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%I #21 Feb 26 2018 00:53:30
%S 1,1,0,1,0,-1,2,0,0,-1,0,2,0,0,0,0,0,-4,5,0,0,0,0,0,1,0,-5,0,5,0,0,0,
%T 0,0,0,0,1,0,0,0,-14,14,0,0,0,0,0,0,0,0,0,0,0,0,7,0,-20,0,14,0,0,0,0,
%U 0,0,0,0,0,0,0,0,-1,0,0,8,0,0,0,-48,42,0,0,0,0,0,0,0,0,0,0,0,0
%N Triangle read by rows giving coefficients of polynomials f_n(q) arising in several different contexts.
%C Coefficients are generalized Catalan numbers.
%C Lengths of successive rows are 1,1,2,3,5,7,10,13,17,21,26,31,... (A033638).
%D A. A. Kirillov, Two more variations on the triangular theme, pp. 243-258 of C. Duval et al., eds., The Orbit Method in Geometry and Physics, Birkhäuser, Basel, 2003.
%H S. B. Ekhad and D. Zeilberger, <a href="http://arxiv.org/abs/math/9512224v1">An Explicit Formula for the Number of Solutions of X^2=0 in Triangular Matrices over a Finite Field</a>, arXiv:math/9512224v1 [math.CO], 1995.
%H A. A. Kirillov and A. Melnikov, <a href="http://www.emis.de/journals/SC/1997/2/html/smf_sem-cong_2_35-42.html">On a remarkable sequence of polynomials</a>, pp. 35-42 of J. Alev et al., eds., Algebre Non-commutative, Groupes Quantiques et Invariants, Rencontre Franco-Belge, Reims 1995, Publications SMF, No. 2, 1996.
%F Let f_n(q) = number of solutions of X^2 = 0 in n X n upper triangular matrices with elements in GF(q) and let f_{n, r}(q) be the number with rank r. Then f_{n+1, r+1}(q) = q^(r+1)*f_{n, r+1}(q) + (q^(n-r)-q^r)*f_{n, r}(q); f_{n+1, 0}(q) = 1; f_n(q) = Sum_{r >= 0} f_{n, r}(q).
%F Ekhad and Zeilberger proved (see link) that:
%F f_(2*n)(q) = Sum{j}(binomial(2*n, n-3*j) - binomial(2*n, n-3*j-1))*q^(n^2-3*j^2-j),
%F f_(2*n+1)(q) = Sum{j}(binomial(2*n+1, n-3*j) - binomial(2*n+1, n-3*j-1))*q^(n^2+n-3*j^2-2*j). - _Michel Marcus_, May 23 2013
%e The first few polynomials are f_0(q) = 1, f_1(q) = 1, f_2(q) = q, f_3(q) = 2q^2-q, f_4(q) = 2q^4-q^2, f_5(q) = 5q^6-4q^5, f_6(q) = 5q^9-5q^7+q^5, ...
%t f[m_?EvenQ, q_] := With[{n = m/2}, Sum[ (Binomial[2*n, n-3*j] - Binomial[2*n, n-3*j-1])* q^(n^2-3*j^2-j), {j, Floor[-(m+1)/3], Floor[(m+1)/3]}]]; f[m_?OddQ, q_] := With[{n = (m-1)/2}, Sum[ (Binomial[2*n+1, n-3*j] - Binomial[2*n+1, n-3*j-1])* q^(n^2+n-3*j^2-2*j), {j, Floor[-(m+1)/3], Floor[(m+1)/3]}]]; Table[ CoefficientList[ f[n, q], q], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Jul 09 2013, after _Michel Marcus_ *)
%o (PARI) apol(n) = {pol = 0; if (n % 2 == 0, ne = n/2; v = (ne+1)/3; for (j = floor(-v), floor(v), pol += (binomial(n, ne-3*j)-binomial(n, ne-3*j-1))*q^(ne^2-3*j^2-j););, no = (n-1)/2; v = (no+1)/3; for (j = floor(-v), floor(v), pol += (binomial(n, no-3*j)-binomial(n, no-3*j-1))*q^(no^2+no-3*j^2-2*j););); return (pol);} \\ _Michel Marcus_, May 23 2013
%K sign,tabf,nice
%O 0,7
%A _N. J. A. Sloane_, Feb 13 2004