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A054444
Even-indexed terms of A001629(n), n >= 2, (Fibonacci convolution).
7
1, 5, 20, 71, 235, 744, 2285, 6865, 20284, 59155, 170711, 488400, 1387225, 3916061, 10996580, 30737759, 85573315, 237387960, 656451269, 1810142185, 4978643596, 13661617195, 37409025455, 102238082976, 278920277425, 759695287349
OFFSET
0,2
COMMENTS
8*a(n) is the number of Boolean (equivalently, lattice, modular lattice, distributive lattice) intervals of the form [s,w] in the Bruhat order on S_n, where s is a simple reflection. - Bridget Tenner, Jan 16 2020
LINKS
Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Honeycombs in the Pascal triangle and beyond, arXiv:2203.13205 [math.HO], 2022. See p. 5.
É. Czabarka, R. Flórez, and L. Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), #15.1.6.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
B. E. Tenner, Interval structures in the Bruhat and weak orders, arXiv:2001.05011 [math.CO], 2020.
FORMULA
a(n) = ((2*n+1)*F(2*(n+1)) + 4*(n+1)*F(2*n+1))/5, with F(n) = A000045(n) (Fibonacci numbers).
a(n) = A060920(n+1, 1).
G.f.: (1 - x + x^2)/(1 - 3*x + x^2)^2.
a(n) = Sum_{k=1..n+1} k*binomial(2*n-2*k+2, k). - Emeric Deutsch, Jun 11 2003
a(n) ~ n*(3 + sqrt(5))^(1+n)*2^(-n)/5. - Stefano Spezia, Mar 29 2022
PROG
(PARI) a(n) = ((2*n+1)*fibonacci(2*(n+1))+4*(n+1)*fibonacci(2*n+1))/5; \\ Jinyuan Wang, Jul 28 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Wolfdieter Lang, Apr 07 2000
STATUS
approved

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Last modified September 22 20:46 EDT 2024. Contains 376140 sequences. (Running on oeis4.)