290(n-2): if distinct a, b, c, d have a ^ b ^ c ^ d = 0, then d = a ^ b ^ c. On the other hand, if a, b, c are distinct, then d = a ^ b ^ c has d != a, d != b, d != c since d = a implies b = c. So the total number of quadruples (a, b, c, d) is 2^n*(2^n-1)*(2^n-2), and the total number of subsets {a, b, c, d} is 2^n*(2^n-1)*(2^n-2)/4! = A016290(n-2).

Peter Luschny, Table of n, a(n) for n = 0..1032
Katrina Honigs and Graham McDonald, Theta characteristics and the fixed locus of [-1] on some varieties of Kummer type, arXiv:2307.13129 [math.AG], 2023. See pp. 6 and 22.
Mathematics Stack Exchange, Count subsets with zero sum of xors
Jianing Song, C Program for A340312, case n = 4

Cf. A000120 (hamming weight of n), A300361 (row sums).
Cf. A340263 (irreducible (?) factor if T(n,k) is seen as representing polynomials).
Cf. A340259(n) = T(n, 2^(n-1)), the central term of row n.
Cf. A340030 (case that only nonzero elements allowed).
Cf. A054669, A058878.
Cf. A006095 (k=3 column), A016290 (k=4 column); cf. also A010080-A010084 and A281123. - Jon E. Schoenfield, Jan 06 2021

/* Generating program for T(4,k), see link. */

A340312_row := proc(n) local a, b, c; c := 2^(n-1);
if n = 0 then return [1, 1] fi;
b := n -> add(binomial(2^n, 2*k)*x^(2*k), k = 0..2^n);
a := n -> x*mul(b(k), k = 0..n);
(x + 1)^c*(b(n-1) - (c-1)*a(n-2));
[seq(coeff(expand(%), x, j), j = 0..2*c)] end:
for n from 0 to 6 do A340312_row(n) od; # Peter Luschny, Jan 06 2021

T[n_, k_] := Binomial[2^n, k]/2^n + If[EvenQ[k], (-1)^(k/2)*(1-1/2^n)* Binomial[2^(n-1), k/2], 0];
Table[T[n, k], {n, 0, 5}, {k, 0, 2^n}] // Flatten (* Jean-François Alcover, Jan 14 2021, after Andrew Howroyd *)

T(n, k)={binomial(2^n, k)/2^n + if(k%2==0, (-1)^(k/2)*(1-1/2^n)*binomial(2^(n-1), k/2))} \\ Andrew Howroyd, Jan 09 2021

def A340312():
    a, b, c = 1, 1, 1
    yield [1, 1]
    yield [1, 1, 0]
    while True:
        c *= 2
        a *= b
        b = sum(binomial(c, 2 * k) * x^(2 * k) for k in range(c + 1))
        p = (x + 1)^c * (b - (c - 1) * x * a)
        yield expand(p).list()
A340312_row = A340312()
for _ in range(6):
    print(next(A340312_row)) # Peter Luschny, Jan 07 2021

T(n, k) = [x^k] p(n; x) where p(n; x) = (x + 1)^c*(b(n-1) - (c-1)*a(n-2)), b(n) = Sum_{k=0..2^n} binomial(2^n, 2*k)*x^(2*k), a(n) = x*Product_{k=0..n} b(k) and c = 2^(n-1), for n >= 1. - Peter Luschny, Jan 06 2021
T(n+1, k) = [x^k] (x+1)^(2^n)*p_n(x) where p_n(x) are the polynomials defined in A340263. - Peter Luschny, Jan 06 2021
From Andrew Howroyd, Jan 09 2021: (Start)
First take any subset of k-1 elements and append the bitwise-xor of the elements. The final element will either be a duplicate or not and consideration of the two cases leads to a formula linking T(n,k) and T(n,k-2) with binomial(2^n,k-1).
T(n, k) = (1/k)*(binomial(2^n,k-1) - (2^n-(k-2))*T(n,k-2)) for k >= 2.
T(n, k) = binomial(2^n, k)/2^n for odd k.
T(n, k) = binomial(2^n, k)/2^n + (-1)^(k/2)*(1-1/2^n)*binomial(2^(n-1), k/2) for even k.
T(n, k) = [x^k] ((1+x)^(2^n) + (2^n-1)*(1-x^2)^(2^(n-1)))/2^n.
T(n, k) = A340030(n,k-1) + A340030(n,k).
(End)

More terms from Andrew Howroyd and Jon E. Schoenfield.

A054669 Triangular array T(n,k) giving the number of labeled even graphs with n nodes and k edges for n >= 0 and 0 <= k <= n*(n-1-[0 == n mod 2])/2 (with no trailing zeros).

Original entry on oeis.org

1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 4, 3, 1, 0, 0, 10, 15, 12, 15, 10, 0, 0, 1, 1, 0, 0, 20, 45, 72, 160, 240, 195, 120, 96, 60, 15, 1, 0, 0, 35, 105, 252, 805, 1935, 3255, 4515, 5481, 5481, 4515, 3255, 1935, 805, 252, 105, 35, 0, 0, 1, 1, 0, 0, 56, 210, 672, 2800, 9320, 24087

Offset: 0

Vladeta Jovovic, Apr 18 2000

From Petros Hadjicostas, Feb 18 2021: (Start)
This is the same as irregular array A058878 but without the (n/2)*[0 == n mod 2] trailing zeros at the end of each row n.
For n odd, we have T(n,n*(n-1)/2) = 1 at end of row n because a complete graph with n nodes and n*(n-1)/2 edges is even and has only one labeling.
For n even, the maximum number of edges an even graph with n nodes can have is n*(n-2)/2 = n*(n-1)/2 - n/2. We have T(n,n*(n-2)/2) = A001147(n/2) because each labeling of an even graph that has n nodes and n*(n-2)/2 edges corresponds to a perfect matching of the complete graph with n vertices (by considering the pairs of vertices that are not connected).
For a discussion about the confusion in defining Euler graphs, see the comments for A058878. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n*(n-1-[0==n mod 2])/2) begins:
  1;
  1;
  1;
  1, 0, 0,  1;
  1, 0, 0,  4,  3;
  1, 0, 0, 10, 15, 12,  15,  10,   0,   0,  1;
  1, 0, 0, 20, 45, 72, 160, 240, 195, 120, 96, 60, 15;
  ...

F. Harary and E. Palmer, Graphical Enumeration, Academic Press, New York, 1973; pp. 13-15.

Andrew Howroyd, Table of n, a(n) for n = 0..1295
P. J. Cameron, Cohomological aspects of two-graphs, Math. Zeit., 157 (1977), 101-119. [See pp. 116-117, where he uses the informal term "labelled Eulerian graph" for "labelled even graph".]
math.stackexchange.com, Is it possible disconnected graph has euler circuit? [sic], August 30, 2015.
Ronald C. Read, Euler graphs on labelled nodes, Canadian Journal of Mathematics, 14 (1962), 482-486.

Row sums are A006125(n-1).
Essentially the same table as A058878.
Cf. A001147, A033678, A340312.

CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
w := p -> factor(2^(-p)*(1 + x)^(p*(p - 1)/2)*
          add(binomial(p, n)*((1 - x)/(1 + x))^(n*(p - n)), n=0..p)):
seq(print(CoeffList(w(n))), n = 0..6); # Peter Luschny, Feb 18 2021

T[n_] := (1/2^n)(1+x)^Binomial[n, 2] Sum[Binomial[n, k] ((1-x)/(1+x))^(k(n-k)), {k, 0, n}] // CoefficientList[#, x]&;
T /@ Range[0, 8] // Flatten (* Jean-François Alcover, Feb 20 2021, after Andrew Howroyd *)

Row(n)=Vecrev(2^(-n)*(1+x)^binomial(n, 2)*sum(k=0, n, binomial(n, k)*((1-x)/(1+x))^(k*(n-k)))) \\ Andrew Howroyd, Jan 05 2021

T(n,k) = [y^k] 2^(-n)*(1+y)^C(n, 2)*Sum_{s=0..n} C(n, s)*((1-y)/(1+y))^(s*(n-s)).
From Petros Hadjicostas, Feb 18 2021: (Start)
T(n,k) = (1/2^n) * Sum_{s=0..n} binomial(n,s) * Sum_{t=0..k} (-1)^t * binomial(s*(n-s), t) * binomial(binomial(s,2) + binomial(n-s, 2), k-t).
T(n, n*(n-1)/2) = 1 if n is odd.
T(n, n*(n-2)/2) = A001147(n/2) if n is even.
T(n,k) = A058878(n,k) for n >= 0 and 0 <= k <= n*(n-1-[0 == n mod 2])/2. (End)

T(0,0) = 1 prepended by Andrew Howroyd, Jan 09 2021

Name edited by Petros Hadjicostas, Feb 18 2021

A341743 T(n,k) is the number of labeled Eulerian graphs with n vertices and k edges (according to Harary and Palmer) or the number of labeled connected Eulerian graphs with n vertices and k edges (according to Mallows and Sloane); irregular triangle T, read by rows (n >= 0 and 0 <= k <= n*(n-1)/2).

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 12, 15, 10, 0, 0, 1, 0, 0, 0, 0, 0, 0, 60, 180, 195, 120, 90, 60, 15, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 360, 1890, 3675, 4830, 5061, 4410, 3255, 1935, 805, 252, 105, 35, 0, 0, 1

Offset: 0

Petros Hadjicostas, Feb 18 2021

The value T(0,0) = 0 has no physical meaning. It is there because it makes the formula for the bivariate e.g.f.-o.g.f. (shown below) work.
Since T(n,k) counts even connected graphs with n vertices and k edges, for n >= 2, each vertex must have at least two edges, so k >= n. Hence, T(n,k) = 0 for 0 <= k < n.
We have T(n,n) = (n-1)!/2 for n >= 3 because T(n,n) counts the different labelings of cyclic graphs with n vertices and n edges, and we have (n-1)! cyclic permutations of the numbers 1, 2, ..., n. We divide by 2 because we get the same labeling if we flip the cyclic graph over (like a bracelet).
We have T(n, n*(n-1)/2) = 1, if n is odd, because the complete graph on n vertices is even (each vertex has degree n-1) and has only one non-isomorphic labeling.
We have T(n, n*(n-1)/2 - s) = 0 for s = 0, 1, 2, ..., (n/2)-1, if n is even, because in an even graph with n vertices we cannot have more than n*(n-2)/2 = n*(n-1)/2 - n/2 edges.
Finally, we have T(n, n*(n-2)/2) = A001147(n/2), if n is even >= 4, because any labeling in an even graph with n vertices and n*(n-2)/2 edges corresponds to a perfect matching in a complete graph with n vertices (by considering the pairs of vertices that are not connected).
See the comments for A058878 about the different (and sometimes confusing) terminology regarding even and (connected or not) Euler graphs.

			Irregular triangle T(n,k) (with rows n >= 0 and columns k = 0..n*(n-1)/2) begins
  0;
  1;
  0, 0;
  0, 0, 0, 1;
  0, 0, 0, 0, 3,  0,  0;
  0, 0, 0, 0, 0, 12, 15,  10,   0,   0,  1;
  0, 0, 0, 0, 0,  0, 60, 180, 195, 120, 90, 60, 15, 0, 0, 0;
  ...
T(5,5) = 12 because we have (5-1)!/2 = 12 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 5 edges:
        *
       /  \
      /    \
     /      \
    *        *
     \      /
      \    /
       *--*
T(5,6) = 15 because we have 5*3 = 15 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 6 edges:
         *______*
        /|\    /
       / | \  /
      *  |  \/
       \ |  *
        \|
         *
In the above graph, we have 5 choices for the vertex that is common to both triangles and using the other 4 numbers 1, 2, 3, 4 we have the following 3 possible labelings of the other 4 vertices: {{1,2},{3,4}}, {{1,3},{2,4}}, {{1,4},{2,3}}.
T(5,7) = 10 because we have C(5,2) = 10 non-isomorphic labelings of the following Eulerian graph with 5 vertices and 7 edges:
V = {a,b,c,d,e} and E = {{a,b}, {a,c}, {a,d}, {a,e}, {b,c}, {b,d}, {b,e}}.
T(5,10) = 1 because all labelings of the complete graph with 5 vertices (and C(5,2) = 10 edges) are isomorphic.
There are no other (unlabeled) Eulerian graphs with 5 vertices: A003049(5) = 4. (In the name of A003049, the phrase "connected Euler graphs" is according to Mallows and Sloane (1975). According to Harary and Palmer (1973), we only need to say "Euler graphs" because, for them, an Euler graph is connected and even.)

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973; see Eqs. (1.4.7), (1.4.18), and (1.4.19) on pp. 11-16.

P. J. Cameron, Cohomological aspects of two-graphs, Math. Zeit., 157 (1977), 101-119.
C. L. Mallows and N. J. A. Sloane (1975), Two-graphs, switching classes and Euler graphs are equal in number, SIAM Journal on Applied Mathematics, 28(4) (1975), 876-880.
math.stackexchange.com, Is it possible disconnected graph has euler circuit? [sic], August 30, 2015.
Ronald C. Read, Euler graphs on labelled nodes, Canadian Journal of Mathematics, 14 (1962), 482-486; see the discussion in Section 4, following Eq. (8) on p. 486.

Cf. A001147, A001710, A003049, A033678, A058878, A062199.

# Slow program based on Eqs. (1.4.7), (1.4.18), and (1.4.19) in Harary and Palmer (1973).
w := proc(n, y) local m: expand(simplify(2^(-n)*(y + 1)^(1/2*n*(n - 1))*add(binomial(n, m)*((1 - y)/(y + 1))^(m*(n - m)), m = 0 .. n))): end proc:
u := proc(x, y, M) local n: add(w(n, y)*x^n/n!, n = 0 .. M): end proc:
T := proc(n, k) coeftayl(log(u(x, y, n + 2)), [x, y] = [0, 0], [n, k])*n!: end proc:
# Another, slightly faster, program based on one of the recurrences:
S := proc(n, k) local s, t: add(binomial(n, s)*add((-1)^t*binomial(s*(n - s), t)*binomial(binomial(s, 2) + binomial(n - s, 2), k - t), t = 0 .. k), s = 0 .. n)/2^n: end proc: # A058878
T := proc(n, k) local x, s, t: option remember: if n = 0 then x := 0: end if: if 1/2*n*(n - 1) < k then x := 0: end if: if 1 <= n and 0 <= k and k <= 1/2*n*(n - 1) then x := S(n, k) - add(add(binomial(n - 1, s)*T(s + 1, t)*S(n - 1 - s, k - t), t = 0 .. k), s = 0 .. n - 2): end if: x: end proc:
# Third program based on another recurrence (the S(n,k) is as above):
T1 := proc(n, k) local x, s, t: option remember: if k = 0 and (n = 0 or 2 <= n) then x := 0: end if: if n = 1 and k = 0 then x := 1: end if; if 1/2*n*(n - 1) < k then x := 0: end if: if 2 <= n and 1 <= k and k <= 1/2*n*(n - 1) then x := S(n, k) - add(add((t + 1)*binomial(n, s)*T1(s, t + 1)*S(n - s, k - 1 - t)/k, t = 0 .. k - 2), s = 0 .. n) - add(binomial(n, s)*T1(s, k), s = 0 .. n - 1): end if: x: end proc:

S[n_, k_] := S[n, k] = Sum[Binomial[n, s]*Sum[(-1)^t* Binomial[s*(n-s), t]*Binomial[Binomial[s, 2] + Binomial[n-s, 2], k-t], {t, 0, k}], {s, 0, n}]/2^n;
T[n_, k_] := T[n, k] = If[n == 0 || k > n(n-1)/2, 0, S[n, k] - Sum[Binomial[n-1, s]*T[s+1, t]*S[n-1-s, k-t], {t, 0, k}, {s, 0, n-2}]];
Table[T[n, k], {n, 0, 8}, {k, 0, n(n-1)/2}] // Flatten (* Jean-François Alcover, Feb 14 2023, after 2nd Maple program *)

Sum_{k=0..n} T(n,k) = A033678(n) for n >= 1.
Bivariate e.g.f.-o.g.f.: Sum_{n,k>=0} T(n,k)*(x^n/n!)*y^k = log(Sum_{n,k>=0} A058878(n,k)*(x^n/n!)*y^k) = log(Sum_{n >= 0} (x^n/n!)*[o.g.f. of n-th row of A058878](y)).
Sum_{s=0..n} Sum_{t=0..k} binomial(n,s) * T(s+1,t) * A058878(n-s,k-t) = A058878(n+1,k) for n >= 0 and 0 <= k <= n*(n+1)/2.
Sum_{s=0..n} Sum_{t=0..k} ((t+1)/(k+1)) * binomial(n,s) * T(s,t+1) * A058878(n-s,k-t) = A058878(n,k+1) for n >= 2 and 0 <= k <= n*(n-1)/2 - 1
T(n,k) = A058878(n,k) - Sum_{s=0..n-2} Sum_{t=0..k} binomial(n-1,s) * T(s+1,t) * A058878(n-1-s,k-t) for n >= 1 and 0 <= k <= n*(n-1)/2, and T(n,k) = 0 otherwise.
T(n,k) = A058878(n,k) - Sum_{s=0..n} Sum_{t=0..k-2} ((t+1)/(k+1)) * binomial(n,s) * T(s,t+1) * A058878(n-s,k-1-t) - Sum_{s=0..n-1} binomial(n,s) * T(s,k) for n >= 2 and 1 <= k <= n*(n-1)/2 (with T(1,0) = 1 and T(n,k) = 0 otherwise).
T(n,k) = 0 for n >= 2 and 0 <= k <= n-1.
T(n,n) = A001710(n-1) = (n-1)!/2 for n >= 3.
Conjecture: T(n,n+1) = n!*(n-4)/8 = 15*A062199(n-5) for n >= 4 (with A062199(-1) = 0).
T(n, n*(n-1)/2) = 1 if n is odd.
T(n, k) = 0 if n is even and n*(n-1)/2 - n/2 + 1 <= k < n*(n-1)/2.
T(n, n*(n-2)/2) = A001147(n/2) if n is even >= 4.

cp's OEIS Frontend

A340030 Triangle read by rows: T(n,k) is the number of hypergraphs on n labeled vertices with k edges and all vertices having even degree, 0 <= k < 2^n.

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A033678 Number of labeled Eulerian graphs with n nodes.

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A340312 Triangle read by rows: T(n,k) is the number of subsets of {0..2^n-1} with k elements such that the bitwise-xor of all the subset members gives zero, 0 <= k <= 2^n.

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A054669 Triangular array T(n,k) giving the number of labeled even graphs with n nodes and k edges for n >= 0 and 0 <= k <= n*(n-1-[0 == n mod 2])/2 (with no trailing zeros).

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A341743 T(n,k) is the number of labeled Eulerian graphs with n vertices and k edges (according to Harary and Palmer) or the number of labeled connected Eulerian graphs with n vertices and k edges (according to Mallows and Sloane); irregular triangle T, read by rows (n >= 0 and 0 <= k <= n*(n-1)/2).

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