OFFSET
0,3
COMMENTS
Let (a_n) be the sequence and (a_(n+1)) the sequence beginning at 1. Let B and iB be the binomial and inverse binomial transforms, respectively. Then B((a_n)) = A001108(n) (a(n)-th triangular number is a square); B((a_(n+1))) = A002315(n) (NSW Numbers); iB((a_(n+1))) = A096980(n). Note: a 2nd sequence generated by the same floretion is A057087 (Scaled Chebyshev U-polynomials evaluated at i. Generalized Fibonacci sequence.). As is often the case with two sequences corresponding to a single floretion, both satisfy the same recurrence relation.
Floretion Algebra Multiplication Program, FAMP Code: (a_n) = 2ibasekseq[A*B] (with initial term zero), (a_(n+1)) = 1tesseq[A*B], A = + .5'i - .5'j + .5'k + .5i' - .5j' + .5k' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; B = - .5'i + .5'j + .5'k - .5i' + .5j' + .5k' - .5'ik' - .5'jk' - .5'ki' - .5'kj'
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (4,4).
FORMULA
a(n+1) = -(1/2)*(2-2*2^(1/2))^n*(-1+2^(1/2))-(1/2)*(2+2*2^(1/2))^n(-1-2^(1/2)); G.f.: x*(1+2*x)/(1-4*x-4*x^2).
a(n) = sum{k=0..n, (-1)^k*C(n-1, k)*(Pell(2n-2k)-Pell(2n-2k-1))}, n>0, where Pell(n) = A000129(n). - Paul Barry, Jun 07 2005
a(n+1) = ((3+sqrt18)(2+sqrt8)^n+(3-sqrt18)(2-sqrt8)^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009, index corrected Jul 11 2012
a(n) = 2^(n-1) * A001333(n), n>0. - Ralf Stephan, Dec 02 2010
MATHEMATICA
CoefficientList[Series[x*(1+2*x)/(1-4*x-4*x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 26 2012 *)
PROG
(Magma) I:=[0, 1, 6]; [n le 3 select I[n] else 4*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Creighton Dement, Jun 01 2005
STATUS
approved