Recursive sufficiency for the Collatz conjecture and computational verification

We define the notion of recursive sufficiency for the Collatz conjecture and we use it to present some results concerning the computational verification of the conjecture. For any integer N ≥ 1 and any recursively sufficient set F, it is proved that all integers in the interval [1, N] satisfy the conjecture if and only if F ∩ [1, N] satisfies the conjecture. We offer a sequence of sieves for which the corresponding sequence of elimination percentages tends to 100\%, and as a result, for any integer P arbitrarily close to 100, we give a sieve whose elimination percentage is at least P%. Also, we prove that if N=2(3^n)+1 is the largest known integer for which all integers 1, 2, ..., N satisfy the conjecture, then all integers N+1, N+2, ..., 2N will satisfy the conjecture as well, and hence, they can be eliminated from the verification process.

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Publication Details

Journal
Notes on Number Theory and Discrete Mathematics
Published
2025-01-01
DOI
https://doi.org/10.7546/nntdm.2025.31.3.471-480
Primary Topic
Benford’s Law and Fraud Detection
Type
article
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article

Recursive sufficiency for the Collatz conjecture and computational verification

Mohammad M. Ansari
Notes on Number Theory and Discrete Mathematics
Benford’s Law and Fraud Detection
article

Recursive sufficiency for the Collatz conjecture and computational verification

Mohammad M. Ansari
article en

Abstract

We define the notion of recursive sufficiency for the Collatz conjecture and we use it to present some results concerning the computational verification of the conjecture. For any integer N ≥ 1 and any recursively sufficient set F, it is proved that all integers in the interval [1, N] satisfy the conjecture if and only if F ∩ [1, N] satisfies the conjecture. We offer a sequence of sieves for which the corresponding sequence of elimination percentages tends to 100\%, and as a result, for any integer P arbitrarily close to 100, we give a sieve whose elimination percentage is at least P%. Also, we prove that if N=2(3^n)+1 is the largest known integer for which all integers 1, 2, ..., N satisfy the conjecture, then all integers N+1, N+2, ..., 2N will satisfy the conjecture as well, and hence, they can be eliminated from the verification process.

Notes on Number Theory and Discrete MathematicsVol. 31(3)
Openalex Percentile: Top 81%
Benford’s Law and Fraud Detection
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