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A094864 a(0)=1, a(1)=2, a(2)=6, a(3)=18; for n >= 4, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). 3
1, 2, 6, 18, 53, 154, 443, 1264, 3582, 10092, 28291, 78962, 219541, 608318, 1680438, 4629414, 12722033, 34882954, 95451407, 260698732, 710802606, 1934955072, 5259642751, 14277467618, 38707663273, 104816737274, 283521290598, 766112145594, 2068131437357 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
LINKS
Elena Barcucci, Renzo Pinzani, and Renzo Sprugnoli , Directed column-convex polyominoes by recurrence relations, Lecture Notes in Computer Science, No. 668, Springer, Berlin (1993), pp. 282-298.
Manosij Ghosh Dastidar and Michael Wallner, Bijections and congruences involving lattice paths and integer compositions, arXiv:2402.17849 [math.CO], 2024. See p. 23.
S. Rinaldi and D. G. Rogers, How the odd terms in the Fibonacci sequence stack up, 8th Nordic Comb. Conf, Aalborg, Denmark, Oct 20 2004. [Broken link]
S. Rinaldi and D. G. Rogers, How the odd terms in the Fibonacci sequence stack up, Math. Gaz. vol 90, no 519 (2006) pp. 431-442.
FORMULA
O.g.f: -(2*x-1)*(x-1)^2/(x^2-3*x+1)^2 = (-1-2*x)/(x^2-3*x+1)+(2-5*x)/(x^2-3*x+1)^2. - R. J. Mathar, Dec 02 2007
a(n) = (2*F(2n+1)+(n-2)*L(2n-3))/5, where F(n) is the n-th Fibonacci number and L(n) is the n-th Lucas number. - Rigoberto Florez, Jul 29 2019
MATHEMATICA
Table[((n - 2)*LucasL[2*n-3] + 2*Fibonacci[2n+1])/5, {n, 1, 20}] (* Rigoberto Florez, Jul 29 2019)
LinearRecurrence[{6, -11, 6, -1}, {1, 2, 6, 18}, 40] (* Vincenzo Librandi, Jul 30 2019 *)
PROG
(PARI) Vec(-(2*x-1)*(x-1)^2/(x^2-3*x+1)^2 + O(x^40)) \\ Michel Marcus, Feb 14 2016
(Magma) I:=[1, 2, 6, 18]; [n le 4 select I[n] else 6*Self(n-1)-11*Self(n-2)+6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jul 30 2019
CROSSREFS
Sequence in context: A052979 A005507 A252822 * A120010 A132790 A358417
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jun 14 2004
STATUS
approved

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Last modified April 25 10:51 EDT 2024. Contains 371967 sequences. (Running on oeis4.)