0
$\begingroup$

Ramanujan's tau function defined over $\mathbb Z^+=\{1,2,3,\ldots\}$ is given by $$q\prod_{n=1}^\infty (1-q^n)^{24}=\sum_{n=1}^\infty\tau(n)q^n\quad \ (|q|<1).$$ It plays an important role in the theory of modular forms. The five initial values of this function are $$\tau(1)=1,\ \tau(2)=-24,\ \tau(3)=252,\ \tau(4)=-1472,\ \tau(5)=4830.$$ Deligne proved in 1974 that $|\tau(n)|\le d(n)n^{5.5}$ for all $n\in\mathbb Z^+$, where $d(n)$ is the number of divisors of $n$. Recently, I conjectured that $|\tau(n)|>2n^4$ for any integer $n>2$ (cf. Question 484763).

Fermat's Last Theorem and Goldbach's Conjecture involve additive properties of powers and primes which come from the multiplicative structure of the integers. As the tau function is multiplicative, it is interesting to investigate its additive properties.

In additive combinatorics, for $A\subseteq \mathbb Z$ people investigate $$A+A=\{a+b:\ a,b\in A\},\ \ A-A=\{a-b:\ a,b\in A\},$$ and $$ A\dotplus A=\{a+b:\ a,b\in A\ \text{and}\ a\not=b\}. $$ Motivated by this, here I pose several conjectures on additive combinatorics for the tau function. For convenience, I set $$T:=\{\tau(n):\ n=1,2,3,\ldots\}.$$

Conjecture 1. We have $$\min\{|x|:\ x\in T+T\}=|\tau(1)+\tau(2)|=23$$ and $$\min\{|\tau(m)-\tau(n)|:\ m,n\in\mathbb Z^+\ \text{and}\ m\not=n\}=\tau(1)-\tau(2)=25.$$

Remark 1. I have verified for $1\le m,n\le 10^5$ that
$$|\tau(m)+\tau(n)|\ge23,\ \ \ \text{and}\ \ |\tau(m)-\tau(n)|\ge 25\ \text{if}\ m\not=n.$$ I think it's interesting to study how close two $\tau$-values can be.

Conjecture 2. The equation $x+y=z$ has no solution with $x,y,z\in T$.

Remark 2. This conjecture is somewhat similar to Schur's theorem in additive combinatorics or Fermat's last theorem for the fourth and fifth powers. Also, it implies Lehmer's conjecture that $0\not\in T$. We have verified that $\tau(k)+\tau(m)=\tau(n)$ for no $k,m,n\le 5000$.

Conjecture 3. For any $k,m,n\in\mathbb Z^+$, if $\tau(k)+\tau(m)=2\tau(n)$ then we must have $k=m=n$. Thus $T$ contains no nontrivial three-term arithmetic progression.

Remark 3. One may compare this with van der Waerden's theoem and Szemeredi's theorem in additive combinatorics. For $k,m,n\le 5000$, we have verified that if $\tau(k)+\tau(m)=2\tau(n)$ then $k=m=n$.

Conjecture 4. (i) The only perfect powers in the set $$\{|x|:\ x\in T+T\}=\{|\tau(m)+\tau(n)|:\ m,n\in\mathbb Z^+\}$$ are $$|\tau(56)+\tau(112)|=133952^2\ \ \text{and}\ \ |\tau(849)+\tau(1054)|=67428732^2.$$

(ii) The only perfect powers in the set $$\{\tau(m)-\tau(n):\ m,n\in\mathbb Z^+\ \text{and}\ m\not=n\}$$ are $$\tau(1)-\tau(2)=5^2\ \ \text{and}\ \ \tau(708)-\tau(438)=43974504^2.$$

Remark 4. This is motivated by Question 484381. I have verified that there are no other solutions for $m,n\le 4000$.

Recall that a set $A\subseteq \mathbb N=\{0,1,2,\ldots\}$ is called an asymptotic base of order $h$ if any sufficiently large integer can be written as a sum of $h$ elements of $A$.

Conjecture 5. There are positive integers $M$ and $N$ such that any integer $m$ with $|m|\ge M$ can be written as at most six elements in the set $T$, and any integer $n\ge N$ can be written as at most six elements in the set $\{|x|:\ x\in T\}$.

Remark 5. This is motivated by Deligne's upper bound for $|\tau(n)|$ as well as Warning's problem in additive number theory.

QUESTION. Any ideas towards solving the above problems? Counterexamples to some conjectures here are also welcome!

$\endgroup$
5
  • 5
    $\begingroup$ I am an outsider to this topic. I would appreciate an explanation as to why the question is getting downvoted. $\endgroup$ Commented Jan 1 at 11:50
  • $\begingroup$ "Warning" should be "Waring". This isn't a substantial enough typo to allow for an edit $\endgroup$ Commented Jan 1 at 11:57
  • 1
    $\begingroup$ I didn't downvote and won't downvote. But I read it and assume it is because it looks like it is asking many questions instead of one. I don't know if the user is asking for a single approach that helps with all of the questions though. $\endgroup$ Commented Jan 1 at 14:23
  • 2
    $\begingroup$ @AndrejBauer I am also surprised, and it happens with other questions by the poster - looks like systematic downvoting. That is especially strange, since from what I heard - Zhi-Wei Sun produced many interesting conjectures attracted attention of other mathematicians. E.g.: arxiv.org/abs/1210.2493 $\endgroup$ Commented Jan 1 at 17:19
  • $\begingroup$ After verification with Maple, $\tau(x)+\tau(y)=\tau(z)$ has no solution for $1\leq x\leq 10^5, 1\leq y,z\leq 10^6$. $|\tau(x)-\tau(y)|\geq 25 $ holds for $1 \leq x\neq y\leq 10^7$. $|\tau(x)+\tau(y)|\geq 23 $ holds for $1 \leq x\neq y\leq 10^7$. $|\tau(x)+\tau(y)|\geq 2$ holds for $1 \leq x, y\leq 10^7$. $\endgroup$ Commented Jan 10 at 7:22

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.