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%I #54 Mar 19 2019 12:55:00
%S 1,1,1,2,2,2,1,1,3,6,10,12,12,10,6,3,1,1,4,12,28,52,84,116,140,150,
%T 140,116,84,52,28,12,4,1,1,5,20,60,150,326,620,1060,1635,2295,2952,
%U 3480,3780,3780,3480,2952,2295,1635,1060,620,326,150,60,20,5,1
%N Triangle read by rows: SM(n,m) is the number of symmetric 0-1 matrices of order n such that the total number of 1's is m (n >= 1, 0 <= m <= n^2).
%H Alois P. Heinz, <a href="/A184948/b184948.txt">Rows n = 1..31, flattened</a>
%H P. Cameron, T. Prellberg and D. Stark, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v13i1r85">Asymptotics for incidence matrix classes</a>, Electron. J. Combin. 13 (2006), #R85, p. 11.
%F SM(n,m) is the sum of binomial(n,k) * binomial(n*(n-1)/2,(m-k)/2) over those k with the same parity as m. To see this consider that k is the number of 1s on the diagonal.
%F From _Robert Israel_, Feb 02 2011: (Start)
%F According to Maple,
%F > simplify(sum(binomial(n,2*j)*binomial(r,M-j),j=0..M)) assuming posint;
%F binomial(r,M)*hypergeom([-M, -1/2*n, 1/2-1/2*n],[1/2, r-M+1],-1)
%F > simplify(sum(binomial(n,2*k+1)*binomial(r,M-k),k=0..M)) assuming posint;
%F n*binomial(r,M)*hypergeom([-M, 1-1/2*n, 1/2-1/2*n],[3/2, r-M+1],-1)
%F If m is even you want the first formula with r=n*(n-1)/2 and M=m/2.
%F If m is odd the second formula with r=n*(n-1)/2 and M=(m+1)/2.
%F Thus for n=5 and m=6,
%F binomial(10,3)*hypergeom([-3,-5/2,-2],[1/2,8],-1) = 620
%F and for n=5 and m=5,
%F 5*binomial(10,3)*hypergeom([-3, -3/2, -2],[3/2, 8],-1) = 1060. (End)
%F G.f. for row n: (1+x)^n*(1+x^2)^(n(n-1)/2 for n>=1. - _Paul D. Hanna_, Feb 03 2011
%F G.f.: A(x,y) = Sum_{n>=1} x^n*(1+y)^n*Product_{k=1..n} (1-x(1+y)(1+y^2)^(2k-2))/(1-x(1+y)(1+y^2)^(2k-1)) due to a q-series identity. - _Paul D. Hanna_, Feb 03 2011
%F Sum_{k>=0..n^2} k*SM(n,k) = n^2/2 * 2^(n(n+1)/2).
%F SM(n,m) = Sum_{k=0..floor(m/2)} C(C(n,2),k)*C(n,m-2*k), from equation (11) in the Cameron et al., reference. - _L. Edson Jeffery_, Feb 29 2012
%e Triangle begins:
%e SM(1, m) = 1, 1
%e SM(2, m) = 1, 2, 2, 2, 1
%e SM(3, m) = 1, 3, 6, 10, 12, 12, 10, 6, 3, 1
%e SM(4, m) = 1, 4, 12, 28, 52, 84, 116, 140, 150, 140, 116, 84, 52, 28, 12, 4, 1
%e SM(5, m) = 1, 5, 20, 60, 150, 326, 620, 1060, 1635, 2295, 2952, 3480, 3780, 3780, 3480, 2952, 2295, 1635, 1060, 620, 326, 150, 60, 20, 5, 1
%e ...
%t row[n_] := CoefficientList[(1+x)^n (1+x^2)^(n(n-1)/2), x];
%t Array[row, 5] // Flatten (* _Jean-François Alcover_, Mar 19 2019 *)
%o (PARI) {SM(n,k)=polcoeff((1+x^2)^(n*(n-1)/2)*(1+x)^n,k)} \\ _Paul D. Hanna_
%o (PARI) {SM(n,k)=local(A); A=sum(m=1, n, x^m*(1+y)^m*prod(k=1, m, (1-x*(1+y)*(1+y^2)^(2*k-2))/(1-x*(1+y)*(1+y^2)^(2*k-1)+x*O(x^n))));polcoeff(polcoeff(A, n,x),k,y)} \\ _Paul D. Hanna_
%Y Row sums give A006125(n+1).
%Y Cf. A262666.
%K nonn,tabf
%O 1,4
%A _N. J. A. Sloane_, Feb 03 2011, based on a posting to the Sequence Fans Mailing List by _Brendan McKay_, Feb 02 2011
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