A111569 a(n) = a(n-1) + a(n-3) + a(n-4) for n > 3, a(0) = -1, a(1) = 1, a(2) = 2, a(3) = 1.

-1, 1, 2, 1, 1, 4, 7, 9, 14, 25, 41, 64, 103, 169, 274, 441, 713, 1156, 1871, 3025, 4894, 7921, 12817, 20736, 33551, 54289, 87842, 142129, 229969, 372100, 602071, 974169, 1576238, 2550409, 4126649, 6677056, 10803703, 17480761, 28284466, 45765225

Offset

0,3

Comments

In reference to the program code given, 4*tesseq[A*H] = A001638 (a Fielder sequence) where A001638(2n) = L(n)^2. Here we have: a(2n+1) = A007598(n+1) = Fibonacci(n+1)^2.

Floretion Algebra Multiplication Program, FAMP Code: 4kbaseiseq[B+H] with B = - .25'i + .25'j - .25i' + .25j' + k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and H = + .75'ii' + .75'jj' + .75'kk' + .75e

First bisection is A260259 (see previous comment for the second bisection). [Bruno Berselli, Nov 02 2015]

Links

Table of n, a(n) for n=0..39.

Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.

Robert Munafo, Sequences Related to Floretions.

Index entries for linear recurrences with constant coefficients, signature (1, 0, 1, 1).

Formula

G.f.: (1-2*x-x^2)/((x^2+x-1)*(1+x^2)).

a(n) = 2*A056594(n+3)/5 - 6*A056594(n)/5 + A000032(n+1)/5. [R. J. Mathar, Nov 12 2009]

Crossrefs

Cf. A000045, A001638, A007598, A111570, A111571, A111572, A111573, A260259.

Sequence in context: A011016 A096540 A277081 * A371926 A213786 A055130

Adjacent sequences: A111566 A111567 A111568 * A111570 A111571 A111572

Keyword

easy,sign

Author

Creighton Dement, Aug 07 2005

Status

approved