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• ## Collatz type functions using Legendre/ Jacobi symbols

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• Has anyone looked at Collatz-type functions, using functions such as the Jacobi/ Legendre symbols? The function would look something like Start with an
Message 1 of 2 , May 9, 2006
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Has anyone looked at "Collatz-type" functions, using functions such as
the Jacobi/ Legendre symbols?

The function would look something like

check its Legendre symbol, base y, a prime
If Legendre[x[1]/y] = 0 then x[2] = some function of x[1] F(x1) such
as x[1]+1
If Legendre[x[1]/y] = 1 then x[2] = some other function F(x2) of x[1]
If Legendre[x[1]/y) = -1 then x[2] = a third function F(x3) of x[1]

Repeat, looking at the Legendre symbols of x[2] to output x[3]. Then
stop at x[n], either a repeat of a previous x[] (loop) or x[n-1]
(constant future values, preferably 1).

There does not appear to be a constant definition of the Legendre and
Jacobi Symbols, some sources, such as Zzmath for Excel, interpet
Jacobi[2/2] as 1, whereas Maple shows this as 0. MathWorld has a
statement that the Legendre symbol definiation is sometimes
generalised such that Legendre[a/p]=0 when a|p

What would be interesting choices of F[x1],F[x2] and F[x3]?

Regards

Robert Smith
• Robert wrote: Has anyone looked at Collatz-type functions, using functions such as the Jacobi/ Legendre symbols? The function
Message 1 of 2 , May 10, 2006
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Robert <rw.smith@...> wrote: Has anyone looked at "Collatz-type" functions, using functions such as
the Jacobi/ Legendre symbols?

The function would look something like

check its Legendre symbol, base y, a prime
If Legendre[x[1]/y] = 0 then x[2] = some function of x[1] F(x1) such
as x[1]+1
If Legendre[x[1]/y] = 1 then x[2] = some other function F(x2) of x[1]
If Legendre[x[1]/y) = -1 then x[2] = a third function F(x3) of x[1]

Repeat, looking at the Legendre symbols of x[2] to output x[3]. Then
stop at x[n], either a repeat of a previous x[] (loop) or x[n-1]
(constant future values, preferably 1).

There does not appear to be a constant definition of the Legendre and
Jacobi Symbols, some sources, such as Zzmath for Excel, interpet
Jacobi[2/2] as 1, whereas Maple shows this as 0. MathWorld has a
statement that the Legendre symbol definiation is sometimes
generalised such that Legendre[a/p]=0 when a|p

What would be interesting choices of F[x1],F[x2] and F[x3]?

Regards

Robert Smith

No I haven't, but I' ve always wondered why 3n+1, in the case of odd numbers and n/2 in the case of even numbers, cannot simply be collapsed into n+((n+1)/2)) (in the case of odd numbers), thus jumping a step.

Regards

Bob

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