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Number of Baxter permutations: A001181/2.
4

%I #19 Jan 04 2024 08:57:07

%S 1,3,11,46,211,1037,5377,29101,163120,941480,5570280,33664996,

%T 207249719,1296670793,8229378293,52895993341,343891293422,

%U 2258771535962,14974619271658,100117092310368,674548712552456,4577138309318008,31261253291922136,214800030086785976

%N Number of Baxter permutations: A001181/2.

%H Alois P. Heinz, <a href="/A046996/b046996.txt">Table of n, a(n) for n = 2..1120</a>

%F a(n) = (1/2)*B(n, 1) where B(n, x) are the Baxter polynomials with coefficients A359363. - _Peter Luschny_, Jan 04 2024

%p a:= proc(n) option remember; `if`(n<4, 2*n-3,

%p ((7*n^2+7*n-2)*a(n-1)+8*(n-1)*(n-2)*a(n-2))/((n+2)*(n+3)))

%p end:

%p seq(a(n), n=2..25); # _Alois P. Heinz_, Jul 29 2022

%t Sum[Binomial[n+1, k-1] Binomial[n+1, k] Binomial[n+1, k+1], {k, 1, n}]/(2 Binomial[n+1, 1] Binomial[n+1, 2])

%o (SageMath)

%o def A046996(n): return PolyA359363(n, 1) // 2

%o print([A046996(n) for n in range(2, 26)]) # _Peter Luschny_, Jan 04 2024

%Y Cf. A359363.

%K nonn,easy

%O 2,2

%A _N. J. A. Sloane_

%E More terms from _Olivier GĂ©rard_, Jun 26 2001