A291660 a(n) = 2*a(n-1) - a(n-2) + a(n-4) for n>3, a(0)=2, a(1)=3, a(2)=5, a(3)=7, a sequence related to Lucas numbers.

2, 3, 5, 7, 11, 18, 30, 49, 79, 127, 205, 332, 538, 871, 1409, 2279, 3687, 5966, 9654, 15621, 25275, 40895, 66169, 107064, 173234, 280299, 453533, 733831, 1187363, 1921194, 3108558, 5029753, 8138311, 13168063, 21306373, 34474436, 55780810, 90255247, 146036057, 236291303

Offset: 0

Author

Jean-François Alcover and Paul Curtz, Aug 31 2017

Keywords

• nonn

Comments

The array of successive differences begins:

2, 3, 5, 7, 11, 18, 30, 49, 79, 127, ... = a(n)

1, 2, 2, 4, 7, 12, 19, 30, 48, 78, ... = b(n)

1, 0, 2, 3, 5, 7, 11, 18, 30, 49, ... = a(n-2)

-1, 2, 1, 2, 2, 4, 7, 12, 19, 30, ... = b(n-2)

3, -1, 1, 0, 2, 3, 5, 7, 11, 18, ... = a(n-4)

...

Main diagonal is 2,2,2,... = A007395.

Adding a(n) and first column with alternating signs, one gets two autosequences: 2*Lucas numbers A000032 (2, 1, 3, 4, 7, 11, 18, ...) or 2*A286350 (0, 2, 2, 3, 4, 7, 12, ...) according to signs.

Links

• OEIS Wiki, Autosequence
• Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 1).

Crossrefs

Cf. A000032, A097924, A286350.

Programs

GAP

L:=[2,3,5,7];; for i in [5..10^3] do L[i]:=2*L[i-1]-L[i-2]+L[i-4]; od; L;  #  Muniru A Asiru, Sep 02 2017

Mathematica

LinearRecurrence[{2, -1, 0, 1}, {2, 3, 5, 7}, 40]

Formula

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

a(3n) = A097924(n).

a(3n) + a(3n+1) = a(3n+2).

a(n) = (1/15)*2^(-n-1)*((30-9*sqrt(5))*(1-sqrt(5))^n + (1+sqrt(5))^n*(30 + 9*sqrt(5)) + 5*2^(n+1)*sqrt(3)*sin(n*Pi/3)).