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A216963
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Triangle read by rows, arising in enumeration of permutations by cyclic peaks, cycles and fixed points.
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5
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1, 0, 1, 1, 1, 4, 5, 11, 28, 5, 41, 153, 71, 162, 872, 759, 61, 715, 5191, 7262, 1665, 3425, 32398, 66510, 29778, 1385, 17722, 211937, 601080, 443231, 60991, 98253, 1451599, 5446847, 5994473, 1642877, 50521, 580317, 10393114, 49940615, 76889330, 35162440, 3249025
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OFFSET
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0,6
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COMMENTS
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See Ma and Chow (2012) for precise definition (cf. Proposition 3).
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LINKS
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EXAMPLE
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Triangle begins:
: 1;
: 0;
: 1;
: 1, 1;
: 4, 5;
: 11, 28, 5;
: 41, 153, 71;
: 162, 872, 759, 61;
: 715, 5191, 7262, 1665;
...
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MAPLE
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p:= proc(n) option remember; expand(`if`(n<4,
[1, 0, x, x*(1+q)][n+1], (n-1)*q*p(n-1)+
2*q*(1-q)*diff(p(n-1), q)+x*(1-q)*
diff(p(n-1), x)+(n-1)*x*p(n-2)))
end:
T:= n-> (t-> seq(coeff(t, q, i), i=0..
max(0, degree(t))))(subs(x=1, p(n))):
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MATHEMATICA
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p[0] = 1; p[1] = 0; p[2] = x; p[3] = (1 + q) x;
p[n_] := p[n] = Expand[(n - 1) q p[n - 1] + 2 q (1 - q) D[p[n - 1], q] + x (1 - q) D[p[n - 1], x] + (n - 1) x p[n - 2]];
T[n_] := CoefficientList[p[n] /. x -> 1 , q]; T[1] = {0};
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PROG
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(PARI) tabf(m) = {P = x; M = subst(P, x, 1); for (d=0, poldegree(M, q), print1(polcoeff(M, d, q), ", "); ); print(""); Q = (1+q)*x; M = subst(Q, x, 1); for (d=0, poldegree(M, q), print1(polcoeff(M, d, q), ", "); ); print(""); for (n=3, m, newP = n*q*Q + 2*q*(1-q)*deriv(Q, q) + x*(1-q)*deriv(Q, x) + n*x*P; M = subst(newP, x, 1); for (d=0, poldegree(M, q), print1(polcoeff(M, d, q), ", "); ); print(""); P = Q; Q = newP; ); } \\ Michel Marcus, Feb 09 2013
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CROSSREFS
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T(2n+1,n) gives A000364(n) for n>0.
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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