Correction to: The first negative Fourier coefficient of an Eisenstein series newform (The Ramanujan Journal, (2024), 63, 3, (687-697), 10.1007/s11139-023-00779-1)

Abstract

In the proof of Theorem 1.4, three lines below Eq. (3.3), I incorrectly claimed that η(D1,D2)=n(D1) when η|D2. For example, if D1=5 and D2=-15, then η(D1,D2)=3. In this case, η=3|-15, but n(5)=2. We will now prove Conjecture 1.3 rather than disproving it. We have (Formula presented.) where (Formula presented.) In order to do so, we rely on two lemmas. Fix ε>0. Then, there exists some ℓ0=ℓ0(ε) such that for all sufficiently large x and ℓ⩾ℓ0, we have (Formula presented.) Let ε>0. Since the series (Formula presented.) converges, there is some ℓ0=ℓ0(ε) such that (Formula presented.) Let ℓ⩾ℓ0 and split the sum as follows: (Formula presented.) The first sum is o(x) by (ii) of Theorem 2.1. For the second sum, let ℓ′ be the integer for which pℓ′ is the largest prime with pℓ′⩽(logx)1/3. Then, we combine (2.4) together with (i) of Theorem 2.1 in order to obtain (Formula presented.) Now, since ℓ′<pℓ′, we have O(ℓ′pℓ′x2/3)=O((xlogx)2/3). This shows that for sufficiently large x, we have (Formula presented.) □ Let χ1,χ2 be Dirichlet characters associated with the fundamental discriminants D1 and D2. For a prime p, (Formula presented.) so that (Formula presented.) We can now prove the following “two-dimensional version” of Lemma 2.2: Fix ε>0. Then, there exists some ℓ1=ℓ1(ε) such that for all sufficiently large x and ℓ⩾ℓ1, we have (Formula presented.) Let ε>0, and choose ℓ1:=ℓ0(ε/4), where ℓ0 is as in Lemma 2.2. Let ℓ⩾ℓ1. First, we claim that η(D1,D2)⩽n(D2). In order to see this, let p2:=n(D2), i.e., p2 is the smallest prime such that χ2(p2)=-1. If p2∤D2, then by (2.5) and the definition of η(D1,D2), we have η(D1,D2)⩽p2. The case where p2|D2 cannot happen because otherwise χ2(p2)=0. This proves the claim, and we immediately obtain the inequality (Formula presented.) Now, by Lemma 2.2, for sufficiently large x, say x⩾N=N(ε), we have (Formula presented.) This suggests that we split the sum on the right hand side of (2.6) into the ranges x/|D1|⩽N and x/|D1|>N: (Formula presented.) For the first sum, since x/|D1|⩾N, we may use (2.7) in order to obtain (Formula presented.) For the second sum, since x/|D1|<N, we can use the trivial bound (Formula presented.) for some positive constant c depending only on N. Since N depends only on ε, we conclude that c will depend only on ε. This shows that (Formula presented.) Finally, since cx<ε2xlogx for x>e2c/ε, we have (Formula presented.) for sufficiently large x, completing the proof. □ We are ready to make Linowitz and Thompson’s heuristic argument rigorous: Using (2.1) and partial summation, we see that (Formula presented.) In particular, this shows that for sufficiently large x, we have (Formula presented.) Now, let ε>0, and let ℓ⩾ℓ1, where ℓ1 is as in Lemma 2.3. We split the sum as follows: (Formula presented.) By Lemma 2.3, the second sum is <εxlogx. This together with (3.8) shows that for sufficiently large x, (Formula presented.) For the first sum in (3.9), since now we are dealing with a sequence of primes {p1,…,pℓ} with ℓ not depending on x, we may employ Theorem 1.2 in order to obtain (Formula presented.) where (Formula presented.) This shows that for any ε>0 with ℓ⩾ℓ1 and for all sufficiently large x, (Formula presented.) Finally, after letting ℓ→∞ and then ε→0, Conjecture 1.3 follows.