By Krassimir T Atanassov

A research of Fibonacci sequences and the well known Fibonacci numbers. it may be of curiosity to investigate mathematicians wishing to boost the tips themselves, and to leisure mathematicians, who should still benefit from the a variety of visible ways and the issues inherent in them. there's a carrying on with emphasis on diagrams, either geometric and combinatorial, which is helping to tie disparate subject matters jointly, weaving round the unifying subject matters of the golden suggest and diverse generalizations of the Fibonacci recurrence relation. little or no earlier mathematical wisdom is thought, except the rudiments of algebra and geometry, so the textual content can be utilized as a resource of enrichment fabric and venture paintings for students. A bankruptcy on video games utilizing goldpoint tiles is integrated on the finish, and it seeks to supply a lot fabric for exciting mathematical actions concerning geometric puzzles of a combinatoric nature

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1 and 2). [10] R. Ackoff, The art of problem solving, A Wiley-Interscience Pu blication, New York, 1978. [11] J. Kleijnen, Statistical techniques in simulation, M Dekker INC, New York, 1974 (Volss. 1 and 2). [12] P. Fishburn, Utility theory for decision making, J. Wiley & Sons, INC, New York, 1970. [13] H Wagner, Principles of operations research, Prectice-Hall, Inc. 1969. [14] J. Kanter, Management-oriented management information systems, Prentice-Hall Inc. , Englewood Clitts, New Jersay, 1977.

False . . false W I false I| I I false ... false W = i y x' 1,2 y' iI r ... -- L' , 1, IL' i-l i, 1-1 Z Z' 1 and W -. " ( 3 T : 1 £ T i |L' l)((f(r ) = 1) & (m > 0) & Z' p, T p, T 1

SMGN In § 9. 1 [x] we have shown that if a fixed UGN had started functioning before a given GN, then the UGN may not to be able present the functioning of rem 1 [17] the given GN. However, to re the proof of Theo is wrong. Following Theorem 2 [17] and Theorem 1 of § 9 in App. l, we construct a UGN for which (a) T = maxfT / (3E |3E € £ £)(T r = pr pr E)i 1 3 (it is possible: T = - oo); -o o -o (b) t - minft / (3E e £|(t = pr pr E)l E)] 2 3 -o (everywhere t > 0) and -K (c) t K -K = maxft / (3E € £)