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A092580
Triangle read by rows: T(n,k) is the number of permutations p of [n] in which exactly the first k terms satisfy the up-down property, i.e., p(1)<p(2), p(2)>p(3), p(3)<p(4), ...
4
1, 1, 1, 3, 1, 2, 12, 4, 3, 5, 60, 20, 15, 9, 16, 360, 120, 90, 54, 35, 61, 2520, 840, 630, 378, 245, 155, 272, 20160, 6720, 5040, 3024, 1960, 1240, 791, 1385, 181440, 60480, 45360, 27216, 17640, 11160, 7119, 4529, 7936, 1814400, 604800, 453600, 272160, 176400, 111600, 71190, 45290, 28839, 50521
OFFSET
1,4
COMMENTS
Row sums are the factorial numbers (A000142). First column is A001710. Second column is A001715. Diagonal is A000111.
LINKS
E. Deutsch and W. P. Johnson, Create your own permutation statistics, Math. Mag., 77, 130-134, 2004.
FORMULA
T(n, k) = n!*[(k+1)*E(k)-E(k+1)]/(k+1)! for k<n and T(n, n) = E(n), where tan(x)+sec(x) = Sum_{n>=0} [E(n)x^n/n!] (i.e., E(n) = A000111(n)).
Sum_{k=0..n} (k+1) * T(n,k) = A230960(n). - Alois P. Heinz, Apr 27 2023
EXAMPLE
T(4,3)=3 because 1432, 2431, 3421 are the only permutations of [4] in which exactly the first 3 entries satisfy the up-down property.
Triangle starts:
1;
1, 1;
3, 1, 2;
12, 4, 3, 5;
60, 20, 15, 9, 16;
360, 120, 90, 54, 35, 61;
...
MAPLE
b:= proc(u, o) option remember;
`if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
end:
E:= n-> b(n, 0):
T:= (n, k)-> `if`(n=k, E(n), n!*((k+1)*E(k)-E(k+1))/(k+1)!):
seq(seq(T(n, k), k=1..n), n=1..10); # Alois P. Heinz, Aug 12 2016
MATHEMATICA
b[u_, o_] := b[u, o] = If[u + o == 0, 1, Sum[b[o - 1 + j, u - j], {j, 1, u}]]; e[n_] := b[n, 0]; T[n_, k_] := If[n == k, e[n], n!*((k + 1)*e[k] - e[k + 1])/(k + 1)!]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)
CROSSREFS
T(2n,n) gives A362603.
Sequence in context: A300930 A109528 A136125 * A004468 A254630 A145463
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch and Warren P. Johnson (wjohnson(AT)bates.edu), Apr 10 2004
STATUS
approved