A138177 - OEIS

%I #43 Nov 17 2018 16:20:46

%S 1,1,2,1,4,4,1,7,15,10,1,10,36,52,26,1,14,74,176,190,76,1,18,132,460,

%T 810,696,232,1,23,222,1060,2705,3756,2674,764,1,28,347,2180,7565,

%U 15106,17262,10480,2620,1,34,525,4204,19013,51162,83440,80816,42732,9496,1,40

%N Triangle T(n,k) read by rows: number of k X k symmetric matrices with nonnegative integer entries and without zero rows or columns such that sum of all entries is equal to n, n>=1, 1<=k<=n.

%C See the Brualdi/Ma reference for the connection to A161126. - _Joerg Arndt_, Nov 02 2014

%C T(n,k) is also the number of semistandard Young tableaux of size n whose entries span the interval 1..k. See also _Gus Wiseman_'s comment in A138178. The T(4,2) = 7 semi-standard Young tableaux of size 4 spanning the interval 1..2 are:

%C    11  122  112  111  1222  1122  1112

%C    22  2    2    2                      . - _Jacob Post_, Jun 15 2018

%H Alois P. Heinz, <a href="/A138177/b138177.txt">Rows n = 1..141, flattened</a>

%H Richard A. Brualdi, Shi-Mei Ma, <a href="http://dx.doi.org/10.1016/j.ejc.2014.08.026">Enumeration of involutions by descents and symmetric matrices</a>, European Journal of Combinatorics, vol.43, pp.220-228, (January 2015).

%H FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/SemistandardTableaux">Semistandard Young tableaux</a>

%H Samantha Dahlberg, <a href="http://arxiv.org/abs/1410.7356">Combinatorial Proofs of Identities Involving Symmetric Matrices</a>, arXiv:1410.7356 [math.CO], (27-October-2014)

%F T(n,k) = Sum_{i=0..k} (-1)^i * binomial(k,i) * A210391(n,k-i). - _Alois P. Heinz_, Apr 06 2015

%e Triangle T(n,k) begins:

%e   1;

%e   1,  2;

%e   1,  4,   4;

%e   1,  7,  15,   10;

%e   1, 10,  36,   52,   26;

%e   1, 14,  74,  176,  190,   76;

%e   1, 18, 132,  460,  810,  696,  232;

%e   1, 23, 222, 1060, 2705, 3756, 2674, 764;

%e   ...

%p gf:= k-> 1/((1-x)^k*(1-x^2)^(k*(k-1)/2)):

%p A:= (n, k)-> coeff(series(gf(k), x, n+1), x, n):

%p T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k):

%p seq(seq(T(n, k), k=1..n), n=1..12);  # _Alois P. Heinz_, Apr 06 2015

%t gf[k_] := 1/((1-x)^k*(1-x^2)^(k*(k-1)/2)); A[n_, k_] := Coefficient[ Series [gf[k], {x, 0, n+1}], x, n]; T[n_, k_] := Sum[(-1)^j*Binomial[k, j]*A[n, k-j], {j, 0, k}]; Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 31 2016, after _Alois P. Heinz_ *)

%Y Cf. (row sums) A138178, A135589, A135588, A161126, A210391.

%Y Main diagonal gives A000085. - _Alois P. Heinz_, Apr 06 2015

%Y T(2n,n) gives A266305.

%Y T(n^2,n) gives A268309.

%K nonn,tabl

%O 1,3

%A _Vladeta Jovovic_, Mar 03 2008