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carpetpool 2018-04-21 16:32

Heptagon Flat
The Lucas Sequences U(P,Q,n) share a property similar to the cyclotomic sequences x^n-1 by the fact that if n does not divide Q and n is prime, then either U(P,Q,n-1) or U(P,Q,n+1). This is the equivalent to Fermat's Little Theorem for cyclotomic numbers x^n-1 where if n is prime and n does not divide x, then n divides x^(n-1)-1, or as we see it, x^(n-1) = 1 modulo n. It is practical to consider other types of sequences which share a property of Fermat's Little Theorem.

is the heptagonal version of the Fibonacci Series, and for its generalized sequences a(x,n), it is the case that prior to some small restrictions (such as n not dividing x) either n divides a(x,n-1) or a(x,n^2+n+1).





How the cyclotomic, Fibonacci, and Lucas polynomials were recovered, I am trying to recover the generalizing polynomials for the heptagonal Fibonacci Sequences. Thanks for help.

Dr Sardonicus 2018-04-22 15:35

The numbers may be formulated exactly in terms of

f = x^3 - x^2 - 2*x + 1

by taking

a = Mod(x,f) and b = Mod(x^2 - 1, f).

The quantity b has characteristic polynomial x^3 - 2*x^2 - x + 1, the reciprocal polynomial to f.

In order to express

a^n * b^m = C1 + C2*a + C3*b,

it suffices to find the dual basis, with respect to the trace, of

[1, Mod(x, f), Mod(x^2 - 1, f)].

The dual basis [t1, t2, t3] with respect to [Mod(1, f), Mod(x, f), Mod(x^2, f)] is easily found (Newton's identities). The required basis is obviously

[w1, w2, w3] = [t1 + t3, t2, t3].

Then we obtain

C1 = trace(w1 * a^n * b^m), C2 = trace(w2 * a^n * b^m), C3 = trace(w3 * a^n * b^m).

We have

[w1, w2, w3] = [Mod(-1/7*x^2 + 4/7, x^3 - x^2 - 2*x + 1), Mod(-1/7*x^2 + 2/7*x + 1/7, x^3 - x^2 - 2*x + 1), Mod(2/7*x^2 - 1/7*x - 3/7, x^3 - x^2 - 2*x + 1)].

In particular, the row of the table in the web page for b^m is given by

trace(w1 * b^m * a^n), n = -2 to 6.

The column for a^n is given by

trace(w[1]* a^n * b^m), m = -1 to 7.

The numbers in any given row satisfy the recurrence

C[sub]n+3[/sub] = C[sub]n+2[/sub] + 2*C[sub]n+1[/sub] - C[sub]n[/sub]

while those in a given column satisfy

C[sub]n+3[/sub] = 2*C[sub]n+2[/sub] + C[sub]n+1[/sub] - C[sub]n[/sub].

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