A162980 Triangle read by rows: T(n,k) is the number of reverse alternating (i.e., up-down) permutations of {1,2,...,n} having k fixed points (n >= 0, 0 <= k <= 1 + floor(n/2)).
1, 0, 1, 0, 0, 1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 22, 22, 12, 4, 1, 102, 102, 51, 15, 2, 506, 506, 264, 88, 19, 2, 2952, 2952, 1476, 458, 89, 9, 18502, 18502, 9504, 3168, 726, 110, 9, 131112, 131112, 65556, 20868, 4479, 621, 44, 991226, 991226, 504864, 168288, 39696, 6672, 749, 44
Offset: 0
Examples
T(5,2)=3 because we have 15243, 14352, and 25341. Triangle starts: 1; 0, 1; 0, 0, 1; 1, 1; 2, 2, 1; 6, 6, 3, 1; 22, 22, 12, 4, 1; 102, 102, 51, 15, 2;
Links
- R. Chapman and L. K. Williams, A conjecture of Stanley on alternating permutations, The Electronic J. of Combinatorics, 14, 2007, #N16.
- R. P. Stanley, Alternating permutations and symmetric functions, J. Comb. Theory A 114 (3) (2007) 436-460.
Programs
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Maple
fo := exp(E*(arctan(q*t)-arctan(t)))/(1-E*t): fe := sqrt((1+q^2*t^2)/(1+t^2))*exp(E*(arctan(q*t)-arctan(t)))/(1-E*t): foser := simplify(series(fo, t = 0, 18)): feser := simplify(series(fe, t = 0, 18)): Q := proc (n) if `mod`(n, 2) = 1 then coeff(foser, t, n) else coeff(feser, t, n) end if end proc: for n from 0 to 16 do Q(n) end do: g := sec(x)+tan(x): gser := series(g, x = 0, 20): for n from 0 to 18 do a[n] := factorial(n)*coeff(gser, x, n) end do: for n from 0 to 15 do P[n] := sort(subs({E = a[1], E^2 = a[2], E^3 = a[3], E^4 = a[4], E^5 = a[5], E^6 = a[6], E^7 = a[7], E^8 = a[8], E^9 = a[9], E^10 = a[10], E^11 = a[11], E^12 = a[12], E^13 = a[13], E^14 = a[14], E^15 = a[15], E^16 = a[16]}, Q(n))) end do: 1; 0, 1; 0, 0, 1; 1, 1; 2, 2, 1; for n from 5 to 13 do seq(coeff(P[n], q, j), j = 0 .. 1+floor((1/2)*n)) end do;
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Mathematica
nmax = 10; fo = Exp[e*(ArcTan[q*t] - ArcTan[t])]/(1 - e*t); fe = Sqrt[(1+q^2 t^2)/(1+t^2)]*Exp[e*(ArcTan[q*t] - ArcTan[t])]/(1-e*t); Q[n_] := If[OddQ[n], SeriesCoefficient[fo, {t, 0, n}], SeriesCoefficient[fe, {t, 0, n}]] // Expand; b[n_] := n!*SeriesCoefficient[Sec[x] + Tan[x], {x, 0, n}]; P[n_] := (Q[n] /. e^k_Integer :> b[k]) /. e :> b[1] // Expand; T[n_, k_] := Coefficient[P[n], q, k]; Table[CoefficientList[P[n], q], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Jul 24 2018, from Maple *)
Formula
The row generating polynomials can be obtained from Proposition 6.1 of the Stanley reference (see the Maple program).
Comments