A001267 One-half the number of permutations of length n with exactly 3 rising or falling successions.

0, 0, 0, 0, 1, 8, 60, 444, 3599, 32484, 325322, 3582600, 43029621, 559774736, 7841128936, 117668021988, 1883347579515, 32026067455084, 576605574327174, 10957672400252944, 219190037987444577, 4603645435776504120, 101292568208941883236, 2329975164242735146316

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

(1/2) times number of permutations of 12...n such that exactly 3 of the following occur: 12, 23, ..., (n-1)n, 21, 32, ..., n(n-1).

References

• F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..200
• J. Riordan, A recurrence for permutations without rising or falling successions, Ann. Math. Statist. 36 (1965), 708-710.

Crossrefs

Cf. A002464, A000130, A086852. Equals A086854/2. A diagonal of A010028.

Programs

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:
a:= n-> coeff(S(n), t, 3)/2:
seq(a(n), n=0..25);  # Alois P. Heinz, Jan 11 2013

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]]]; a[n_] := Coefficient[S[n], t, 3]/2; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Mar 11 2014, after Alois P. Heinz *)

Formula

Coefficient of t^3 in S[n](t) defined in A002464, divided by 2.

a(n) ~ 2/(3*exp(2)) * n!. - Vaclav Kotesovec, Aug 10 2013

Recurrence: (n-4)*(2*n^6 - 52*n^5 + 557*n^4 - 3136*n^3 + 9740*n^2 - 15727*n + 10242)*a(n) = + (n-4)*(2*n^7 - 50*n^6 + 511*n^5 - 2693*n^4 + 7450*n^3 - 9041*n^2 - 157*n + 6666)*a(n-1) - (2*n^8 - 58*n^7 + 735*n^6 - 5289*n^5 + 23430*n^4 - 64575*n^3 + 106105*n^2 - 92312*n + 30900)*a(n-2) - (2*n^7 - 54*n^6 + 615*n^5 - 3795*n^4 + 13554*n^3 - 27681*n^2 + 29473*n - 12330)*(n-2)*a(n-3) + (2*n^6 - 40*n^5 + 327*n^4 - 1388*n^3 + 3184*n^2 - 3675*n + 1626)*(n-2)^2*a(n-4). - Vaclav Kotesovec, Aug 10 2013