Erdős Problem 243: exact modular dynamics and arbitrarily long flat transients

Date checked: 2026-08-12

Outcome

This is rigorous partial progress, not a proof or counterexample to Problem 243.

For every prescribed length L and every nonzero integer h, there is a positive rational number whose pseudo-greedy gap numerators are

e_1 = e_2 = ... = e_L = h,   e_{L+1} = e_{L+2} = ... = 0.

The construction works with either sign of h, has no rounding ties, and gives a strictly increasing integer sequence satisfying all the hypotheses of Problem 243. By taking its scale large, the first L nonzero gaps can be made arbitrarily small individually and in total.

Consequently, no uniform fixed-window argument based only on small gaps can force a zero, a sign change, an exit from the small-gap band, or (for negative h) a decrease of the integer numerator c_n. Any complete proof needs a genuinely global or scale-sensitive ingredient.

There is also a simple global sufficient condition:

If the negative gap mass sum_n max(-epsilon_n, 0) is finite, then the gap sequence is eventually zero.

In particular, absolute summability of the gaps implies termination.

Status and source boundary

• The current public page <https://www.erdosproblems.com/243> lists the problem

as open. The live collaboration gate checked on 2026-08-12 showed 0 proof claims, no current worker or collaborator, and 7 comments. None of those visible comments contained this construction.

• Junya Koizumi, *Irrationality of the reciprocal sum of doubly exponential

sequences*, <https://arxiv.org/abs/2504.05933>, Theorem 16, reformulates the problem as the eventual vanishing of the pseudo-greedy gap sequence under the hypothesis epsilon_n -> 0.

• The state variables below are those of Koizumi's Lemma 15. The modular

one-step formula and the flat-transient construction were not located in a text search of that paper or in the currently visible page comments. That is only a search report, not a novelty or priority claim.

• Local source snapshot used for checking:

koizumi_2504.05933.pdf, SHA-256 b27a16281cfe52b2c76264e088f2b1c61623e1caba6b23e5506d19cfaa929190. The third-party PDF need not be redistributed with a verification bundle.

Problem and pseudo-greedy state

Problem 243 asks whether every strictly increasing sequence of positive integers (a_n) satisfying

a_{n+1}/a_n^2 -> 1,                    sum_n 1/a_n is rational

must eventually satisfy

a_{n+1} = a_n^2 - a_n + 1.

For a positive rational remainder x_n=c_n/d_n, keep the unreduced positive integers c_n,d_n and define e_n to be the centered residue

e_n == d_n (mod c_n),       -c_n/2 <= e_n < c_n/2.

The half-open interval agrees with nearest-integer rounding in which a half integer rounds upward. Then

a_n     = (d_n-e_n)/c_n + 1,
c_{n+1} = c_n-e_n,
d_{n+1} = d_n a_n,
epsilon_n = e_n/c_n.

These are exact integer identities; no reduction of c_n/d_n is needed.

Exact one-step modular dynamics

Put m=c_{n+1}=c_n-e_n. Since

d_n = c_n(a_n-1)+e_n

and c_n == e_n (mod m), one has

d_n == e_n a_n (mod m),
d_{n+1}=d_n a_n == e_n a_n^2 (mod m).

Therefore the next centered residue is exactly

e_{n+1} = centered_residue(e_n a_n^2, c_n-e_n).              (1)

If

q_n = (e_n a_n^2-e_{n+1})/(c_n-e_n),

then the quotient is integral and direct substitution gives

a_{n+1} = a_n(a_n-1)+q_n+1.                                 (2)

Equations (1)-(2) permit exact modular exploration without materializing the doubly exponential denominators.

Flat-transient theorem

Theorem. Let L>=1 and let h be any nonzero integer. There is a reduced positive rational r=c_1/d_1 whose pseudo-greedy state satisfies

e_j=h  for 1<=j<=L,             e_j=0 for every j>=L+1.

More explicitly, choose a prime

p > (L+2)|h|

and define

c_j = p+(L+1-j)h       (1<=j<=L),
c_{L+1}=p,
P_j = product_{k=j}^L c_k,
P=P_1.

Because p>L and p does not divide h, the coefficient h^(L-1)L! is invertible modulo p. Choose a positive, sufficiently large integer T in the residue class

h^(L-1)L! T == -1 (mod p),                                  (3)

and put

d_1=PT+h,                    r=c_1/d_1.

Taking a sufficiently large representative of the class in (3), for example one with T>=2, is harmless.

Proof of the constant prefix

We prove by induction that

d_j == h (mod P_j)                  (1<=j<=L).                (4)

The case j=1 follows from d_1=PT+h. Suppose (4) holds. Since c_j divides P_j, the residue of d_j modulo c_j is h. Moreover,

c_j >= min(p-L|h|,p) > 2|h|,

so h is strictly inside the centered interval and is the unique centered residue: there is no tie. Thus e_j=h. Also

a_j=(d_j-h)/c_j+1 == 1 (mod P_{j+1}),

and consequently

d_{j+1}=d_j a_j == h (mod P_{j+1}).

Finally c_{j+1}=c_j-h, giving the displayed formula for every c_j and completing the induction.

Proof of the zero tail

Modulo p,

P = product_{t=1}^L (p+t h) == h^L L!,

so (3) gives

d_1=PT+h == h(h^(L-1)L!T+1) == 0 (mod p).

Every later d_j is a multiple of d_1. At state L+1 the numerator is c_{L+1}=p, hence the centered residue of d_{L+1} modulo p is zero. Once e_j=0, both c_j=p and divisibility p|d_j persist, so every later residue is zero. This proves the theorem.

Admissibility for Problem 243

All side conditions can be checked exactly.

• Positive state and no ties. The choice of p gives

c_j>2|h|>0 throughout the nonzero prefix and at the terminal state. Also d_1=PT+h>0 for T>=2; all later denominators are positive.

• Reduced initial rational. Since c_1 divides P,

gcd(c_1,d_1)=gcd(c_1,h)=gcd(p,h)=1.

• Strict increase. Writing P_2=product_{j=2}^L c_j, with the empty

product equal to one, gives a_1=P_2T+1>=3. Every centered gap satisfies -1/2<=epsilon_n<1/2, and Koizumi's exact recurrence is

  a_{n+1}=a_n^2/(1-epsilon_n)-a_n+1-epsilon_{n+1}.
  

Therefore, for a_n>=3,

  a_{n+1} > (2/3)a_n^2-a_n+1/2 > a_n.
  

Induction proves strict increase.

• Required asymptotic. From state L+1 onward, epsilon_n=0, so

a_{n+1}=a_n^2-a_n+1 exactly and a_{n+1}/a_n^2 -> 1.

• Rational reciprocal sum. At the first zero state,

d_n=c_n(a_n-1), so the remaining rational is 1/(a_n-1). The zero-gap tail telescopes via

  1/a_n = 1/(a_n-1)-1/(a_{n+1}-1).
  

Hence the full reciprocal sum is exactly the initial rational c_1/d_1.

Thus these are genuine admissible Problem 243 sequences which eventually obey the claimed recurrence. They are deliberately long transients, not counterexamples.

Quantitative corollary and the obstruction to fixed windows

Corollary. Given L>=1, eta>0, delta>0, and either sign sigma in {-1,+1}, there is an admissible sequence with

sign(e_j)=sigma,      0<|epsilon_j|<eta             (1<=j<=L),
sum_{j=1}^L |epsilon_j| < delta,
epsilon_j=0                                         (j>=L+1).

Take h=sigma in the theorem and then take p so large that

min_{1<=j<=L} c_j > max(1/eta, L/delta).

The two inequalities follow immediately from |epsilon_j|=1/c_j.

This rules out any uniform finite-window assertion whose only hypotheses are smallness of the gaps (even small total mass), convergence to zero, absolute summability, or weak eventual one-sidedness, and whose conclusion is that within that fixed window there must be:

• a zero gap;
• a sign change;
• an exit from the prescribed small-gap band; or
• a decrease of c_n (choose h<0, when c_{j+1}=c_j+|h|).

The word “uniform” matters. A bound depending on the arithmetic scale or on additional state information is not excluded.

The examples are eventually zero and hence satisfy both weak conditions epsilon_n>=0 eventually and epsilon_n<=0 eventually. This says only that those conditions do not bound the transient length; it says nothing about an eventually negative, never-zero orbit.

A global termination criterion

Write

epsilon_n^- = max(-epsilon_n,0).

Proposition. If

sum_n epsilon_n^- < infinity,

then e_n=epsilon_n=0 for all sufficiently large n.

Proof. The exact numerator recurrence gives

c_{n+1}=c_n(1-epsilon_n).

A nonnegative gap can only decrease c_n. At a negative gap, log(1-epsilon_n)=log(1+epsilon_n^-)<=epsilon_n^-. Consequently

c_N <= c_1 exp(sum_n epsilon_n^-) =: M

for every N. If e_n<0, integrality gives -e_n>=1, hence

epsilon_n^- = (-e_n)/c_n >= 1/M.

Finite negative mass therefore permits only finitely many negative e_n. After the last one, (c_n) is a nonincreasing sequence of positive integers, so it eventually becomes constant. Since e_n=c_n-c_{n+1}, this forces e_n=0 eventually. QED.

Consequences:

• sum_n |epsilon_n|<infinity implies eventual zero.
• An eventually nonnegative gap sequence terminates; this is also Koizumi's

Proposition 19(2), corresponding to the known Badea regime.

• Any counterexample to the pseudo-greedy reformulation must have

sum_n epsilon_n^-=infinity. Because a counterexample also has epsilon_n->0 and can never hit the absorbing value zero, its positive integer c_n must in fact tend to infinity: integrality gives |epsilon_n|=|e_n|/c_n >= 1/c_n whenever e_n is nonzero.

• No conclusion about eventual nonpositivity alone is claimed.

Exact verification

Run the self-contained, standard-library checker:

python3 runs/erdos243_wavew054_verify.py

The site repository also freezes the same checker at verification/243/flat-transients/check_243.py together with a hostile audit.

It performs the following independent exact checks.

1. It compares the modular recurrence with direct Fraction/integer

pseudo-greedy simulation for 1,000 random rationals and eight steps each.

1. It verifies (1)-(2) for 320,760 valid finite states.
2. It verifies the construction for 80 combinations of both signs,

1<=L<=100, including direct rational replay, strict increase, the first zero, and the exact telescoping sum for the tractable cases.

1. It verifies 72 rational (eta,delta) bounds with both signs.
2. It constructs and independently replays a 1,000-step certificate with

h=-1, every nonzero |epsilon_j|<10^-6, followed by zero.

Expected output:

random_direct_vs_modular=PASS trials=1000 steps=8
finite_transition_box=PASS states=320760
construction_grid=PASS cases=80
eta_delta_grid=PASS cases=72
large_certificate: L=1000 h=-1 p=1001003 c0=1000003 digits(P)=6001 trace_sha256=54814b022d6fa9ae977a7e40937296ddeaf03804dbfa706dbbf93605b32a8481
ALL EXACT CHECKS PASSED

The large trace digest hashes newline-separated records index,c,e,a_mod_(c-e) for states 0,...,1000. Its huge denominator is never trusted implicitly: a second replay carries d_n modulo the remaining product c_n c_{n+1}...c_{1001} and does not use formula (1).

Exploratory hostile searches are recorded in search.py and the two JSON logs. Among 5,000 random initial states constrained by e_1=-1, and 10,000 random initial states with arbitrary nonzero centered e_1, all hit zero within 100 tested steps (observed maxima 15 and 18). This finite experiment is not evidence of termination at scale and is not used in any proof.

Exact claim boundary

Proved here:

• the one-step centered-residue recurrence;
• arbitrary finite constant-residue prefixes of either sign, with an

explicitly engineered zero tail;

• arbitrarily long, arbitrarily small, arbitrarily low-mass nonzero gap

windows inside fully admissible sequences;

• termination under finite negative gap mass.

Not proved here:

• Problem 243 or Koizumi's Conjecture 6;
• existence of an infinite nonzero small-gap orbit;
• termination under eventual nonpositivity;
• any uniform bound that is allowed to depend on the arithmetic scale;
• novelty or historical priority of the construction.

PARTIAL: Exact dynamics produce arbitrarily long constant nonzero gap transients of either sign (then an absorbing zero tail), refuting uniform small-gap fixed-window strategies, while finite total negative gap mass is proved sufficient for eventual zero.