Lychrel Numbers in Base 10

Lychrel Numbers in Decimal Notation

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Let $\operatorname{rev}_{10}(n)$ be the integer obtained by reversing the decimal digits of $n$, dropping leading zeros in the reversed expression. Starting from $a_0=n$, put $a_{j+1}=a_j+\operatorname{rev}_{10}(a_j)$. A decimal palindrome reads identically in both directions. The question asks for a positive starting integer for which no later iterate $a_j$, $j\geq1$, is a palindrome. [S1] [S2]

Statement recorded in the supplied source.

Do Lychrel numbers exist in base 10? [S1]

Short English statement

Is there a positive integer whose repeated decimal reverse-and-add process never produces a palindrome?

Research attempt

Scope and route.

The first tested palindrome is a later iterate, so $a_0$ is not itself tested as a stopping condition. Reversal drops zeros that become leading zeros. The p196 research project [E1] documents long finite computations; these do not certify an infinite nonpalindromic orbit. The present attempt analyzes the exact carry process and identifies what an invariant proof would have to control.

The orbit increases and cannot form a finite cycle.

For every positive integer $a$, its reversal is positive. Therefore $a_{j+1}>a_j$, and in fact $a_j\ge a_0+j$. Every positive-start orbit is unbounded and has no repeated integer. Thus the usual strategy of finding a finite cycle avoiding palindromes cannot work for this iteration. It also means that a fixed digit-length finite-state model cannot describe the whole orbit: decimal lengths are unbounded.

The exact digit and carry equations.

Let $a=\sum_{i=0}^{L-1}d_i10^i$ with $d_{L-1}\ne0$ and $0\le d_i\le9$. The reversal equals $\sum_{i=0}^{L-1}d_{L-1-i}10^i$, including harmless leading zero coefficients when $d_0=0$. With incoming carry $c_0=0$, compute \[ s_i=d_i+d_{L-1-i}+c_i,\qquad e_i=s_i-10c_{i+1},\qquad c_{i+1}=\lfloor s_i/10\rfloor. \] Every carry is zero or one. The output digits are $e_0,\ldots,e_{L-1}$, and a leading digit one is appended exactly when $c_L=1$. These equations handle both length-preserving and length-increasing steps.

If every mirrored digit sum $d_i+d_{L-1-i}$ is at most nine, induction gives all carries zero, and $e_i=d_i+d_{L-1-i}=e_{L-1-i}$. The next iterate is then a palindrome. Consequently every step on a putative Lychrel orbit must have at least one carry. The converse is false: a carry can still yield a palindrome, as $59+95=154$ followed by $154+451=605$ and $605+506=1111$ illustrates. Carry occurrence is an obstruction to the simple no-carry argument, not a certificate of permanent failure.

Congruences do not give a global obstruction here.

Modulo nine, reversal preserves the digit sum, so \[ a_j\equiv2^j a_0\pmod9. \] Every residue modulo nine is represented by a positive palindrome (the one-digit numbers one through nine suffice). Thus this invariant alone never excludes all palindromes. Modulo eleven, using $10\equiv-1$ gives \[ \operatorname{rev}_{10}(a)\equiv(-1)^{L-1}a\pmod{11}. \] An even-length input therefore produces a multiple of eleven. This is compatible with the fact that every even-length palindrome is a multiple of eleven: mirrored positions cancel in its alternating digit sum. Divisibility by eleven is not an obstruction to the very palindromes one needs to avoid.

These calculations illustrate why a modulus must be paired with length and carry information. A residue restriction that excludes only some palindrome lengths cannot establish that no later iterate is palindromic, because the length changes along the orbit.

What a finite invariant proof would require.

One possible approach is to describe arbitrarily long digit strings by a family of patterns closed under the exact carry transducer. To prove nontermination, three assertions would be required: the chosen start belongs to the family, every permitted transition remains in it, and no string in it is a palindrome. Checking the transition on finitely many representative lengths is insufficient unless the rule explicitly proves closure for every possible length and every boundary carry. The incoming carry at the low end is fixed while a high-end carry can append a digit, so mirror symmetry of the input addition does not make the carry evolution symmetric automatically.

No such closed family containing $196$ is established here. In particular the repeated presence of carries during an experiment does not imply their permanent presence. It is a necessary condition along the observed prefix only.

Executed experiment and exact stopping rule.

Starting with $196$, the script performs $1000$ reverse-and-add steps using integer arithmetic and tests the complete decimal string after each step. It sees no palindrome. The final integer has $411$ digits; its SHA-256 digest is stored in the result file to make the run reproducible without printing the entire integer. This checksum identifies the computed output and is not a mathematical certificate about subsequent iterates. Files are checks_second.py and results_second.json under _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. The implemented reversal correctly sends $120$ to $21$.

Remaining gap and outcome.

The no-cycle result, digit equations, no-carry sufficient condition, and modular identities are proved. None gives an invariant preventing all future palindromes for any specific start. Outcome: UNRESOLVED. The start $196$ remains a finite-tested candidate in this notebook, not a proved decimal Lychrel number.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1256 (NT-049), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 Eric W. Weisstein, Lychrel Number, Wolfram MathWorld. https://mathworld.wolfram.com/LychrelNumber.html. Bibliographic information from the supplied record.

• S3 Jason Doucette, The 196 Palindrome Quest. https://p196.org/. Bibliographic information from the supplied record.

• E1 The p196 project, 196 and Other Lychrel Numbers, project research and computation site. https://p196.org/. Inspected 1 October 2026. The local thousand-step experiment is independent of its larger reported computations.