For each odd \(n\leq10^8\), the computation set \(m=\operatorname{bitlength}(n)\), \(B=2^{m-1}+1\), and tested \(k=1,2,\ldots,B\) in order. The first \(k\) for which \(kn\) had even bit length and half of its bits equal to 1 was recorded as \(k(n)\).

This loop gives a finite stopping proof for each input. Finding a balanced product at \(k\) proves minimality because every smaller multiplier has been rejected. Exhausting \(1\leq k\leq B\) would exhibit a counterexample to the bound. A hit at \(B\) is compared directly with \(n=2^m-1\), so either direction of the proposed equality characterization would stop the run on its first failure.

The run checked 50,000,000 odd values of \(n\) and 724,052,131 multiplier candidates. It found no failure. Equality occurred 26 times, at \(n=2^m-1\) for \(1\leq m\leq26\). The largest minimum in range was \(k(67108863)=33554433\). The FNV-1a checksum of the little-endian 64-bit words \((n\ll32)\mathbin{|}k(n)\), in increasing \(n\), is `6955bbea71c91c4c`.

This record certifies the stated finite range. The universal bound and the converse assertion about equality remain open in this entry.

Fix \(m\geq1\), put \(n=2^m-1\), and let \(u=k-1\). The identity \[ k(2^m-1)=u2^m+(2^m-1-u) \] expresses the product as two binary blocks of width \(m\): the padded \(m\)-bit expansion of \(u\), followed by its bitwise complement. The padded word has length \(2m\) and exactly \(m\) ones.

If \(1\leq k\leq2^{m-1}\), then \(u<2^{m-1}\). Its padded block begins with at least one zero. Ordinary binary notation removes those leading zeros while retaining all \(m\) ones in the two-block word. The resulting word has fewer than \(2m\) bits and still has \(m\) ones, so it is unbalanced.

For \(k=2^{m-1}+1\), one has \(u=2^{m-1}\). The first block now begins with 1, so the ordinary expansion has all \(2m\) bits. The two complementary blocks contain \(m\) ones in total, making the product balanced. Every smaller positive multiplier has already been excluded, which proves \[ k(2^m-1)=2^{m-1}+1. \] This proves the sharp family in the candidate and covers the forward direction of its equality statement.

Write an odd \(m\)-bit input as \(n=2^m-a\), where \(a\) is odd. For \(k\geq2\) with \(ka\leq2^m\), direct subtraction gives \[ kn=(k-1)2^m+(2^m-ka). \] The low block is the \(m\)-bit complement of \(ka-1\). Put \(t=\operatorname{bitlength}(k-1)\) and let \(s_2\) denote popcount. The ordinary product has \(m+t\) bits and \[ s_2(kn)=s_2(k-1)+m-s_2(ka-1). \] Consequently, \(kn\) is balanced exactly when \[ 2\bigl(s_2(ka-1)-s_2(k-1)\bigr)=m-t. \] For \(a=1\), the two popcounts agree and the criterion yields the Mersenne proof. For general odd \(a\), the digit-sum difference has no monotonicity that would force a solution before \(2^{m-1}+1\). Multipliers with \(ka>2^m\) also introduce a quotient carry into the high block. These two gaps prevent this criterion from proving the universal claim.

Spiegelhofer and Stoll study the binary sum-of-digits function \(s_2\) along arithmetic progressions. Their Theorem 1.1 proves full complexity, up to a shift, for finite tuples of digit-sum differences. Their elementary discussion also points out the carry difficulty introduced by multiplication by 3.

Focused searches for balanced binary multiples, prescribed binary weight in a residue class, and the exact bound \(2^{m-1}+1\) found no primary source stating this candidate. The cited paper concerns eventual digit-sum patterns and gives no multiplier of the required size. The novelty and universal status of the candidate remain unverified.

The program uses 64-bit products, hardware-equivalent exact popcount, and the exact bit length of each positive product. The largest possible product in this run is below \(2^{63}\). Its output was `DONE max_n=100000000 odd_n=50000000 tested_k=724052131 equality=26 record_n=67108863 record_k=33554433 fnv1a=6955bbea71c91c4c`.

A separate Python 3 implementation repeated the same increasing search through \(n=100000\). It checked 50,000 odd inputs and 587,572 multipliers, found 16 equality cases, ended with record \((65535,32769)\), and produced the matching prefix checksum `68d628a3847dc259`.