

A092942


A Fibonacci sequence with "corrections" at every third step: ++++++++++... i.e. at every 3rd step there is a subtraction instead of addition.


3



0, 1, 1, 2, 3, 1, 4, 5, 1, 6, 7, 1, 8, 9, 1, 10, 11, 1, 12, 13, 1, 14, 15, 1, 16, 17, 1, 18, 19, 1, 20, 21, 1, 22, 23, 1, 24, 25, 1, 26, 27, 1, 28, 29, 1, 30, 31, 1, 32, 33, 1, 34, 35, 1, 36, 37, 1, 38, 39, 1, 40
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OFFSET

0,4


COMMENTS

The sequence is rather simple. It becomes more interesting if you apply other periodic correction patterns. What is also interesting that it (and related sequences like 0,1,1,0,1,1,0,1,1,0,...) was used to cryptanalyse RC5 blockcipher since it describes the Hamming weight of a difference if at every 3rd step there is no data rotation. Since the attacker has to pay in probability to cause no rotations, the related question was how many corrected Fibonacci sequences with up to m corrections are there. The paper contains a recursive program that enumerates all "corrected" Fibonacci sequences of length N, with up to m corrections (in that case we do not restrict the locations of the corrections).
0, 1, 1, 2, 3, 5, 2, 7, 9, 16, 7, 23, 30, 53... = Fibonacci with corrections at every 4th step.


LINKS

Table of n, a(n) for n=0..60.
A. Biryukov home page
A. Biryukov and E. Kushilevitz, Improved Cryptanalysis of RC5, Lecture Notes in Computer Science 1403, Proceedings of EUROCRYPT'98, pp. 8599, 1998.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,1).


FORMULA

a(n) = a(n1) + a(n2); if n = 3k, n=3k+1, for k=1, 2, 3, .. a(n) = a(n1)  a(n2); if n = 3k+2, for k=0, 1, 2, 3, ... a(0) = 0, a(1) = 1;
G.f.: x*(1+x)*(x^32*x^21) / ( (x1)^2*(1+x+x^2)^2 ).  R. J. Mathar, Dec 15 2014


CROSSREFS

Sequence in context: A104706 A094137 A038802 * A229137 A282510 A131225
Adjacent sequences: A092939 A092940 A092941 * A092943 A092944 A092945


KEYWORD

nonn,easy


AUTHOR

Alex Biryukov, Apr 19 2004


STATUS

approved



