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A349479
Irregular triangle read by rows: T(n,k) = S1(n,k)*2^k, where S1(n,k) is the associated Stirling number of the first kind (cf. A008306) (n >= 0, 0 <= k <= floor(n/2)).
0
1, 0, 0, 2, 0, 4, 0, 12, 12, 0, 48, 80, 0, 240, 520, 120, 0, 1440, 3696, 1680, 0, 10080, 29232, 19040, 1680, 0, 80640, 256896, 211456, 40320, 0, 725760, 2493504, 2429280, 705600, 30240, 0, 7257600, 26547840, 29430720, 11285120, 1108800, 0, 79833600, 307992960, 378595008, 177580480, 27720000, 665280
OFFSET
0,4
COMMENTS
T(n,k) is the number of cycle-colored n-derangements possessing exactly k cycles; two colors are available.
LINKS
Steven Finch, Rounds, Color, Parity, Squares, arXiv:2111.14487 [math.CO], 2021.
EXAMPLE
Triangle begins:
[0] 1;
[1] 0;
[2] 0, 2;
[3] 0, 4;
[4] 0, 12, 12;
[5] 0, 48, 80;
[6] 0, 240, 520, 120;
[7] 0, 1440, 3696, 1680;
[8] 0, 10080, 29232, 19040, 1680;
[9] 0, 80640, 256896, 211456, 40320;
...
MAPLE
b:= proc(n) option remember; expand(`if`(n=0, 1, add(
2*x*b(n-j)*binomial(n-1, j-1)*(j-1)!, j=2..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..floor(n/2)))(b(n)):
seq(T(n), n=0..14); # Alois P. Heinz, Nov 19 2021
MATHEMATICA
S1[0, 0] = 1; S1[_, 0] = 0; S1[n_, k_] /; k > Quotient[n, 2] = 0;
S1[n_, k_] := S1[n, k] = (n-1)*(S1[n-1, k] + S1[n-2, k-1]);
T[n_, k_] := S1[n, k]*2^k;
Table[T[n, k], {n, 0, 14}, {k, 0, Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Dec 28 2021 *)
CROSSREFS
Row sums give A087981.
Sequence in context: A137449 A056946 A225740 * A111757 A286122 A286776
KEYWORD
nonn,tabf
AUTHOR
Steven Finch, Nov 19 2021
STATUS
approved