sums of powers, a probabilistic approachybilu/fraza/deshouillers_slides.pdf · sums of powers, a...

Sums of powers, a probabilistic approach

Jean-Marc Deshouillers

Institut Mathematique de Bordeaux

Journee arithmetique FraZa 2017IMB, Bordeaux, November 8th, 2017

En memoria de Javier Cilleruelo

Jean-Marc Deshouillers (IMB) French-South African GANDA meetingJournee arithmetique FraZa 2017 IMB, Bordeaux, November 8th, 2017 1

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1 Sums of k numbers which are s-th powers of integers (2 ≤ s ≤ k)

2 Squares and cubes

3 The Erdos-Renyi probabilistic modelRandom sequencesPseudo s-th powersPoisson model for sums of s pseudo s-th powersSums of (s + 1) pseudo s-powers

4 Javier Cilleruelo’s contributionGaps between consecutive sums of s pseudo s-powersAdditive complements to sums of s pseudo s-powersSums of (s + ε) pseudo s-powersTwo words about the proofs

5 Arithmetical model and numerical tests


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Sums of k numbers which are s-th powers of integers (2 ≤ s ≤ k)

We are interested in representing a non-negative integer n as the sum ofk numbers which are s-th powers of non-negative integers xi ,

n = x s1 + x s

2 + · · ·+ x sk ; (1)

we letRs,k (n) be the total number of those representations,

rs,k (n) be the number of representations with x s1 ≤ x s

2 ≤ · · · ≤ x sk and

r ′s,k (n) the number of representations with x s1 < x s

2 < · · · < x sk .

As,k = n : rs,k (n) > 0 = n : Rs,k (n) > 0We further count the number of integers n up to the real number x

which are sums of k s-th powers by the counting functions

As,k (x) = CardAs,k ∩ [1, x ] = Card n ≤ x : rs,k (n) > 0and their densitieslim infx→∞ As,k (x)/x and lim supx→∞ As,k (x)/x .


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What we can prove

For k ≥ s

As,s(x) ≤ (1 + o(1))x/s!.

Rs,k (n) has order n(k−s)/s on average, i.e.∑

n≤x Rs,k (n) ∼ Cs,k xk/s .

For k large with respect to s

As,k (x) = x + O(1) (Hilbert, answering Waring).

Rs,k (n) ∼ Ss,k (n)n(k−s)/s (Hardy-Littlewood).

What would we like to know

Is lim inf As,s(x)/x > 0? or even ¿does lim As,s(x)/x exist and ispositive?

Is Rs,s+1(n) > 0 as soon as Ss,s+1(n) > 0 and n large enough?


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Squares

What we know

4 squares. Every integer is a sum of 4 squares, i.e. A2,4(x) = bxc.Lagrange (1770).

3 squares. Every integer which is not of the shape 4a(8b + 7) is a sumof 3 squares. Gauß, Legendre (ca 1800).

2 squares. A2,2(x) ∼ Cx/√

log x , as x →∞. Landau (1909).The representation function of an integer as a sum of two squares (R2,2) ismultiplicative and thus rather erratic, taking very large values.

Let us write A2,2 = b1 < b2 < · · · < bn < · · · . We have

bn+1 − bn < b1/4n . Folklore (¡greedy algorithm!).

What we expect

For the consecutive sums of two squares, we expect∀ε > 0: bn+1 − bn < bεn.


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Cubes

What we know.

7 cubes. Every large integer is a sum of 7 cubes, i.e. A3,7 = x + O(1).Linnik (1942).

R3,7(n) n4/3. Vaughan (1989).

Every integer larger than 454 is a sum of 7 cubes. Siksek (2016).

What we expect.

4 cubes. Is every large integer a sum of 4 cubes?

Is 7, 373, 170, 279, 850 the largest integer n for which R3,4(n) = 0?D.-Hennecart-Landreau (2000).

3 cubes. Is lim inf A3,3(x)/x > 0 ?


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The indicator (or characteristic) function of a subsequence of N0

Let A be a sequence of N0, say A = a0 < a1 < · · · < an < · · · .We can also represent it by a function χA from N0 to 0, 1 whichassociates to each integer n the value 1 if and only if n ∈ A. This gives abijection between the subsequences of the integers and the sequences withvalues in 0, 1.

Random sequences

Let Ω be a probability space and (ξn)n≥0 be a sequence of independentrandom variables with values in 0, 1. To each ω ∈ Ω, we can associatethe sequence (ξn(ω))n≥0. It is a sequence of 0 and 1 and we can associateto it a sequence of integers Aω such that n ∈ Aω if and only if ξn(ω) = 1,or in other words

χAω(n) = ξn(ω).

We say that A is the random sequence associated to the sequence (ξn)n≥0.


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Pseudo s-th powersIf one considers two consecutive s-th powers around x , their difference(thanks to Taylor) is something like(x1/s + 1)s − (x1/s)s ∼ sx1−1/s .We can thus say that, among sx1−1/s consecutive integers around x thereis generally one s-th power, or in other words, the heuristic probability thatan integer n is an s-th power isn/(sn1−1/s) which is equal to 1/sn1/s .

Using this remark, Erdos and Renyi, defined random sequence of pseudos-th powers by asking the randon variables ξn to satisfy the condition

Prob(ξn) =1

sn1/sfor n ≥ 1. (2)

We shall denote with a tilda the quantities related to pseudo s-th powers,for example rs,k (n), As,k , As,k (x), . . . and remember that those expressions

are random variables. As such, everything is possible, e.g. As,1ω = ∅ for

some ω ∈ Ω or As,1ω = N for some (other) ω, but we have

Almost surely : As,1(x) ∼ x1/s , as x →∞. (3)


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Poisson model for sums of s pseudo s-th powersAs in the real case, the number of representations can be easily handledon average:

a.s.∑n≤x

rs,s(n) ∼ λsx where λs = Γ(1/s)/ (sss!) . (4)

Since rs,s(n) is constant on average, it is natural to expect a Poissondistribution for its values. It is the aim of Erdos and Renyi to show

a.s. ∀d ≥ 0 :1

xCardn ≤ x : rs,s(n) = d → e−λs

λds

d!. (5)

The proof has been completed by Goguel (1975), using the method ofmoments.


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The model versus the realityFrom the previous relation (5), we get

a.s.1

xCardn ≤ x : rs,s(n) 6= 0 → 1− e−λs (6)

ora.s. As,s(x) ∼ (1− e−λs )x , as x →∞,

which tells us that the sums of s pseudo s-th powers have a density.In the real life, the only case we know is that of squares, where we knowthat sums of two squares have a zero density...However, we shall see in the last part that some arithmetic may beplugged into the Erdos-Renyi model, leading to a probabilistic model• with zero density for sums of two pseudo-squares,• with positive density for s ≥ 3,• that density being consistant with numerical evidence for s = 3 and

s = 4.


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Sums of (s + 1) pseudo s-powers

Iosifescu and D. proved in 2000 that the pseudo s-th powers are a.s.. Themain result is that there exist two positive constants u = u(s) andv = v(s) such that

a.s. exp(−vn1/s) ≤ Prob(Rs,s+1(n) = 0) ≤ exp(−un1/s).

Question 1. Do we have Prob(Rs,s+1(n) = 0) ≤ exp(−wn1/s)?

The fact that the pseudo s-th powers are a basis of order s + 1 simplyfollows by Borel-Cantelli.

The largest integer which is not a sum of at most s + 1 pseudo-s powers isthus a random variable which is a.s. finite.Question 2 What is the law of the largest integer which is not a sum ofat most s + 1 pseudo-s powers?

Answer to this question would confirm our belief that 7,373,170,279,850 isthe largest integer which is not a sum of 4 cubes (already mentioned).


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Gaps between consecutive sums of s pseudo s-powersLet us write

As,s = b1 < b2 < · · · < bn < · · ·

Theorem (Cilleruelo and D., 2016)

We have almost surely

lim supn→∞

bn+1 − bn

log bn=

1

λs=

sss!

Γs(1/s).

This is consistant with the fact that the density of As,s is 1− e−λs .


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Additive complements to sums of s pseudo s-powers

Theorem (Cilleruelo, Lambert, Plagne and D., 2016)

Let C be a fixed sequence of integers satisfying

lim infx→∞

C (x)

log x> λ−1

s .

Then, a.s. every sufficiently large integer is a sum of s pseudo s-th powersand an element of C.

Let D be a fixed sequence of integers satisfying

lim infx→∞

D(x)

log x< λ−1

s .

Then, a.s., there exist infinitely integers which cannot be represented asthe sum of s pseudo s-th powers and an element of D.


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Sums of (s + ε) pseudo s-powers

Theorem (Cilleruelo, Lambert, Plagne and D.)

Let c > (λs(1− 2λs))−1. Almost surely, a sequence of pseudo s-th powersA has the following property: any large enough integer n can be written inthe form

n = a1 + · · ·+ as+1, with ai ∈ A and an+1 < (c log n)s .

Question 3. Can one replace c > (λs(1− 2λs))−1 by c > (λs)−1?

Question 4. Show that the statement of the theorem does not hold anylonger if c < λ−1

s .


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A word about the proofsIf r ≥ t + 1 and r ′ ≥ t + 1, we have

∗∑ω∼ω′

P(Eω ∩ Eω′ ) ≤∑

1≤x1,...,xta1x1+···+at xt<z

b1x1+···+bt xt<z′

(x1 · · · xt )−1+1/s ×

∑

1≤ut+1,...,ur ,at+1ut+1+···+ar ur

=z−(a1x1+···+at xt )

(ut+1 · · · ur )−1+1/s

×

∑1≤vt+1,...,vr′ ,

bt+1vt+1+···+br′ vr′=z′−(b1x1+···+bt xt )

(vt+1 · · · vr′ )−1+1/s

.

By Lemma 2.1 i) we have

∗∑ω∼ω′

P(Eω ∩ Eω′ ) ∑

x1,...,xta1x1+···+at xt<z


(x1 · · · xt )−1+1/s ×

×(

z − (a1x1 + · · ·+ at xt )) r−t

s−1(

z ′ − (b1x1 + · · ·+ bt xt )) r′−t

s−1.

∑

x1,...,xta1x1+···+at xt<z


(x1 · · · xt )−1+1/s (z − (a1x1 + · · ·+ at xt ))−t/s (z ′ − (b1x1 + · · ·+ bt xt ))−t/s


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Another word about the proof of “Gaps”Let µ > 0 be given. We define the events

F(µ)i = sA ∩ [i , i + µ log i ] = ∅, and wish to prove

(*) Prob(

F(µ)i

)= i−µλs +o(1).

We think of Borel-Cantelli and understand why 1/λs is a threshold.

The expression F(µ)i itself is the intersection of all the events stating that

a1 + · · ·+ as is not in the interval [i , i + µ log i ]. If those events wereindependent, some combinatorics would be enough to establish (*).

Fortunately, F(µ)i is the intersection of complementary events which are

natural in our model, and a magic inequality due to Jansson permits us toconsider them as almost independent... at the cost of some acrobaticmanipulations. Again, this quasi-independence permits us to apply amodified version of Borel-Cantelli.

By the way, this correlation inequality can be used at no cost in differentquestions (gaps, complementary sequences...,) where we have a clearthreshold, but is responsible for some approximations in other questions(sums of s + 1 or s + ε pseudo s-th powers).


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Arithmetical model and numerical tests

The model for the pseudo s-th powers only takes care of the “infinitevaluation”. Around 2000, Hennecart, Landreau and D. have developedprobabilistic models which also take care of the distribution of the pseudos-th powers in arithmetic progressions. For fixed K , we forced the pseudos-th powers to be distributed as the real s-th powers in all the arithmeticprogressions modulo any integer less than K. For example, this leads a.s. toa density δs(K ) for the sums of s pseudo s-th powers. Moreover, when Ktends to infinity, δs(K ) tends to 0 when s = 2 and to a positive limit whens ≥ 3. We also made some numerical experiments for s = 3 and s = 4.

Question 5. Find a probabilistic model for squares in which Landauasymptotics for sums of two squares is a.s. valid for sums of 2pseudo-squares.


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comparing the probabilistic density with the experimentaldensity

ng at Figure 1, it is impressive to observe how well our model of sums ofudobiquadrates fits with the ordina,ry sums of 4 biquadrates. The general

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References

1. P. Barrucand, *Sur la distributi.ncrgËbicarres", Note aux C. R- Acad. Sc. H

2. J-M. Deshouillers, F. Hennecart, B. Irfinement to the probabilistic modd dEmetica.

3. P. Erdôs et A. Rényi, "Additirrc propcrfrActa Arith. 6 (.1960), 8&rr0.

4. J.H. Goguel, uUber Summen von rEAngew.-M arh. 278 / 279 (1975),6&.n.

5. C. Hooley "On some topics conndd rMath. 369 (1986), 1lG'153.

6. E. tandau, "Uber die Einteilung der ..,Phys. (3) 13 (1908), 30F312.

7. B. Landreau, "Modèle probabiliste porr Ipositio Math. 99 (1995), f-31.

Jean-Marc DnsHouILLERsMathématiques StochastiquesUniversité Victor Segalen Bordeaux 2F-33076 Bordeaux Cedexe.mail: J-M.Deshouillers@u-bordeaux2.Ê

Fbançois HnNNsclnr et Berna,rd L.l,xonsarLaboratoire d'AlgorithmiqueArithmétique ExpérimentaleUniversité Bordeaux IF-33405 TALENCE Cedexe'mail: [email protected] , landlq


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Fig. 1. The experimental and probabilistic densities


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