%I #21 Dec 25 2024 15:50:01
%S 0,10,58,236,836,2766,8814,27472,84472,257522,780770,2358708,7108908,
%T 21392278,64307926,193185944,580082144,1741295034,5225982282,
%U 15682141180,47054812180,141181213790,423577195838,1270798696416,3812530307016,11437859356546,34314114940594
%N Number of permutations of [n] with 3 sequences.
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 261.
%H Harry J. Smith, <a href="/A060157/b060157.txt">Table of n, a(n) for n = 3..200</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7,-17,17,-6).
%F a(n) = 11/2 - n - 2^(n+1) + (1/2)*3^n.
%F G.f.: 2*x^4*(5-6*x)/((1-x)^2*(1-2*x)*(1-3*x)). - _Colin Barker_, Feb 17 2012
%e a(4)=10 because each of the 5 (=A000111(4)) up-down permutations and 5 down-up permutations has 3 sequences. For example, the 3 sequences of 2413 are 24, 41, and 13. - _Emeric Deutsch_, Jul 11 2009
%p n3 := n->11/2-n-2^(n+1)+1/2*3^n; seq(n3(i),i=3..30);
%t Table[11/2-n-2^(n+1)+3^n/2,{n,3,30}]
%o (PARI) a(n) = { (3^n + 11)/2 - 2^(n + 1) - n } \\ _Harry J. Smith_, Jul 02 2009
%Y Cf. A028399, A060158.
%Y Cf. A000111. - _Emeric Deutsch_, Jul 11 2009
%K nonn,easy
%O 3,2
%A Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Mar 12 2001