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A092303
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Triangle read by rows giving coefficients of polynomials f_n(q) arising in several different contexts.
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0
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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
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OFFSET
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0,7
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COMMENTS
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Coefficients are generalized Catalan numbers.
Lengths of successive rows are 1,1,2,3,5,7,10,13,17,21,26,31,... (A033638).
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REFERENCES
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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.
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LINKS
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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.
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FORMULA
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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
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EXAMPLE
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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, ...
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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sign,tabf,nice
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AUTHOR
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STATUS
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approved
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