OPEN PROBLEMS from the FLAGSTAFF CONFERENCE-- June 2002

 Proposer Brief Statement of Problem Further notes Curtis Cooper & Michael Wiemann Fix m: Define x-1 = 0,    xn = xm - xn - 1 Let Pm(n) be a solution to the above Find a closed formula for Pm(n) Glenn Hurlbert & Rob Hochberg Inscribe n-gon in semi-circle Require integer sides If n=3==>Pythagoreean triples If n=4=>Pythagoreean Quadruple Are there infinitely many n with nested Pythagoreean n-tuples? Eric Egge & Toufik Mansour Fix n. Let Sn be all n-permutations Let R be a set of Permutations Let Sn(R)=(s in Sn: s avoids R) Find formula for  |Sn(R)| Find |Sn(R)| when R [avoids/contains] 123 and [contains/avoids] r>1 other patterns. Arthur Benjamin & Jennifer Quinn Lemma: Fn and Ln count the number of domino-square tilings of n x 1 boards/bracelets. Using this lemma find proofs for all formula in Vajda Heiko Harborth Take an empty Pascals triangle Fill the 2 diagonal borders with finite # 1,0 Fill rest of triangle by addition modulo 2 Is every n equal to the number of 1s in triangle for some initial borders? Given n, Find minimal borders such that # 1 in Pascal triangle = n? Clark Kimberling 1, 1, 2, 3, 5, 8, 4, 12, 6,..... Start sequence with 1,1 If   Sn /2 not in list then Sn+1=Sn / 2; else Sn+1 = Sn+Sn-1 Is every natural number in this list? Heiko Harborth Pick a base b  to represent numbers in Let S1 = a   be arbitrary. Let Sn+1   =  Sn + reversal of Sn  Find a and b such that no palindrom occurs in the sequence Sn Heiko Harborth Develop algorithms for magic n-gons (Generalize magic square theory) Daniel Fielder A Deception problem Russell Hendel Given k, Solve   Fn = Pk,m for (n,m) Solve generalizations Larry Somer Let un+2 = a un+1 + b un ;  Let u0=2 and u1  be arbitrary (For u1=1, a=b=1, we obtain the Lucas numbers) Discuss the density of u­n with prime factors congruent to 1 modulo 4

Notes

*1 See section 8 of Divisibility of an F-L Type Convolution, section 8 by the authors RETURN

*2 It is known that

P0(n)=1 -iff(n is odd,1,0), P1(n)=n/2 + if(n odd,1,0)/2, P2(n)=n2/2 +n/2

Pm(n) = 0.5 *( (n+1)m -S C(m,i) Pm-i (n))

Where C(m,i) is the binomial coefficient, m chose 1, and  iff(Condition,value1,value2)    equals value1 if condition is true and equals value2 otherwise RETURN

*3 Let Sn denote the set of all permutations of {1,...,n} written in one-line notation. Suppose p1, p2 in Sn. We say

p1 avoids p2 if p1 contains no subsequence with all the same pairwise comparisons of p2.

For example, 214538769 avoids 312 and 2413 but does not avoid 1243 because of the subsequence 2586   RETURN

*4. Pattern avoidance has applicability to the areas of Singularities in Shubert varieties, Chebychev Polynomials of the 2nd kind, rook polynomials for a rectangular board and various sorting algorithms RETURN

*5. Egge and Mansour in their paper 132-avoiding Two-stack Sortable Permutations, Fibonacci Numbers and Pell Numbers, found explicit formula for |Sn(R)| (Where | | denotes cardinality) when R [avoids|contains] 123 but [contains\avoids] r=1 other patterns. The proofs however involved generating functions.

Hence we have two problems: (a) Find alternate proofs using combinatoric interpretations, (b) find explicit formula for |Sn(R)| for r >1. RETURN

*6. Define a domino and square respectively as a 2 x 1 and 1 x 1 board. Then Fn is the number of distinct domino-square tilings of an n x 1 board and Ln is the number of distinct tilings of an n x 1 bracelet (Where a bracelet is an n x 1 board whose first and last squares are adjacent). RETURN

*7. Roughly 88% of the formulae in Vajda have been proved combinatorically. See the followin references

Benjamin A.T. and Quinn J.J.  Recounting Fibonacci and Lucas Identities, College Math Journal 30.5, 1999 Benjamin A.T. and Quinn J.J.  Random approaches to  Fibonacci Identities, Amer Math Mon;107.6,2000

Benjamin A.T. & Quinn J.J. & F.E. Su Counting on continued Fractions, Math Magazine, 73.2, 2000

Benjamin A.T. & Quinn J.J & F.E. Su.  General Fibonacci Identities thru Phased Tilings, Fibonacci       Quarterly 38.3, 2000

Vajda, S Fibonacci & Lucas Numbers & the Golden Section: Theory and Applications. Wiley & Sons, NY 1989 RETURN

*8. The following similar problem appeared in Crux Mathematicorum

Consider the following sequence 1,3,9,4,2,5,18....

Let   a(n) = Int(a(n-1)  /   2) if this member is not already  there

Let  a(n) = 3a(n-1) otherwise

Show that every natural number occurs in the sequence a(n) RETURN

*9. There is no solution for base 10. Solutions are known for bases that are powers of 2. RETURN

*10 The following magic hexagon was presented

 16 9 3 12 2 17 7 10 4 5 1 18 13 8 6 11 15 14 9

Some elementary necessary conditions for magic n-gons are the following

If an n-gon, with r rows,  is filled with 1,...,n then each row must sum to
n(n+1) / (2r) In particular n(n+1) / (2r) must be integral.  For each shape (hexagon, triangles, octagons) this places restrictions. RETURN

*11. Here is the full statement of the problem

- Throughout let   n    vary over the set (1,2,3)

- Let p be a prime

- Let fn(x) be irreducible polynomials modulo p

- Let rn be positive integers

- Let tn be the smallest solutions of the simultaneous eq.  1 - x tn = 0 (mod fn(x) rn)

Show that the above implies

1  -    x LCM(tn) = 0 mod (Product fn(x) rn) RETURN

*12. This problem was motivated by one of the opening remarks of the Dean of Northern Arizona University that

“ This is the 10th Fibonacci conference

1+2+...+10=55

F10 = 55"

An obvious generalization is

Find all pairs (n,m) such that Fn = Tm, where Tm is the m-th triangular number

More generally if Gn and Hm are arbitrary sequences defined by a recursive equation

find all (n,m) such that Gn = Hm. RETURN

*13. Special cases of the problem have been solved. Florian Luca was kind enough to supply a Bibliography of papers

Ming, L. On triangular Fibonacci numbers, Fibo. Quart. 27 (1989),

98

Ming, L. On triangular Lucas numbers, Applications of Fibonacci Numbers

the Netherlands, 1991, 98

Ming, L. Pentagonal numbers in the Fibonacci sequence, Appl. of

Fibonacci numbers, vol. 6, Kluwer, Dordrecht, the Netherlands, 1994,

349-354.

Ming L., Pentagonal numbers in the Lucas sequence, Portugaliae Math. 53

(1996), 352-329.

What these papers have in common is that the proof if Jacobi symbol based

and they provide succesive refinements of an idea originated in

J.H.E. Cohn, On square Fibonacci numbers, J. London Math. Soc. 39

(1964), 537-540.

Other papers are

Wayne McDaniel, Pronic Fibonacci and Lucas numbers, Fibo. Quart. 1998, 56-59 &

60-62;

Luca, Florian, Appl. of Fibonacci numbers, vol. 8, Kluwer,

Dordrecht, the Netherlands, 1999, 241-249).

General finiteness results for F_n=P(x) or L_n=P(x) with P a polynomial of

degree >1 appear in

Nemes, Petho: Polynomial values in Linear Recurrences,

II, Journal of Number Theory 24 (1986), 47-53

I found the following relevant problem  B-875 FQ 37.2 1999; 3 is the only solution to Tn = a Fermat Number RETURN

*14. Somer proved that If b is a square and a has prime factors congruent to 1 modulo 4 then

all un, n>2, have prime factors congruent to 1 modulo 4.

Divisibility of the Lucas numbers by prime factors congruent to 1 modulo 4  has been studied extensively and there are many known numerical results. The above problem is an attempt to generalize. RETURN

Web Page prepared by Russell Jay Hendel.; Comments/Corrections/References?