A292521 a(n) = a(n-2) - 2a(n-3) + a(n-4) for n>3, with a(0)=2, a(1)=0, a(2)=1, a(3)=-1, a sequence related to Pellian numbers.

2, 0, 1, -1, 3, -3, 6, -10, 15, -25, 41, -65, 106, -172, 277, -449, 727, -1175, 1902, -3078, 4979, -8057, 13037, -21093, 34130, -55224, 89353, -144577, 233931, -378507, 612438, -990946, 1603383, -2594329, 4197713, -6792041, 10989754, -17781796, 28771549, -46553345

Offset: 0

Author

Jean-François Alcover and Paul Curtz, Sep 18 2017

Keywords

• sign

Comments

Successive differences begin:

2, 0, 1, -1, 3, -3, 6, -10, 15, -25, ... = a(n)

-2, 1, -2, 4, -6, 9, -16, 25, -40, 66, ... = b(n)

3, -3, 6, -10, 15, -25, 41, -65, 106, -172, ... = a(n+4)

-6, 9, -16, 25, -40, 66, -106, 171, -278, 449, ... = b(n+4)

15, -25, 41, -65, 106, -172, 277, -449, 727, -1175, ... = a(n+8)

...

Main diagonal [2] 1, 6, 25, 106, 449, ... (omitting first term) is A048875 (Pellian numbers with second term 6).

Links

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

Crossrefs

Cf. A048875, A291660.

Programs

Mathematica

LinearRecurrence[{0, 1, -2, 1}, {2, 0, 1, -1}, 40]

PARI

x='x+O('x^99); Vec((2-x^2+3*x^3)/(1-x^2+2*x^3-x^4)) \\ Altug Alkan, Sep 18 2017

Formula

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

a(n) = A291660(-n) (negative indices computed using A291660 sequence function).

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