|
%I #25 Apr 28 2023 19:55:46
%S 1,1,1,3,1,2,12,4,3,5,60,20,15,9,16,360,120,90,54,35,61,2520,840,630,
%T 378,245,155,272,20160,6720,5040,3024,1960,1240,791,1385,181440,60480,
%U 45360,27216,17640,11160,7119,4529,7936,1814400,604800,453600,272160,176400,111600,71190,45290,28839,50521
%N 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), ...
%C Row sums are the factorial numbers (A000142). First column is A001710. Second column is A001715. Diagonal is A000111.
%H Alois P. Heinz, <a href="/A092580/b092580.txt">Rows n = 1..150, flattened</a>
%H E. Deutsch and W. P. Johnson, <a href="http://www.jstor.org/stable/3219101">Create your own permutation statistics</a>, Math. Mag., 77, 130-134, 2004.
%F 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)).
%F Sum_{k=0..n} (k+1) * T(n,k) = A230960(n). - _Alois P. Heinz_, Apr 27 2023
%e 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.
%e Triangle starts:
%e 1;
%e 1, 1;
%e 3, 1, 2;
%e 12, 4, 3, 5;
%e 60, 20, 15, 9, 16;
%e 360, 120, 90, 54, 35, 61;
%e ...
%p b:= proc(u, o) option remember;
%p `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
%p end:
%p E:= n-> b(n, 0):
%p T:= (n, k)-> `if`(n=k, E(n), n!*((k+1)*E(k)-E(k+1))/(k+1)!):
%p seq(seq(T(n, k), k=1..n), n=1..10); # _Alois P. Heinz_, Aug 12 2016
%t 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_ *)
%Y Cf. A000111, A000142, A001710, A001715, A092582, A230960.
%Y T(2n,n) gives A362603.
%K nonn,tabl
%O 1,4
%A _Emeric Deutsch_ and Warren P. Johnson (wjohnson(AT)bates.edu), Apr 10 2004
|
