A373418 Triangle read by rows: T(n,k) is the number of permutations in symmetric group S_n with (n-k) fixed points and an odd number of non-fixed point cycles. Equivalent to the number of cycles of n items with cycle type defined by non-unity partitions of k <= n that contain an odd number of parts.

0, 0, 0, 0, 0, 1, 0, 0, 3, 2, 0, 0, 6, 8, 6, 0, 0, 10, 20, 30, 24, 0, 0, 15, 40, 90, 144, 135, 0, 0, 21, 70, 210, 504, 945, 930, 0, 0, 28, 112, 420, 1344, 3780, 7440, 7420, 0, 0, 36, 168, 756, 3024, 11340, 33480, 66780, 66752, 0, 0, 45, 240, 1260, 6048, 28350, 111600, 333900, 667520, 667485

Offset: 0

Author

Julian Hatfield Iacoponi, Jun 04 2024

Keywords

• nonn
• tabl

Comments

a(n) + A343417(n) = A098825(n) = partial derangement "rencontres" triangle.

a(n) - A343417(n) = (k-1) * binomial(n,k) = A127717(n-1,k-1).

Difference of 2nd and 1st leading diagonals (n > 0):

T(n,n-1) - T(n,n) = 0,-1,1,2,6,9,15,20,28,35,45,54,...

= (0-1) + (2+1) + (4+3) + (6+5) + (8+7) + (10+9) + ...

Cf. A084265(n) with 2 terms 0,-1 prepended (moving its offset from 0 to -2).

Examples

			Triangle begins:
   n: {k<=n}
   0: {0}
   1: {0, 0}
   2: {0, 0,  1}
   3: {0, 0,  3,   2}
   4: {0, 0,  6,   8,    6}
   5: {0, 0, 10,  20,   30,   24}
   6: {0, 0, 15,  40,   90,  144,   135}
   7: {0, 0, 21,  70,  210,  504,   945,    930}
   8: {0, 0, 28, 112,  420, 1344,  3780,   7440,   7420}
   9: {0, 0, 36, 168,  756, 3024, 11340,  33480,  66780,  66752}
  10: {0, 0, 45, 240, 1260, 6048, 28350, 111600, 333900, 667520, 667485}
T(n,0) = 0 because the sole permutation in S_n with n fixed points, namely the identity permutation, has 0 non-fixed point cycles, not an odd number.
T(n,1) = 0 because there are no permutations in S_n with (n-1) fixed points.
Example:
T(3,3) = 2 since S_3 contains 3 permutations with 0 fixed points and an odd number of non-fixed point cycles, namely the derangements (123) and (132).
Worked Example:
T(7,6) = 945 permutations in S_7 with 1 fixed point and an odd number of non-fixed point cycles;
T(7,6) = 945 possible 6- and (2,2,2)-cycles of 7 items.
N(n,y) = possible y-cycles of n items;
N(n,y) = (n!/(n-k)!) / (M(y) * s(y)).
y = partition of k<=n with q parts = (p_1, p_2, ..., p_i, ..., p_q) such that k = Sum_{i=1..q} p_i.
Or:
y = partition of k<=n with d distinct parts, each with multiplicity m_j = (y_1^m_1, y_2^m_2, ..., y_j^m_j, ..., y_d^m_d) such that k = Sum_{j=1..d} m_j*y_j.
M(y) = Product_{i=1..q} p_i = Product_{j=1..d} y_j^m_j.
s(y) = Product_{j=1..d} m_j! ("symmetry of repeated parts").
Note: (n!/(n-k)!) / s(y) = multinomial(n, {m_j}).
Therefore:
T(7,6) = N(7,y=(6)) + N(7,y=(2^3))
       = (7!/6) + (7!/(2^3)/3!)
       = 7! * (1/6 + 1/48)
       = 5040 * (3/16);
T(7,6) = 945.
		

Crossrefs

Cf. A373417 (even case), A373340 (row sums), A216779 (main diagonal).

Programs

Maple

b:= proc(n, t) option remember; `if`(n=0, t, add(expand(`if`(j>1, x^j, 1)*
      b(n-j, irem(t+signum(j-1), 2)))*binomial(n-1, j-1)*(j-1)!, j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 0)):
seq(T(n), n=0..10);

Mathematica

Table[Table[n!/(n-k)!/2 * (Sum[(-1)^j/j!, {j, 0, k}] - ((k - 1)/k!)),{k,1,n}], {n,1,10}]

Formula

T(n,k) = (n!/(n-k)!/2) * ((Sum_{j=0..k} (-1)^j/j!) + (k-1)/k!). Cf. Sum_{j=0..k} (-1)^j/j! = A053557(k) / A053556(k).