A086856 Triangle read by rows: T(n,k) = one-half number of permutations of length n with exactly k rising or falling successions, for n >= 1, 0 <= k <= n-1. T(1,0) = 1 by convention.
1, 0, 1, 0, 2, 1, 1, 5, 5, 1, 7, 20, 24, 8, 1, 45, 115, 128, 60, 11, 1, 323, 790, 835, 444, 113, 14, 1, 2621, 6217, 6423, 3599, 1099, 183, 17, 1, 23811, 55160, 56410, 32484, 11060, 2224, 270, 20, 1, 239653, 545135, 554306, 325322, 118484, 27152, 3950, 374, 23, 1, 2648395
Offset: 1
Examples
Triangle T(n,k) begins: 1; 0, 1; 0, 2, 1; 1, 5, 5, 1; 7, 20, 24, 8, 1; 45, 115, 128, 60, 11, 1; 323, 790, 835, 444, 113, 14, 1; ...
References
- F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- J. Riordan, A recurrence for permutations without rising or falling successions, Ann. Math. Statist. 36 (1965), 708-710.
Crossrefs
Programs
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Maple
S:= proc(n) option remember; `if`(n<4, [1, 1, 2*t, 4*t+2*t^2] [n+1], expand((n+1-t)*S(n-1) -(1-t)*(n-2+3*t)*S(n-2) -(1-t)^2*(n-5+t)*S(n-3) +(1-t)^3*(n-3)*S(n-4))) end: T:= (n, k)-> ceil(coeff(S(n), t, k)/2): seq(seq(T(n, k), k=0..n-1), n=1..10); # Alois P. Heinz, Jan 11 2013
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Mathematica
S[n_] := S[n] = If[n < 4, {1, 1, 2*t, 4*t+2*t^2}[[n+1]], Expand[(n+1-t)*S[n-1] - (1-t)*(n-2+3*t)*S[n-2] - (1-t)^2*(n-5+t)*S[n-3] + (1-t)^3*(n-3)*S[n-4]]]; T[n_, k_] := Ceiling[Coefficient[S[n], t, k]/2]; Table[Table[T[n, k], {k, 0, n-1}], {n, 1, 11}] // Flatten (* Jean-François Alcover, Jan 14 2014, translated from Alois P. Heinz's Maple code *)
Comments