A033190 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(Fibonacci(k)+1,2).

0, 1, 3, 9, 28, 90, 297, 1001, 3431, 11917, 41820, 147918, 526309, 1881009, 6744843, 24244145, 87300092, 314765506, 1135980801, 4102551897, 14823628015, 53581222773, 193724727804, 700551945014, 2533702591613, 9164618329825, 33151607475987, 119927166988761

Offset

0,3

Comments

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 3. - Herbert Kociemba, Jun 14 2004

Links

Michael De Vlieger, Table of n, a(n) for n = 0..1792

László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.

Index entries for linear recurrences with constant coefficients, signature (8,-21,20,-5).

Formula

G.f.: (-x^4+6x^3-5x^2+x)/((1-3x+x^2)*(1-5x+5x^2)).

From Herbert Kociemba, Jun 14 2004: (Start)

a(n) = (1/5)*Sum_{r=1..9} sin(r*Pi/10)*sin(3*r*Pi/10)*(2*cos(r*Pi/10))^(2*n), n >= 1.

a(n) = 8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4), n >= 5. (End)

From Greg Dresden, Jan 24 2021: (Start)

a(2n) = (5*Fibonacci(4*n) + (5^n)*Lucas(2*n))/10 for n > 0.

a(2n+1) = (Fibonacci(4*n+2) + (5^n)*Fibonacci(2*n+1))/2 for n >= 0.

(End)

Maple

A033190 := proc(n)
    add(binomial(n, k)*binomial(combinat[fibonacci](k)+1, 2), k=0..n) ;
end proc: # R. J. Mathar, Feb 18 2016

Mathematica

LinearRecurrence[{8, -21, 20, -5}, {0, 1, 3, 9, 28}, 30] (* Harvey P. Dale, Jan 24 2019 *)

Crossrefs

Sequence in context: A094164 A094803 A094826 * A370074 A071724 A000245

Adjacent sequences: A033187 A033188 A033189 * A033191 A033192 A033193

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved