

A111569


a(n) = a(n1) + a(n3) + a(n4) for n>3, a(0) = 1, a(1) = 1, a(2) = 2, a(3) = 1.


6



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


REFERENCES

Daniel C. Fielder, Special integer sequences controlled by three parameters. Fibonacci Quart 6, 1968, 6470.


LINKS

Table of n, a(n) for n=0..39.
Robert Munafo, Sequences Related to Floretions


FORMULA

G.f. (12*xx^2)/((x^2+x1)*(1+x^2)).
a(n) = 2*A056594(n+3)/56*A056594(n)/5+A000032(n+1)/5. [R. J. Mathar, Nov 12 2009]


PROG

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


CROSSREFS

Cf. A000045, A001638, A007598, A111570, A111571, A111572, A111573, A260259.
Sequence in context: A011016 A096540 A277081 * A213786 A055130 A051292
Adjacent sequences: A111566 A111567 A111568 * A111570 A111571 A111572


KEYWORD

easy,sign


AUTHOR

Creighton Dement, Aug 07 2005


STATUS

approved



