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A001267 One-half the number of permutations of length n with exactly 3 rising or falling successions.
(Formerly M4550 N1934)
8

%I M4550 N1934 #23 Dec 19 2021 09:54:41

%S 0,0,0,0,1,8,60,444,3599,32484,325322,3582600,43029621,559774736,

%T 7841128936,117668021988,1883347579515,32026067455084,576605574327174,

%U 10957672400252944,219190037987444577,4603645435776504120,101292568208941883236,2329975164242735146316

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

%C (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).

%D F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Alois P. Heinz, <a href="/A001267/b001267.txt">Table of n, a(n) for n = 0..200</a>

%H J. Riordan, <a href="http://www.jstor.org/stable/2238176">A recurrence for permutations without rising or falling successions</a>, Ann. Math. Statist. 36 (1965), 708-710.

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

%F a(n) ~ 2/(3*exp(2)) * n!. - _Vaclav Kotesovec_, Aug 10 2013

%F 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

%p S:= proc(n) option remember; `if`(n<4, [1, 1, 2*t, 4*t+2*t^2]

%p [n+1], expand((n+1-t)*S(n-1) -(1-t)*(n-2+3*t)*S(n-2)

%p -(1-t)^2*(n-5+t)*S(n-3) +(1-t)^3*(n-3)*S(n-4)))

%p end:

%p a:= n-> coeff(S(n), t, 3)/2:

%p seq(a(n), n=0..25); # _Alois P. Heinz_, Jan 11 2013

%t 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_ *)

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

%K nonn

%O 0,6

%A _N. J. A. Sloane_

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