A240847 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

0, 0, 1, 0, 1, 0, 0, -2, -5, -12, -25, -50, -96, -180, -331, -600, -1075, -1908, -3360, -5878, -10225, -17700, -30509, -52390, -89664, -153000, -260375, -442032, -748775, -1265832, -2136000, -3598250, -6052061, -10164540

Offset: 0

Author

Paul Curtz, Apr 13 2014

Keywords

• sign
• easy

Comments

F1(m, n) is the difference table of a(n):

0, 0, 1, 0, 1, 0, 0, -2, ...

0, 1, -1, 1, -1, 0, -2, -3, ...

1, -2, 2, -2, 1, -2, -1, -4, ...

-3, 4, -4, 3, -3, 1, -3, -2, ...

7, -8, 7, -6, 4, -4, 1, -4, ...

-15, 15, -13, 10, -8, 5, -5, 1, ...

30, -28, 23, -18, 13, -10, 6, -6, ...

The recurrence holds for every row and every signed column.

Main diagonal: F1(n, n) = A001477(n).

First upper diagonal: F1(n, n+1) = -A001477(n).

F1(m, n) = F1(m, n-1) + F1(m+1, n-1).

Inverse binomial transform: 0, 0, 1, -3, 7, -15, 30, ... = 0, 0, followed by (-1)^n*A023610(n). Without signs: F2(0, n) = 0, 0, 1, 3, 7, 15, 30, ... = b(n) has the same recurrence.

F1(0, n) + F2(0, n) = 0, followed by A099920(n).

a(n) and b(n) are reciprocal by their inverse binomial transform.

0, followed by A001629(n) is an autosequence.

F1(m, 1) = (-1)^n*A029907(n).

F1(1, n) = 0, 1, -1, 1, -1, followed by -A226432(n+3).

F1(m, 2) = (-1)^n*A208354(n).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Crossrefs

Cf. A000032, A000045, A001629 (main sequence for the recurrence), A067331.

Programs

GAP

List([0..40], n-> (6*Fibonacci(n-3) - (n-3)*Lucas(1,-1,n-3)[2])/5 ); # G. C. Greubel, Feb 06 2020

Magma

[(6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5: n in [0..40]]; // G. C. Greubel, Feb 06 2020

Maple

with(combinat): seq( ((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5, n=0..40); # G. C. Greubel, Feb 06 2020

Mathematica

a[n_]:= a[n]= 2*a[n-1] +a[n-2] -2*a[n-3] -a[n-4]; a[0]= a[1]= a[3]= 0; a[2]= 1; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Apr 17 2014 *)
CoefficientList[Series[x^2*(1-2*x)/(1-x-x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+c-2b-a}; NestList[nxt,{0,0,1,0},40][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)

PARI

Vec(x^2*(1-2*x)/(1-x-x^2)^2 + O(x^100)) \\ Colin Barker, Apr 13 2014

PARI

vector(41, n, my(m=n-1); ((m+3)*fibonacci(m-3) - 2*(m-3)*fibonacci(m-2) )/5 ) \\ G. C. Greubel, Feb 06 2020

Sage

[((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5 for n in (0..40)] # G. C. Greubel, Feb 06 2020

Formula

a(n) = 0, 0, 1, 0, 1, 0, 0, followed by -A067331.

G.f.: x^2*(1-2*x)/(1-x-x^2)^2. - Colin Barker, Apr 13 2014

a(n) = ( (10*n + (3-5*n)*t)*(1+t)^n + (10*n-(3-5*n)*t)*(1-t)^n )/(25*2^n), where t=sqrt(5). - Bruno Berselli, Apr 17 2014

a(n) = (6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5 = ((n+3)*Fibonacci(n-3) - 2*(n-3)*Fibonacci(n-2))/5. - G. C. Greubel, Feb 06 2020