not solved problem

Lallouet philip.lallouet at
Sat Aug 2 14:35:44 CEST 2008

Dear seqfans

I found, but i am not probably the first, that starting fron any pair of real OR complex numbers a(1) and b(1), the two associated sequences defined by the following genarating functions :

(1)    a(n) = sqrt[a(n-1)]+sqrt[b(n-1]

(2)    b(n) = sqrt[(a(n-1)+b(n-1)]

have always the same limits A and B where

A=3,394547695. . . .= B*(B-1)

B=2,409069851 , the greater of the two real roots of y^6-4*y^5+4*y^4-4*y+4=0

To remain in the domain of integers , let us start with a pair of integer numbers c(1) and d(1)<=c(1) and replace the genarating functions by

(3)    c(n) = floor{sqrt[a(n-1)]+sqrt[b(n-1)]}

(4)    d(n) =floor{sqrt[a(n-1)]+sqrt[b(n-1)]}

We find that for any starting pair, the two associated sequences have always the same limits C=3 and D=2

Starting in this way with big enough values of c(1) and d(1)<= c(1), and reversing the sequences we get two finite sequences, as long as we want , beginning with the pair  e(1)=3 and f(1)=2  and whose terms satisfy 

the functions and inegality

(5)   e(n-1)=floor{sqrt[e(n)] + sqrt[f(n)]}

(6)   f(n-1)=floor{sqrt[e(n)+f(n)]} 

(7)   e(n) >= f(n)

but these functions don't define univocally e(n) and f(n) as successors of e(n-1) and f(n-1)

(for example the pair (3 3) has only one successor pair = (8 1) but 25 different pairs, from (5 4) to (14 1), have for ancestors the pair (4 3) 

I wonder if it's possible to define generating functions enabling to build a pair of infinite sequence of this type (different of the sequences of constant terms 3,3,3 . . . and,2,2,2 . . . ), i.e starting with the pair (3 2), 

and satisfying the conditions (5), (6) ans (7) (maybe by adding a rule fixing the choice among all the pairs e(n) and f(n) satisfying these relations)

Is somebody interested by this problem?

Best regards to each of you.

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