PADUA AND PISA ARE EXPONENTIALLY FAR APART

We answer the question posed by Ian Stewart which Padovan numbers are at the same time Fibonacci numbers. We give a result on the difference between Padovan and Fibonacci numbers, and on the growth of Padovan numbers with negative indices.


Introduction
1.1.What this paper is not about.This paper has nothing to do with Italian topography.It is about Padovan numbers and their distances to Fibonacci numbers.Briefly said the main result of this paper is an explicit lower bound for these distances, which grows exponentially.The problem solved here is a generalization of a question asked by Ian Stewart in his Scientific American Mathematical Recreations column [S].In this column Stewart described the similarities between Padovan and Fibonacci numbers, and remarked that, appropriately, Pisa, the city of Fibonacci, and Padua, the city with Italian name Padova, are roughly only 100 miles apart.As we show that Padovan and Fibonacci numbers are far apart, this might serve as an explanation for our title.

Padovan and Fibonacci numbers.
The Padovan numbers P m , named after Richard Padovan, are defined by P 0 = 1, P 1 = 0, P 2 = 0, P m+1 = P m−1 + P m−2 , and the Fibonacci numbers F n , named after Leonardo 'Fibonacci' Pisano, are defined by Note that in our definition the indices of the Padovan numbers are shifted by 5 compared to Stewart's definition.We list a few of these numbers: These short tables already illustrate the growth behaviour of the sequences.Whereas P m and F n grow very regularly, indeed exponentially, with F n showing the faster growth rate, the P −m show oscillating behaviour, with still exponentially but slowly growing amplitude.These observations are warranted by the following lemma, which is easy to prove, e.g. by mathematical induction.

Lemma 1.
(i) Let γ be the real root of x 3 − x − 1, and let δ be the non-real root of Then for all m ∈ Z Then for all n ∈ Z Because |β| < 1 the term 1 √ 5 β n tends to 0 as n grows, so that this lemma implies at once that with m positive.Now it's the term λγ −m that tends to 0 as m → ∞, whereas the other two terms in the above expression for P −m grow exponentially in absolute value.Notice that where the number µδ −m µδ −m is on the unit circle (and travels around it as m varies).This explains the oscillating behaviour.The amplitude is |δ −m |, which is equal to γ m/2 , since γδδ = 1, and thus |δ| = δδ Thus the growth rate of the amplitude is only γ 1/2 = 1.15 . . . .

Stewart's first question.
Stewart in his column [S] asked two specific questions on Padovan numbers with nonnegative indices.First, he remarks that some Padovan numbers are Fibonacci numbers too, namely and asks whether there are other solutions, and whether the number of solutions is finite or infinite.In this paper we answer this question in showing that there are no other solutions.
Stewart did not care about Padovan numbers with negative indices.Had he done so, he certainly would have noticed (also counting P m = −F n as solution) that A natural question is whether these are all the solutions of P −m = ±F n .It follows from results of Evertse [E] and van der Poorten and Schlickewei [PS] that the number of solutions is finite, and even an explicit upper bound for the number of solutions can be given.We conjecture that the ones mentioned above are all the solutions, but we have no idea how to attack this problem.It follows from a result of Mignotte [M2] that an equation of the type u m = v n has only finitely many solutions, and can be solved effectively and practically, when {u m } and {v n } both are recurrence sequences with one dominating root, i.e. of the characteristic roots (such as γ, δ, δ for {P m }) one is in absolute value strictly larger than the others (γ).Indeed, the very first paper in which a diophantine problem was explicitly solved by the methods that we use in this paper, was a problem of this type, cf.Baker and Davenport [BD].It seems that our present result is the first explicit equation of this type involving a second and a third order recurrence sequence.

Stewart's second question.
Stewart's second question is which Padovan numbers are squares.There obviously are the solutions We have no idea how to prove anything in this direction.Stewart might also have noticed that but he didn't.Again we have no clue whatsoever how to proceed with this problem.

Results I.
However, we can answer two other questions Stewart did not ask.The first one we get almost for free from our method for solving P m = F n .Namely we can prove the following result.
An answer to Stewart's original question to find the solutions of P m = F n follows immediately from (i), upon inspection of the tables in Section 1.2.
In (i) we could in principle have derived a similar result with the exponent 1/2 replaced by any δ < 1.A similar remark holds for (ii), with the constant 0.24174 adjusted accordingly.In (ii), when m is larger than approximately 10 353605 , the first term in the max expression is the best one.It shows that asymptotically for an arbitrarily small > 0, and thus that Padua and Pisa are indeed exponentially far apart (note that P m ∼ λγ m ).For m smaller than approximately 1.1549 × 10 13 , the third term in the max expression is the best one.
As an immediate application we now are able to instantaneously solve problems of the type |P m − F n | ≤ 10 6 , say.Namely, (i) tells us that there are no solutions with P m ≥ 10 12 , and if P m < 10 12 then, by the fact that P m is the nearest integer to λγ m for m ≥ 5, we immediately have m ≤ 104.The solutions now are easy to determine.

Results II.
The other problem we'll address in this paper is the growth behaviour of P −m .As we saw above this sequence oscillates with exponentially growing amplitude, thus shows complicated behaviour, getting small compared to the amplitude when the number µδ −m µδ −m on the unit circle happens to come near to −1.This happens when Arg µ + m Arg δ is near to an odd integer times 1 2 π.Nevertheless there is a result of Mignotte [M1] that yields a lower bound for |P −m |, and in the following theorem we will make this explicit.This might serve as a meager substitute for our inability to solve P −m = ±F n , but this seems more or less to be at the limit of the available methods.
(ii) If m is a nonnegative integer not equal to 3 or 12 then From (i) it follows that the sequence {P m } m∈Z has exactly 5 zeroes, namely at m = −12, −3, 1, 2, 4.This result is due to Beukers [B], with a different proof.
Similar remarks as those immediately following the statement of Theorem 2 can be made here.Especially we want to remark that asymptotically |P −m | γ (m/2)(1− ) for an arbitrarily small > 0, and thus although in the oscillation P −m can become a bit smaller than the amplitude γ m/2 , this never becomes really dramatic.Also, an application such as finding all solutions to |P −m | ≤ 10 6 is now instantaneous: from (i) we obtain γ m < 10 24 , hence m ≤ 196.

Perrin numbers.
Stewart in his column also considers the Perrin numbers A , defined like the Padovan numbers by Our methods will certainly be able to prove the following assertions, or in case they are false, to prove similar assertions with the constants replaced by the correct ones.We leave details of such proofs to the interested reader.

Assertions 4.
( We may assume without loss of generality that m ≥ 11, so that P m ≥ 4. From Lemma 1 we have so that by m ≥ 11 and P m ≥ 4 we find inequalities that we will use repeatedly: Next we want to estimate the error term in (4).A rough estimate is But we can do much better.First we remark that if in some cases we can prove that |P m − F n | > cλγ m for some constant c > 0, then we are essentially done, as it immediately follows that |P m − F n | > c P m for some other constant c > 0. In the case that we have 1 so by ( 4) and ( 7) and with (5) this yields So in this case P m = F n , and |P m − F n | ≤ P 1/2 m at once implies P m ≤ 12, and also the inequality ( 2) is obvious in this case.
So we may assume that α −n < γ −m/2 , and then we find for the error term in (4) a much better estimate than (7), namely To deal with the main term of (4) we introduce a linear form in logarithms of algebraic numbers: Notice that It follows that Λ = 0.In the case Λ ≤ −1 we have e Λ − 1 = 1 − e Λ ≥ 1 − e −1 , so by ( 4), ( 7) and ( 10) we find which is covered by ( 8).So we may assume that Λ > −1.Then e Λ − 1 > (1 − e −1 )|Λ|, and with (10) this gives us an important estimate: The last special case we have to deal with is the case n > m.Then With (4), ( 7) and ( 12) this yields (13 which again is covered by ( 8).So we may assume that n ≤ m.

Application of transcendence theory.
Now we are ready to apply the main ingredients: explicit lower bounds for |Λ|, that transcendence theory provides.We use the results of Baker and Wüstholz [BW], which give the asymptotically best results, and of Voutier [V], which gives the best results for small m.Notice that our linear form has three terms, and the field Q(α, γ) is of degree 6.By Λ = 0 the theorem of [BW] and Theorem 3 of [V] give where C BW and C V are large absolute constants that we can compute explicitly.Indeed, for C BW we have Here the function h on our sextic field is defined by where h is the absolute logarithmic Weil height defined by where the sum runs over all d embeddings σ of the field Q(ξ) into C, d is the degree of this field, and a 0 is the leading coefficient of the minimal polynomial of ξ.It is easy to compute the required values for h (on noting that for λ √ 5 we have a 0 = 23 2 ), and we find So for the constant C BW in ( 14) this leads to (15) C BW < 1.4613 × 10 15 .
And we have where the function h on our sextic field is defined by Notice that in applying Voutier's Theorem 3 we have taken B = m 2 (whence the factor 4), which by n ≤ m is larger than .

Finishing the proof.
Now we combine ( 4), ( 9), ( 12), ( 14) and ( 16), to obtain ( 17) Note that e.g. when m = 17, where there is a solution of P m = F n , the right hand side of ( 17) is negative, so the inequality is still true, but useless.Of course ( 17) is useful only if the right hand side is positive, and that happens if m ≥ 3.5537 × 10 12 .Thus at this point we have proved that if P m = F n then m < 3.5537 × 10 12 .That's at least something.
Finally it follows by ( 5) that which is a major step towards the proof of (ii).
In the next subsection we will show that if m < 1.1549 m has only the solutions mentioned in the statement (i) in the theorem.
Assuming this for the moment, we can now finish the proof of (ii).Namely, from inspection of the solutions of we find that there are no solutions with 0 m occurs for P 29 = 616 and F 15 = 610, which explains the number 0.24174 < 0.241746 . . .= 6 √ 616 .This proves that Further, by m < 1.1549 × 10 13 it follows that √ λγ m/2 < 0.23527e 1.7950×10 9 (log m) 2 , which by (5) implies that and thus by (20) we have 18), ( 19), ( 20) and ( 21) together imply two of the three bounds of (ii), and the third one follows by a similar reasoning, of which we do not give the details.

Application of computational diophantine approximation.
It remains to prove that if m < 1.1549 × 10 13 then the inequality m has no solutions with m ≥ 30 (notice that the solutions with m ≤ 29 are very easy to find).This we do by a computational diophantine approximation technique, known as the Baker-Davenport method.Here the linear form Λ will play a crucial rôle.See [dW] for a more complete description of these techniques.
From ( 12), ( 4) and ( 7) we find and thus using our inequality and ( 5) and m ≥ 30 we have  Finally, finding the solutions with 30 ≤ m ≤ 78 can simply be done by enumeration.This completes the proof of Theorem 2.

Proof of Theorem 3
3.1.Preparations.This proof follows to a large extent the line of argument set out in the previous proof 1 , and is in details a bit simpler.For m ≥ 0 we write (cf.( 1 If Λ ≥ π 3 then e iΛ − 1 = 2 sin 1 2 Λ ≥ 1, hence by ( 24) we have 1 As my colleague Henk Hoogland would say: this is going to be a proof by texteditor.The reader interested in writing out a proof of the Assertions 4 can obtain the L a T E X-code of this paper from the author upon request.
In this case the statements of the theorem are immediate.

Application of transcendence theory.
A lower bound for |λ| is again furnished by transcendence theory.Indeed, [BW] and [V] give (by ( 23) and on noting that Λ = 0) where C BW and C V are large absolute constants that can be computed explicitly.
In fact, for C BW we find where −1 is to be interpreted as e iπ .Now we have the function h on the sextic field Q(δ, δ).Using h ξ ξ ≤ 2h(ξ) we find And for C V we have and we find h µ µ < 2.0904, h δ δ < 1.5955, h (−1) = 3.4π 3 < 3.5605.

Application of computational diophantine approximation.
It remains to prove that if m < 1.2335 × 10 16 then the inequality |P −m | ≤ γ m/4 has no solutions with m ≥ 31 (notice that the solutions with m ≤ 30 are very easy to find).This we do by the same computational diophantine approximation technique that we used in the proof of Theorem 2, using again the linear form Λ. This time we omit some numerical details.Finally, finding the solutions with 31 ≤ m ≤ 139 can simply be done by enumeration.This completes the proof of Theorem 3.
If m and n are nonnegative integers such that |P m − F n | ≤ P 1/2 m , then m ≤ 29 and n ≤ 15. (ii) If m and n are nonnegative integers such that P 22) |Λ| < 3.8663γ −m/2 .Consider the lattice Γ = {Cx | x ∈ Z 2 } defined by the matrix So the columns of the matrix C form a basis of the lattice.Further, For a possible solution (m, n) we now look at the distance d between the lattice point C −m n and the point y.This distance is the length of the vector C We have by the definition of Λ and the upper bound for m that |Λ − 10 30 Λ| ≤ 1 + m +
From (25) and our inequality and m ≥ 31 we have |Λ| < 2.1130γ −m/4 .Consider the lattice Γ = {Cx | x ∈ Z 2 } defined by the matrix For a possible solution m the distance d to look at now is the distance between the lattice point C m 2 −1 and the point y, being the length of the 10 34 ψ + m 10 34 φ + (2 − 1) 10 34 π .The distance d between y and the nearest lattice point is larger than 8.9716 × 10 16 (we omit numerical details behind this fact).It follows as in Section 2.4 that |Λ| > 6.4192 × 10 −18 , and now together with (34) we immediately find that m ≤ 573.Again we repeat the game, with 10 34 replaced by 10 7 .Now we have d > 2377.3, and we derive |Λ| > 1.1602 × 10 −4 , and now together with (34) we immediately find that m ≤ 139.

Proof of Theorem 2 2.1. Preparations.
If and n are nonnegative integers such thatA = F n then i) If and n are nonnegative integers such that |A − F n | ≤ A 1/2 , then ≤ 29 and n ≤ 18. (ii) (iii) If is a nonnegative integer such that |A − | ≤ γ /4 , then m ≤ 29.(iv) If is a nonnegative integer then 2.