OFFSET
0,8
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).
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
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 *)
PROG
(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
(Magma) [(6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5: n in [0..40]]; // 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
(GAP) List([0..40], n-> (6*Fibonacci(n-3) - (n-3)*Lucas(1, -1, n-3)[2])/5 ); # G. C. Greubel, Feb 06 2020
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Paul Curtz, Apr 13 2014
STATUS
approved