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A162979
Triangle read by rows: T(n,k) is the number of alternating (i.e., down-up) permutations of {1,2,...,n} having k fixed points (n >= 0, 0 <= k <= ceiling(n/2)).
4
1, 0, 1, 1, 0, 1, 1, 0, 2, 2, 1, 6, 6, 3, 1, 24, 24, 11, 2, 102, 102, 51, 15, 2, 528, 528, 252, 68, 9, 2952, 2952, 1476, 458, 89, 9, 19008, 19008, 9240, 2728, 493, 44, 131112, 131112, 65556, 20868, 4479, 621, 44, 1009728, 1009728, 495360, 152448, 31182, 4054, 265, 8271792, 8271792, 4135896, 1334928, 300954, 47670, 4959, 265
OFFSET
0,9
COMMENTS
Sum of entries in row n is the Euler (up-down) number A000111(n).
T(n,0) = T(n,1) = A129817(n) (n>=1).
T(2n,n) = T(2n+1,n+1) = d(n), where d(n) = A000166 is a derangement number (see the Chapman & Williams reference).
Sum_{k>=0} k*T(n,k) = A162978(n).
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.
FORMULA
The row generating polynomials can be obtained from Proposition 6.1 of the Stanley reference (see the Maple program).
EXAMPLE
T(5,2)=3 because we have 32415, 41325, and 52314.
Triangle starts:
1;
0, 1;
1, 0;
1, 1, 0;
2, 2, 1;
6, 6, 3, 1;
24, 24, 11, 2;
102, 102, 51, 15, 2;
MAPLE
fo := exp(E*(arctan(q*t)-arctan(t)))/(1-E*t): fe := sqrt((1+t^2)/(1+q^2*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^14 = a[14], E^15 = a[15], E^16 = a[16], 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]}, Q(n))) end do: for n from 0 to 13 do seq(coeff(P[n], q, j), j = 0 .. ceil((1/2)*n)) end do;
MATHEMATICA
nmax = 13;
fo = Exp[e*(ArcTan[q*t] - ArcTan[t])]/(1 - e*t);
fe = Sqrt[(1+t^2)/(1+q^2*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;
a[n_] := n!*SeriesCoefficient[Sec[x] + Tan[x], {x, 0, n}];
P[n_] := (Q[n] /. e^k_Integer :> a[k]) /. e :> a[1] // Expand;
Table[Switch[n, 0, {1}, 1, {0, 1}, 2, {1, 0}, 3, {1, 1, 0}, _, CoefficientList[P[n], q]] , {n, 0, nmax}] // Flatten (* Jean-François Alcover, Jul 23 2018, from Maple *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Aug 06 2009
STATUS
approved