cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A003024 Number of acyclic digraphs (or DAGs) with n labeled nodes.

Original entry on oeis.org

1, 1, 3, 25, 543, 29281, 3781503, 1138779265, 783702329343, 1213442454842881, 4175098976430598143, 31603459396418917607425, 521939651343829405020504063, 18676600744432035186664816926721, 1439428141044398334941790719839535103, 237725265553410354992180218286376719253505
Offset: 0

Views

Author

Keywords

Comments

Also the number of n X n real (0,1)-matrices with all eigenvalues positive. - Conjectured by Eric W. Weisstein, Jul 10 2003 and proved by McKay et al. 2003, 2004
Also the number of n X n real (0,1)-matrices with permanent equal to 1, up to permutation of rows/columns, cf. A089482. - Vladeta Jovovic, Oct 28 2009
Also the number of nilpotent elements in the semigroup of binary relations on [n]. - Geoffrey Critzer, May 26 2022
From Gus Wiseman, Jan 01 2024: (Start)
Also the number of sets of n nonempty subsets of {1..n} such that there is a unique way to choose a different element from each. For example, non-isomorphic representatives of the a(3) = 25 set-systems are:
{{1},{2},{3}}
{{1},{2},{1,3}}
{{1},{2},{1,2,3}}
{{1},{1,2},{1,3}}
{{1},{1,2},{2,3}}
{{1},{1,2},{1,2,3}}
These set-systems have ranks A367908, subset of A367906, for multisets A368101.
The version for no ways is A368600, any length A367903, ranks A367907.
The version for at least one way is A368601, any length A367902.
(End)

Examples

			For n = 2 the three (0,1)-matrices are {{{1, 0}, {0, 1}}, {{1, 0}, {1, 1}}, {{1, 1}, {0, 1}}}.
		

References

  • Archer, K., Gessel, I. M., Graves, C., & Liang, X. (2020). Counting acyclic and strong digraphs by descents. Discrete Mathematics, 343(11), 112041.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 310.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 19, Eq. (1.6.1).
  • R. W. Robinson, Counting labeled acyclic digraphs, pp. 239-273 of F. Harary, editor, New Directions in the Theory of Graphs. Academic Press, NY, 1973.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P Stanley, Enumerative Combinatorics I, 2nd. ed., p. 322.

Crossrefs

Cf. A086510, A081064 (refined by # arcs), A307049 (by # descents).
Cf. A055165, which counts nonsingular {0, 1} matrices and A085656, which counts positive definite {0, 1} matrices.
Cf. A188457, A135079, A137435 (acyclic 3-multidigraphs), A188490.
For a unique sink we have A003025.
The unlabeled version is A003087.
These are the reverse-alternating sums of rows of A046860.
The weakly connected case is A082402.
A reciprocal version is A334282.
Row sums of A361718.

Programs

  • Maple
    p:=evalf(solve(sum((-1)^n*x^n/(n!*2^(n*(n-1)/2)), n=0..infinity) = 0, x), 50); M:=evalf(sum((-1)^(n+1)*p^n/((n-1)!*2^(n*(n-1)/2)), n=1..infinity), 40); # program for evaluation of constants p and M in the asymptotic formula, Vaclav Kotesovec, Dec 09 2013
  • Mathematica
    a[0] = a[1] = 1; a[n_] := a[n] = Sum[ -(-1)^k * Binomial[n, k] * 2^(k*(n-k)) * a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 13}](* Jean-François Alcover, May 21 2012, after PARI *)
    Table[2^(n*(n-1)/2)*n! * SeriesCoefficient[1/Sum[(-1)^k*x^k/k!/2^(k*(k-1)/2),{k,0,n}],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 19 2015 *)
    Table[Length[Select[Subsets[Subsets[Range[n]],{n}],Length[Select[Tuples[#],UnsameQ@@#&]]==1&]],{n,0,5}] (* Gus Wiseman, Jan 01 2024 *)
  • PARI
    a(n)=if(n<1,n==0,sum(k=1,n,-(-1)^k*binomial(n,k)*2^(k*(n-k))*a(n-k)))
    
  • PARI
    {a(n)=polcoeff(1-sum(k=0, n-1, a(k)*x^k/(1+2^k*x+x*O(x^n))^(k+1)), n)} \\ Paul D. Hanna, Oct 17 2009

Formula

a(0) = 1; for n > 0, a(n) = Sum_{k=1..n} (-1)^(k+1)*C(n, k)*2^(k*(n-k))*a(n-k).
1 = Sum_{n>=0} a(n)*exp(-2^n*x)*x^n/n!. - Vladeta Jovovic, Jun 05 2005
a(n) = Sum_{k=1..n} (-1)^(n-k)*A046860(n,k) = Sum_{k=1..n} (-1)^(n-k)*k!*A058843(n,k). - Vladeta Jovovic, Jun 20 2008
1 = Sum_{n=>0} a(n)*x^n/(1 + 2^n*x)^(n+1). - Paul D. Hanna, Oct 17 2009
1 = Sum_{n>=0} a(n)*C(n+m-1,n)*x^n/(1 + 2^n*x)^(n+m) for m>=1. - Paul D. Hanna, Apr 01 2011
log(1+x) = Sum_{n>=1} a(n)*(x^n/n)/(1 + 2^n*x)^n. - Paul D. Hanna, Apr 01 2011
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is 1/E(-x) = Sum_{n >= 0} a(n)*x^n/(n!*2^C(n,2)) = 1 + x + 3*x^2/(2!*2) + 25*x^3/(3!*2^3) + 543*x^4/(4!*2^6) + ... (Stanley). Cf. A188457. - Peter Bala, Apr 01 2013
a(n) ~ n!*2^(n*(n-1)/2)/(M*p^n), where p = 1.488078545599710294656246... is the root of the equation Sum_{n>=0} (-1)^n*p^n/(n!*2^(n*(n-1)/2)) = 0, and M = Sum_{n>=1} (-1)^(n+1)*p^n/((n-1)!*2^(n*(n-1)/2)) = 0.57436237330931147691667... Both references to the article "Acyclic digraphs and eigenvalues of (0,1)-matrices" give the wrong value M=0.474! - Vaclav Kotesovec, Dec 09 2013 [Response from N. J. A. Sloane, Dec 11 2013: The value 0.474 has a typo, it should have been 0.574. The value was taken from Stanley's 1973 paper.]
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 10*x^3 + 146*x^4 + 6010*x^5 + ... appears to have integer coefficients (cf. A188490). - Peter Bala, Jan 14 2016

A188457 G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1 + 3^n*x)^(n+1).

Original entry on oeis.org

1, 1, 5, 109, 9449, 3068281, 3586048685, 14668583277349, 205716978569685329, 9737002299093315531121, 1536239893108209683958428885, 799846636937376803320381186364509, 1362900713950636674946135205457794784569
Offset: 0

Views

Author

Paul D. Hanna, Mar 31 2011

Keywords

Comments

G.f. satisfies a variant of an identity involving A003024:
1 = Sum_{n>=0} A003024(n)*x^n/(1 + 2^n*x)^(n+1),
where A003024(n) is the number of acyclic digraphs with n labeled nodes.
a(n) is the number of acyclic 2-multidigraphs. Cf. A137435, A339768. - Geoffrey Critzer, Feb 21 2021

Examples

			Illustration of the generating functions.
E.g.f.: 1 = exp(-x) + exp(-3*x)*x + 5*exp(-9*x)*x^2/2! + 109*exp(-27*x)*x^3/3! +...
L.g.f.: log(1+x) = x/(1+3*x) + 5*(x^2/2)/(1+9*x)^2 + 109*(x^3/3)/(1+27*x)^3 +...
G.f.: 1 = 1/(1+x) + 1*x/(1+3*x)^2 + 5*x^2/(1+9*x)^3 + 109*x^3/(1+27*x)^4 +...
G.f.: 1 = 1/(1+x)^2 + 1*2*x/(1+3*x)^3 + 5*3*x^2/(1+9*x)^4 + 109*4*x^3/(1+27*x)^5 +...
G.f.: 1 = 1/(1+x)^3 + 1*3*x/(1+3*x)^4 + 5*6*x^2/(1+9*x)^5 + 109*10*x^3/(1+27*x)^6 +...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = If[n == 0, 1, Sum[-(-1)^k Binomial[n, k] 3^(k(n-k)) a[n-k], {k, 1, n}]];
    a /@ Range[0, 12] (* Jean-François Alcover, Nov 02 2019 *)
  • PARI
    {a(n)=polcoeff(1-sum(k=0, n-1, a(k)*x^k/(1+3^k*x+x*O(x^n))^(k+1)), n)}
    for(n=0,20, print1(a(n),", "))
    
  • PARI
    /* Holds for m>=1: */
    {a(n)=local(m=1);polcoeff(1-sum(k=0, n-1, a(k)*binomial(m+k-1,k)*x^k/(1+3^k*x+x*O(x^n))^(k+m)), n)/binomial(m+n-1,n)}
    for(n=0,20, print1(a(n),", "))
    
  • PARI
    /* Recurrence: */
    {a(n)=if(n<1, n==0, sum(k=1, n, -(-1)^k*binomial(n, k)*3^(k*(n-k))*a(n-k)))}
    for(n=0,20, print1(a(n),", "))
    
  • PARI
    /* E.g.f.: */
    {a(n)=n!*polcoeff(1-sum(k=0, n-1, a(k)*exp(-3^k*x+x*O(x^n))*x^k/k!), n)}
    for(n=0,20, print1(a(n),", "))

Formula

G.f.: 1 = Sum_{n>=0} a(n)*C(n+m-1,n)*x^n/(1 + 3^n*x)^(n+m) for m>=1.
L.g.f.: log(1+x) = Sum_{n>=1} a(n)*(x^n/n)/(1 + 3^n*x)^n.
E.g.f.: 1 = Sum_{n>=0} a(n)*exp(-3^n*x)*x^n/n!.
a(n) = Sum_{k=1..n} (-1)^(k+1)*C(n, k)*3^(k*(n-k))*a(n-k) for n>0 with a(0)=1.
From Peter Bala, Apr 01 2013: (Start)
Let E(x) = sum {n >= 0} x^n/(n!*3^C(n,2)). Then a generating function for this sequence is 1/E(-x) = sum {n >= 0} a(n)*x^n/(n!*3^C(n,2)) = 1 + x + 5*x^2/(2!*3) + 109*x^3/(3!*3^3) + 9449*x^4/(4!*3^6) + .... (End)
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 39*x^3 + 2403*x^4 + 616131*x^5 + ... appears to have integer coefficients. - Peter Bala, Jan 14 2016

A339768 Square array read by descending antidiagonals. T(n,k) is the number of acyclic k-multidigraphs on n labeled vertices, n>=0,k>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 25, 1, 1, 1, 7, 109, 543, 1, 1, 1, 9, 289, 9449, 29281, 1, 1, 1, 11, 601, 63487, 3068281, 3781503, 1, 1, 1, 13, 1081, 267249, 69711361, 3586048685, 1138779265, 1
Offset: 0

Views

Author

Geoffrey Critzer, Feb 21 2021

Keywords

Comments

Here, a k-multidigraph is a directed graph where up to k arcs (directed edges) are allowed to join vertex pairs. The arcs have no identity, i.e., they are indistinguishable except for the ordered pair of distinct vertices that they join.

Examples

			  1,     1,       1,        1,         1,          1, ...
  1,     1,       1,        1,         1,          1, ...
  1,     3,       5,        7,         9,         11, ...
  1,    25,     109,      289,       601,       1081, ...
  1,   543,    9449,    63487,    267249,     849311, ...
  1, 29281, 3068281, 69711361, 742650001, 5004309601, ...
		

Crossrefs

Cf. A003024 (column k=1), A188457 (column k=2), A137435 (column k=3).
Main diagonal gives A354962.

Programs

  • Mathematica
    nn = 5; Table[g[n_] := q^Binomial[n, 2] n!; e[z_] := Sum[z^k/g[k], {k, 0, nn}];
       Table[g[n], {n, 0, nn}] CoefficientList[Series[1/e[-z], {z, 0, nn}], z], {q, 1, nn + 1}] //Transpose // Grid

Formula

Let E(x) = Sum_{n>=0} x^n/(n!*(k+1)^binomial(n,2)). Then 1/E(-x) = Sum_{n>=0} T(n,k)x^n/(n!*(k+1)^binomial(n,2)).
T(0,k) = 1 and T(n,k) = Sum_{j=1..n} (-1)^(j+1) * (k+1)^(j*(n-j)) * binomial(n,j) * T(n-j,k) for n > 0. - Seiichi Manyama, Jun 13 2022

A355071 G.f.: Sum_{n>=0} a(n)*x^n/(n!*4^(n*(n-1)/2)) = log( Sum_{n>=0} x^n/(n!*4^(n*(n-1)/2)) ).

Original entry on oeis.org

0, 1, -3, 81, -13311, 11688705, -51334027263, 1082183686000641, -106464672910860746751, 47880898685034024043741185, -96901748928702482338511172665343, 871602415363671863767026450797790494721
Offset: 0

Views

Author

Seiichi Manyama, Jun 18 2022

Keywords

Crossrefs

Programs

  • PARI
    a(n) = n!*4^(n*(n-1)/2)*polcoef(log(sum(k=0, n, x^k/(k!*4^(k*(k-1)/2)))+x*O(x^n)), n);
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=0; for(i=1, n, v[i+1]=1-sum(j=1, i-1, 4^(j*(i-j))*binomial(i-1, j)*v[i-j+1])); v;

Formula

a(0) = 0; a(n) = 1 - Sum_{k=1..n-1} 4^(k*(n-k)) * binomial(n-1,k) * a(n-k).
Showing 1-4 of 4 results.