A133265 -id:A133265 - cp's OEIS frontend

A107373 a(n) = (n/2)*binomial(n-1, floor((n-1)/2)) - 2^(n-2).

0, 0, 1, 2, 7, 14, 38, 76, 187, 374, 874, 1748, 3958, 7916, 17548, 35096, 76627, 153254, 330818, 661636, 1415650, 2831300, 6015316, 12030632, 25413342, 50826684, 106853668, 213707336, 447472972, 894945944, 1867450648, 3734901296, 7770342787, 15540685574

Offset: 1

N. J. A. Sloane, Jul 20 2007

nonn

Total number of descents in all faro permutations of length n-1. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. See also A340567, A340568 and A340569. - Sergey Kirgizov, Jan 11 2021

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Jean-Luc Baril, Alexander Burstein, and Sergey Kirgizov, Pattern statistics in faro words and permutations, arXiv:2010.06270 [math.CO], 2020. See Table 1.
F. Disanto and S. Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A. 22:1 (2011), 39-60.
Igor Pak, The area of cyclic polygons: Recent progress on Robbins' Conjectures, Adv. Applied Math. 34 (2005), 690-696. Special issue in memory of David Robbins.

Cf. A131019, A131020, A131021.

[(n/2)*Binomial(n-1, Floor((n-1)/2)) - 2^(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 01 2013

A056040 := n -> n!/iquo(n,2)!^2:
A133265 := n -> (n+2+(n-2)*(-1)^n)/2:
A107373 := n -> (A056040(n)*A133265(n)-2^n)/4:
seq(A107373(n),n=1..34); # Peter Luschny, Aug 30 2011

Table[(n/2) Binomial[n-1, Floor[(n-1)/2]] - 2^(n-2), {n, 1, 40}] (* Vincenzo Librandi, Oct 01 2013 *)

a(2*n) = 2*A000531(n-1); a(2*n+1) = A000531(n). - Max Alekseyev, Sep 30 2013

(1-n)*a(n) + 2*(n-1)*a(n-1) + 4*(n-2)*a(n-2) + 8*(-n+2)*a(n-3) = 0. - R. J. Mathar, May 26 2013

A275365 a(1)=2, a(2)=2; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

Original entry on oeis.org

0, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14, 2, 16, 2, 18, 2, 20, 2, 22, 2, 24, 2, 26, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2, 72, 2, 74

Offset: 0

Nathan Fox, Jul 24 2016

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = 2, a(2) = 2.
Starting with n=1, a(n) is A005843 interleaved with A007395.
This sequence is the same as A133265 with the leading 2 changed to a 0.

Nathan Fox, Table of n, a(n) for n = 0..1000
Nathan Fox, Finding Linear-Recurrent Solutions to Hofstadter-Like Recurrences Using Symbolic Computation, arXiv:1609.06342 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A005185, A133265, A188670, A244477, A264756, A264757, A264758, A268368, A275153, A275361, A275362.

Join[{0}, LinearRecurrence[{0, 2, 0, -1}, {2, 2, 4, 2}, 73]] (* Jean-François Alcover, Feb 19 2019 *)

a(0) = 0; thereafter, a(2n) = 2, a(2n+1) = 2n+2.
a(n) = 2*a(n-2) - a(n-4) for n>4.
G.f.: -(2*x^3 -2*x -2)/((x-1)^2*(x+1)^2).

A384932 Decimal expansion of tan(1) + sec(1).

Original entry on oeis.org

3, 4, 0, 8, 2, 2, 3, 4, 4, 2, 3, 3, 5, 8, 2, 7, 8, 4, 8, 4, 1, 8, 7, 2, 8, 0, 4, 8, 8, 5, 7, 0, 1, 0, 3, 6, 6, 5, 5, 7, 6, 4, 7, 4, 2, 7, 4, 7, 5, 5, 2, 9, 3, 3, 7, 2, 1, 9, 1, 0, 4, 8, 8, 3, 5, 5, 7, 6, 7, 6, 8, 0, 8, 4, 1, 3, 3, 2, 3, 9, 9, 5, 4, 7, 6, 9, 4

Offset: 1

Kritsada Moomuang, Jun 12 2025

The continued fraction expansion of this constant - 1 is A133265.

			3.408223442335827848481...

Cf. A049471, A073448, A133265, A248617.

RealDigits[Tan[1] + Sec[1], 10, 100, 0][[1]]

tan(1) + 1/cos(1) \\ Amiram Eldar, Jun 13 2025

Equals A049471 + A073448.
Equals 4 + Integral_{x=0..1} sin(x)/(sin(x) - 1) dx.
Equals exp(Integral_{x=0..1} sec(x) dx).
Equals exp(A248617). - Hugo Pfoertner, Jun 13 2025

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A107373 a(n) = (n/2)*binomial(n-1, floor((n-1)/2)) - 2^(n-2).

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A275365 a(1)=2, a(2)=2; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

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A384932 Decimal expansion of tan(1) + sec(1).

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