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A092303
Triangle read by rows giving coefficients of polynomials f_n(q) arising in several different contexts.
0
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, 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, 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
OFFSET
0,7
COMMENTS
Coefficients are generalized Catalan numbers.
Lengths of successive rows are 1,1,2,3,5,7,10,13,17,21,26,31,... (A033638).
REFERENCES
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.
LINKS
S. B. Ekhad and D. Zeilberger, An Explicit Formula for the Number of Solutions of X^2=0 in Triangular Matrices over a Finite Field, arXiv:math/9512224v1 [math.CO], 1995.
A. A. Kirillov and A. Melnikov, On a remarkable sequence of polynomials, 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.
FORMULA
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).
Ekhad and Zeilberger proved (see link) that:
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_(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
EXAMPLE
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, ...
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Sequence in context: A029430 A321912 A329921 * A329343 A063725 A084888
KEYWORD
sign,tabf,nice
AUTHOR
N. J. A. Sloane, Feb 13 2004
STATUS
approved