A241477 - OEIS

%I #20 Mar 05 2020 07:05:17

%S 1,0,2,2,2,2,0,4,0,2,6,12,4,2,6,0,12,0,4,0,4,20,60,12,12,12,4,20,0,40,

%T 0,12,0,8,0,10,70,280,40,60,36,24,40,10,70,0,140,0,40,0,24,0,20,0,28,

%U 252,1260,140,280,120,120,120,60,140,28,252,0,504,0

%N Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

%C For the combinatorial definitions see A232500. An orbital w over n sectors has its first zero crossing at k if k is the smallest j such that the partial sum(1<=i<=j, w(i))) = 0, where w(i) are the jumps of the orbital represented by -1, 0, 1.

%F If n is even and k is odd then T(n, k) = 0 else if k = 1 then T(n, 1) = A056040(n-1) else T(n, k) = 2*A057977(k-2)*A056040(n-k).

%F T(n, n) = A241543(n).

%F T(n+1, 1) = A126869(n).

%F T(2*n, 2*n) = |A002420(n)|.

%F T(2*n+1, 1) = A000984(n).

%F T(2*n+1, n+1) = A241530(n).

%F T(2*n+2, 2) = A028329(n).

%F T(4*n, 2*n) = |A010370(n)|.

%F T(4*n, 4*n) = |A024491(n)|.

%F T(4*n+1, 1) = A001448(n).

%F T(4*n+1, 2*n+1) = A002894(n).

%e [1], [ 1]

%e [2], [ 0,  2]

%e [3], [ 2,  2,  2]

%e [4], [ 0,  4,  0,  2]

%e [5], [ 6, 12,  4,  2,  6]

%e [6], [ 0, 12,  0,  4,  0, 4]

%e [7], [20, 60, 12, 12, 12, 4, 20]

%p A241477 := proc(n, k)

%p   if n = 0 then 1

%p elif k = 0 then 0

%p elif irem(n, 2) = 0 and irem(k, 2) = 1 then 0

%p elif k = 1 then (n-1)!/iquo(n-1,2)!^2

%p else 2*(n-k)!*(k-2)!/iquo(k,2)/(iquo(k-2,2)!*iquo(n-k,2)!)^2

%p   fi end:

%p for n from 1 to 9 do seq(A241477(n, k), k=1..n) od;

%t T[n_, k_] := Which[n == 0, 1, k == 0, 0, Mod[n, 2] == 0 && Mod[k, 2] == 1,  0, k == 1, (n-1)!/Quotient[n-1, 2]!^2, True, 2*(n-k)!*(k-2)!/Quotient[k, 2]/(Quotient[k-2, 2]!*Quotient[n-k, 2]!)^2];

%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jun 20 2018, from Maple *)

%o (Sage)

%o def A241477_row(n):

%o     if n == 0: return [1]

%o     Z = [0]*n; T = [0] if is_odd(n) else []

%o     for i in (1..n//2): T.append(-1); T.append(1)

%o     for p in Permutations(T):

%o         i = 0; s = p[0]

%o         while s != 0: i += 1; s += p[i];

%o         Z[i] += 1

%o     return Z

%o for n in (1..9): A241477_row(n)

%Y Row sums: A056040.

%Y Cf. A232500.

%K nonn,tabl

%O 1,3

%A _Peter Luschny_, Apr 23 2014