OFFSET
1,1
COMMENTS
-a(n+3) = 2*n^2 - 4, n >= 0, [-4,-2, 4, 14, ...] appears as the first member of the quartet for the square of [n, n+1, n+2, n+3], for n >= 0, in the Clifford algebra Cl_2. The other members are given in A046092(n), A054000(n+1) and A139570(n). The basis of Cl_2 is <1, s1, s2, s12> with s1.s1 = s2.s2 = 1, s12.s12 = -1, s1.s2 = -s2.s1 = s12. See e.g., pp. 5-6, eqs. (2.4)-(2.13) of the S. Gull et al. reference. - Wolfdieter Lang, Oct 15 2014
Related to the previous comment: if one uses the exterior (Grassmann) product with s1.s1 = s2.s2 = s12.s12 = 0 and s1.s2 = -s2.s1 = s12, then the four components of the square of [n, n+1, n+2, n+3] are [A000290(n), A046092(n), A054000(n+1), A139570(n)], n >= 0. - Wolfdieter Lang, Nov 13 2014
2 - a(n)/2 is a square. - Bruno Berselli, Apr 10 2018
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
S. Gull, A. Lasenby and C. Doran, Imaginary Numbers are not Real - the Geometric Algebra of Spacetime, Found. Phys., Vol. 23(9) (1993), pp. 1175-1201.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 10 2012
a(n) = -2*A008865(n-3). - J. M. Bergot, Jun 25 2018
G.f.: -2*x*(2 - 7*x + 7*x^2)/(1 - x)^3. - Colin Barker, Feb 12 2019
E.g.f.: -2*(exp(x)*(x^2 - 5*x + 7) - 7). - Elmo R. Oliveira, Nov 17 2024
MAPLE
[-2*n^2+12*n-14$n=1..50]; # Muniru A Asiru, Feb 12 2019
MATHEMATICA
lst={}; Do[k=n^2-((n-1)^2+(n-2)^2+(n-3)^2); AppendTo[lst, k], {n, 5!}]; lst
Table[-2n^2+12n-14, {n, 1, 50}] (* Vincenzo Librandi, Jul 10 2012 *)
LinearRecurrence[{3, -3, 1}, {-4, 2, 4}, 50] (* Harvey P. Dale, Mar 02 2020 *)
PROG
(Magma) [-2*n^2+12*n-14: n in [1..50]]; // Vincenzo Librandi, Jul 10 2012
(PARI) a(n)=-2*n^2+12*n-14 \\ Charles R Greathouse IV, Sep 24 2015
(PARI) Vec(-2*x*(2 - 7*x + 7*x^2) / (1 - x)^3 + O(x^40)) \\ Colin Barker, Feb 12 2019
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Nov 18 2008
STATUS
approved