A000986 Number of n X n symmetric matrices with (0,1) entries and all row sums 2.
1, 0, 1, 4, 18, 112, 820, 6912, 66178, 708256, 8372754, 108306280, 1521077404, 23041655136, 374385141832, 6493515450688, 119724090206940, 2337913445039488, 48195668439235612, 1045828865817825264, 23826258064972682776, 568556266922455167040
Offset: 0
References
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
- R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.8.
- Herbert S. Wilf, Generatingfunctionology, p. 104.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..445 (first 101 terms from T. D. Noe)
- H. Gupta, Enumeration of symmetric matrices, Duke Math. J., 35 (1968), vol 3, 653-659.
- H. Gupta, Enumeration of symmetric matrices (annotated scanned copy)
- Zhonghua Tan, Shanzhen Gao, and H. Niederhausen, Enumeration of (0,1) matrices with constant row and column sums, Appl. Math. Chin. Univ. 21 (4) (2006) 479-486.
Programs
-
Maple
a:= proc(n) option remember; `if`(n<2, 1-n, add(binomial (n-1, k-1) *(k! +`if`(k>2, (k-1)!, 0))/2 *a(n-k), k=2..n)) end: seq(a(n), n=0..30); # Alois P. Heinz, Feb 24 2011
-
Mathematica
a=1/(2(1-x))-1/2-x/2; b=(Log[1/(1-x)]-x-x^2/2)/2; Range[0, 20]! CoefficientList[Series[Exp[a + b], {x, 0, 20}], x] (* Second program: *) a[n_] := a[n] = If[n<2, 1-n, Sum[Binomial[n-1, k-1]*(k! + If[k>2, (k-1)!, 0])/2*a[n-k], {k, 2, n}]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 20 2017, after Alois P. Heinz *)
Formula
E.g.f.: (1-x)^(-1/2)*exp(-x-x^2/4 + x/((2*(1-x)))).
Sum_{a_1=0..n} Sum_{c=0..min(a_1, n - a_1)} Sum_{b=0..floor((n - a_1 - c)/2)} ((-1)^((n - a_1 - 2b - c) + b)*n!*(2a_1)!) / (2^(n + a_1 - 2c)*a_1!*(n - a_1 - 2b - c)!*b!*(2c)!*(a_1 - c)!). - Shanzhen Gao, Jun 05 2009
Conjecture: 2*a(n) +2*(-2*n+1)*a(n-1) +2*(n^2-2*n-1)*a(n-2) -2*(n-2)*(n-4)*a(n-3) +(n-1)*(n-2)*(n-3)*a(n-4) -(n-2)*(n-3)*(n-4)*a(n-5)=0. - R. J. Mathar, Aug 04 2013
Recurrence: 2*a(n) = 4*(n-1)*a(n-1) - 2*(n-3)*(n-1)*a(n-2) - (n-3)*(n-2)*(n-1)*a(n-4). - Vaclav Kotesovec, Feb 13 2014
a(n) ~ n^n * exp(sqrt(2*n)-n-3/2) / sqrt(2) * (1 + 43/(24*sqrt(2*n))). - Vaclav Kotesovec, Feb 13 2014
Comments