Recall from part 1 that a sequence is said to be of Fibonacci type if it’s given by the recursion relation *a*(*n* + *t*) = *k*(*t*)*a*(*n* + *t* – 1) + *k*(*t* – 1)*a*(*n* + *t* – 2) + … + *k*1a(*n*), with set initial conditions on *a*1, *a*2, …, and *a*(*t*). Recall also that every such sequence is given by a linear combination of sequences of the form *a*(*n*) = (*n*^*k*)(*r*^*n*), where *r* is a root of the associated polynomial *x*^*t* – *k*(*t*)*x*^(*n* + *t* – 1) – … – *k*1 = 0 and *k* is a number between 0 and one less than the multiplicity of *r*.

A Carnival of Mathematics submission that notes how a convoluted sequence that generates all integers not divisible by 2, 3, or 5 has just enough material that Fibonnaci-type sequences are relevant to to prompt me to write a follow up, showing a few additional properties of those sequences.

1. Trivially, a sequence is Fibonacci-type iff it can be written as a sum *c*1(*n*^*k*1)(*r*1^*n*) + … + *c*(*m*)(*n*^*k*(*m*))(*r*(*m*)^*n*). The only if direction is clear from the calculation in the previous post, while the if direction follows by creating a polynomial for which every *r*(*i*) is a root of multiplicity at least *k*(*i*) + 1, such as the least common multiple of (*x* – *r*(*i*))^(*k*(*i*) + 1) over all *i*.

2. Every periodic sequence is Fibonacci-type. This is because a periodic sequence is defined by the relation *a*(*n* + *t*) = *a*(*n*) for some *t*, which provides a suitable recursion relation.

3. The sum and difference of two Fibonacci-type sequences are themselves Fibonacci-type. This follows directly from #1, because there is no restriction on the possible values of *r*(*i*) and *k*(*i*).

4. The product of two Fibonacci-type sequences is Fibonacci-type. To see this, multiply elements of the form (*n*^*k*(*i*))(*r*(*i*)^*n*) pointwise. We have (*n*^*k*1)(*r*1^*n*)(*n*^*k*2)(*r*2^*n*) = (*n*^(*k*1 + *k*2))((*r*1*r*2)^*n*).

5. If the associated polynomial of *a*(*n*) is *f*(*x*) and this of *b*(*n*) is *g*(*x*), then this of *a*(*n*) + *b*(*n*) divides the lowest common multiple of *f* and *g*. This follows from the fact that if *r* is a root of the new associated polynomial of multiplicity *k* + 1, then (*n*^*k*)(*r*^*n*) appears somewhere in *a*(*n*) + *b*(*n*), so it must appear in *a*(*n*) or *b*(*n*).

6. A Fibonacci-type sequence can be continued to zero and negative values of *n* by reversing the recursion relation to *a*(*n*) = (*a*(*n* + *t*) – *k*(*t*)*a*(*n* + *t* – 1) – … – *k*2*a*(*n* + 1))/*k*1.

7. Specifying any *t* distinct points of a Fibonacci-type sequence and its recursion relation is enough to determine it. Usually the points specified are *a*1, *a*2, …, and *a*(*t*), but any *t* points will suffice, since they will provide *t* linear equations using the basis elements (*n*^*k*)(*r*^*n*) that allow recovering all values of *c*(*i*). These points can of course correspond to negative values of *n*.

8. Given any Fibonacci-type sequence *a*(*n*), the shifted sequence *b*(*n*) = *a*(*n* + 1) is Fibonacci-type. This is obvious from the recursion relations, which are invariant under shifting. Changing *n* to *n* + 1 changes (*n*^*k*)(*r*^*n*) to ((*n* + 1)^*k*)(*r*^(*n* + 1)) = *r*((*n* + 1)^*k*)(*r*^*n*) = *r*(*n*^*k* + *kn*^(*k* – 1) + (*k*(*k* – 1)/2)*n*^(*k* – 2) + … + *kn* + 1)(*r*^*n*), which corresponds to the same polynomial as (*n*^*k*)(*r*^*n*), (*x* – *r*)^(*k* + 1). Using #6, this also applies to the shifted sequence *b*(*n*) = *a*(*n* – 1), except of course with a slightly modified change to the basis elements.

9. It makes sense to define the derivative of *a*(*n*), *a*‘(*n*), as *a*(*n*) – *a*(*n* – 1); it shares some characteristics with the derivative of a function. If *a*(*n*) is Fibonacci-type, then so is *a*‘(*n*) from #3 and #8. By observing the action of differentiation on each (*n*^*k*)(*r*^*n*), it follows that *a*‘(*n*) obeys the same recursion relations as *a*(*n*).

10. The general root *r* of the associated polynomial of *a*(*n*) appears in the associated polynomial of *a*‘(*n*) with the same multiplicity. That the multiplicity in *a*‘(*n*) is no higher than in *a*(*n*) follows from #9. For the other direction, if the multiplicity of *r* in *a*(*n*) is *k* + 1, and the coefficient of (*n*^*k*)(*r*^*n*) in *a*(*n*) is the nonzero number *c*1, then the coefficient of (*n*^*k*)(*r*^*n*) in *a*‘(*n*) is *c*1(*r* – 1), from #6. But if *r* = 1, then the multiplicity goes down by 1, since *r* – 1 = 0. Indeed, observe that *n*^*k* – (*n* – 1)^*k* = *kn*^(*k* – 1) – (*k*(*k* – 1)/2)*n*^(*k* – 2) + … + (-1)^(*k* + 1) is a polynomial of degree *k* – 1.

11. It makes sense to define the inverse differentiation operator, integration, by int(*a*)(*n*) = *a*1 + *a*2 + *a*3 + … + *a*(*n*). It’s not difficult to see that int(*a*)'(*n*) = *a*(*n*). If *a*(*n*) is Fibonacci-type then so is int(*a*)(*n*), with the same associated polynomial except that if 1 is a root then its multiplicity goes up by 1. Proving it is complicated, so I’ll do it in stages:

11a. If *a*(*n*) =* n*^*k*, then int(*a*)(*n*) is a polynomial of degree *k* + 1. For that, let *p*(*n*) = *c*(*k* + 1)*n*^(*k* + 1) + *c*(*k*)*n*^*k* + … + *c*0. We want the *n*^*k*-coefficient of *p*(*n* – 1) to be 1 less than this of *p*(*n*), so that *c*(*k* + 1)(*n*^(*k* + 1) – (*n* – 1)^(*k* + 1)) = *n*^*k* + *q*(*n*) where deg(*q*) < *k*; for that, we expand (*n* – 1)^(*k* + 1) binomially to get *c*(*k* + 1) = 1/(*k* + 1). By a similar process, we equate the *c*(*k*) coefficients to get *c*(*k*) = 1/2, and so on until we get *c*0 = 0. That polynomial was constructed to have the recursion relation *p*(*n* + 1) = *p*(*n*) + *n*^*k* and the initial condition *p*(0) = 0, so it’s indeed int(*a*)(*n*).

11b. If *a*(*n*) = *r*^*n* and *r* != 1, then int(*a*)(*n*) is a multiple of *r*^*n* plus a constant. To see why, note that (*r* – 1)(*r*^*n* + *r*^(*n* – 1) + … + 1) telescopes to *r*^(*n* + 1) – 1, so that int(*r*^*n*) = (*r*^(*n* + 1) – 1)/(*r* – 1) = (*r*/(*r* – 1))*r*^*n* – 1/(*r* – 1). That turns *a*(*n*) into a Fibonacci-type sequence whose associated polynomial is (*x* – *r*)(*x* – 1). To get rid of the extra root, we need to sum from negative infinity when |*r*| > 1, i.e. look at (*r* – 1)(*r*^*n* + *r*^(*n* – 1) + …) = *r*^(*n* + 1). When |*r*| < 1, we need to instead look at the negative of the sum from *n* to positive infinity, which will be (*r* – 1)(-*r*^*n* – *r*^(*n* + 1) – …) = -*r*^(*n* + 1). When |*r*| = 1, this sum does not converge and is therefore meaningless.

11c. If *a*(*n*) = (*n*^*k*)(*r*^*n*) and *r* != 1, then int(*a*)(*n*) is Fibonacci-type. For that, assume that this is true for all exponents smaller than *k*, and in particular for all polynomials of degree at most *k* – 1. Then write *r*int(*a*)(*n*) = *r*^2 + (2^*k*)*r*^3 + … + (*n*^*k*)(*r*^(*n* + 1)) = *r* + (2^*k*)*r*^2 + (3^*k*)*r*^3 + … + ((*n* + 1)^*k*)*r*^(*n* + 1) – *r* – (2^*k* – 1)*r*^2 – (3^*k* – 2^*k*)*r*^3 – … – ((*n* + 1)^*k* – *n*^*k*)*r*^(*n* + 1). By the definition of int(*a*)(*n*), we get (*r* – 1)int(*a*)(*n*) = ((*n* + 1)^*k*)*r*^(*n* + 1) – *r* – (2^*k* – 1)*r*^2 – (3^*k* – 2^*k*)*r*^3 – … – ((*n* + 1)^*k* – *n*^*k*)*r*^(*n* + 1). The first term is Fibonacci-type by #9, and the sum of the rest is by assumption.

11d. The associated polynomial of int((*n*^*k*)(*r*^*n*)) is (*x* – 1)(*x* – *r*)^*k*. The first term in #11c has the associated polynomial (*x* – *r*)^(*k* + 1) from #9; the remainder has (*x* – 1)(*x* – *r*)^*k* by assumption. From #5, int((*n*^*k*)(*r*^*n*)) is Fibonacci-type for all *k* with associated polynomial dividing (*x* – 1)(*x* – *r*)^(*k* + 1). That division can’t be proper, because only the first term

has the term (*n*^*k*)(*r*^*n*) while only the remainder has a constant term.

12. The trigonometric functions sin and cos are Fibonacci-type. Note that because pi is irrational, sin(*n*) is not periodic. However, the identity *e*^*ix* = cos(*x*) + *i*sin(*x*), derived from considering the Maclaurin expansions *e*^*x* = 1 + *x* + *x*^2/2 + *x*^3/3! + …, sin(*x*) = *x* – *x*^3/3! + *x*^5/5! – …, cos(*x*) = 1 – *x*^2/2 + *x*^4/4! – …, allows us to express sin and cos in terms of exponentials. We get cos(*x*) = (*e*^*ix* + *e*^(-*ix*))/2, sin(*x*) = (*e*^*ix* – *e*^(-*ix*))/2*i*. But ((*e*^*i*)^*x*) and ((*e*^(-*i*))^*x*) are Fibonacci-type; hence, so are sin and cos, by #3.

13. Separating a Fibonacci-type sequence into parts results in Fibonacci-type sequences. That is, if *a*(*n*) is Fibonacci-type, and *b*(*n*) = *a*(*cn* + *d*), then *b*(*n*) is Fibonacci-type. To see why, note that this turns (*n*^*k*)(*r*^*n*) into ((*cn* + *d*)^*k*)(*r*^(*cn* + *d*)) = (*c*^*k*)(*r*^*d*)((*n* + *d*/*c*)^*k*)((*r*^*c*)^*n*) where ((*n* + *d*/*c*)^*k*) is a polynomial in *n* of degree *k*.

14. Weaving several Fibonacci-type sequences into one results in a Fibonacci-type sequence. I’ll only prove it when “several” means “two”; the generalization is straightforward enough to be left as an exercise. When *a*(*n*) and *b*(*n*) are Fibonacci-type and *c*(*n*) = *a*(*n*/2) for even *n* and *b*((*n* + 1)/2) for odd *n*, we can use the fact that (-1)^*n* + 1^*n* = 2 when *n* is even and 0 when *n* is odd to alternatively activate or deactivate the sequence. More precisely, given (*n*^*k*)(*r*^*n*), note that (((*n*/2)^*k*)(SQRT(*r*)^*n*) + ((*n*/2)^*k*)((-SQRT(*r*))^*n*))/2 = ((*n*/2)^*k*)(*r*^(*n*/2)) when *n* is even and 0 when *n* is odd.

15. If *a*(*n*) is periodic of period *t*, then all of its basis elements are of the form *r*^*n* with *r*^*t* = 1 for all *r*. This follows from the definition, since the associated polynomial of the sequence is *x*^*t* – 1. Now, if *r*^*t* = 1, then *e*^(2pi**i*) = 1 implies *r*^*n* = *e*^(2pi**in*/*t*) = cos(2pi**n*/*t*) + *i*sin(2pi**n*/*t*).

The carnival submission takes a sequence with periodic differences and constructs an explicit formula for it with trigonometric, constant, and linear terms. That is immediately a Fibonacci-type sequence from #1 and #11; #15 also shows that the associated polynomial has 1 as a double root, corresponding to the constant and linear terms, and every other 8th root of unity as a simple root, corresponding to each trigonometric term, where the argument is indeed a multiple of pi**n*/4.

I think I understand.

Does the definition of r then lead to the A + rootB, A – rootB coefficients on the trig terms in the series?

Or is that more likely a product of this series in particular?

Universities across US and Canada are implementing increasingly oppressive speech codes in order to censor politically incorrect speeches. see

http://www.thefire.org/index.php/

Alon, have you heard the great news? WordPress now allows Latex :-)

http://wordpress.com/blog/2007/02/17/math-for-the-masses/

Foxy, the A (+/-) SQRT(B) coefficients in the series aren’t exactly due to r. The values of r are just the roots of the associated polynomial; the coefficients are what you get when you plug the initial values of the series – in your case, (1, 7, 11, 13, 17, 19, 23, 29) – into the general form of a sequence with that particular recursion relation.

Thanks for pointing out the response. It’s easy to lose track of these things. You post so often it’s just *szchwoop*down the page.

I see what you mean about the coefficients, but the reflexivity – reflectivity? reflectedness? Still intrigues me. Of course, that’s probably just because it’s pretty.

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Have you read anything by Honsberger on this topic?

He gives exp(sigma((L(r)*x^r)/r){r=1 to r=infinity}

= sigma(F(r)*x^(r-1)){r=1 to r=infinity}

where L(n) is the nth in the Lucas series and F(n) is the nth in the Fibonacci series.

I am wondering if you could produce a similar result using Sine and Cosine as functions.

Thank you for some other wonderful article. The place else may

anyone get that type of info in such a perfect method of writing?

I have a presentation next week, and I’m on the look for such information.

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