A000349 One-half the number of permutations of length n with exactly 2 rising or falling successions.

0, 0, 0, 1, 5, 24, 128, 835, 6423, 56410, 554306, 6016077, 71426225, 920484892, 12793635300, 190730117959, 3035659077083, 51371100102990, 920989078354838, 17437084517068465, 347647092476801301, 7280060180210901232, 159755491837445900120, 3665942433747225901707

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

(1/2) times number of permutations of 12...n such that exactly 2 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.
• J. Riordan, A recurrence for permutations without rising or falling successions. Ann. Math. Statist. 36 (1965), 708-710.
• 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

Crossrefs

Cf. A002464, A000130, A086852. Equals A086853/2. 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-> ceil(coeff(S(n), t, 2)/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_] := Ceiling[Coefficient[S[n], t, 2]/2]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Mar 11 2014, after Alois P. Heinz *)

Formula

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

Recurrence: (n-3)*(n-2)*(n-4)^3*a(n) = (n-3)*(n^4-9*n^3+23*n^2-4*n-29)*(n-4)*a(n-1) - (n-1)*(n^4-12*n^3+57*n^2-125*n+104)*(n-4)*a(n-2) - (n-2)*(n-1)*(n^4-15*n^3+83*n^2-198*n+169)*a(n-3) + (n-3)^3*(n-2)^2*(n-1)*a(n-4). - Vaclav Kotesovec, Aug 10 2013

a(n) ~ sqrt(2*Pi)*n^(n+1/2)/exp(n+2). - Vaclav Kotesovec, Aug 10 2013