0,3

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.

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 (pbarry(AT)wit.ie), Jun 07 2005

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'

Cf. A057087, A001108, A002315, A096980.

Sequence in context: A002693 A117423 A084778 this_sequence A001599 A074247 A053783

Adjacent sequences: A108048 A108049 A108050 this_sequence A108052 A108053 A108054

easy,nonn

Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jun 01 2005