Running least common multiples: distinct values and multiplicities

The exact growth restriction on a dominating bound.

A bound need not be small at every late endpoint. For example, set

At an odd endpoint $a=\ell +h$, no segment with $\ell \ge 1$ can pass: its residue is at most $W_{\ell ,h}=P_a/P_{\ell }\le 8^a/2<G(B,a)$. At even endpoints the test is unchanged. If $S$ is irrational, the proof above therefore supplies successful segments ending at every sufficiently large even index, from each fixed start. Thus $G(B,a)/8^a\to 0$ is not necessary for the equivalence.

More precisely, for every integer-valued $G\ge K$,

Suppose first that $\mathsf{E}(G)$ holds. The earlier rationality contradiction uses only $G\ge K$. For a successful segment starting at $\ell \ge a_0$, put $a=\ell +h$. Since a least positive residue is at most its modulus,

Such endpoints satisfy $a>a_0$, and $P_{a_0}\to \mathrm{\infty }$; hence the lower limit is zero. Conversely, suppose $S$ is irrational and these lower limits vanish. Fix $B$ and $\ell \ge 1$. By the fixed-start residue limit, with $\delta =\lceil B{X}_{\ell }\rceil -B{X}_{\ell }>0$,

The bounds $8^a/15<P_a\le 8^a$ imply $\underset{a}{lim inf}G(B,a)/P_a=0$. Along a sufficiently late subsequence of endpoints the residue therefore exceeds $G$. This gives arbitrarily long successful segments from each fixed start, but need not give success at every sufficiently large length, as the parity example shows.

In particular, $G(B,a)=K(B,a)+8^a$ allows no segment at all. More generally, an eventual lower bound $G(B,a)\ge c{8}^a$, for one eligible $B$ and some $c>0$, excludes every sufficiently late start: choose $P_{\ell }\ge 1/c$ and compare the bound with $W_{\ell ,h}\le 8^{\ell +h}/P_{\ell }$. These are restrictions on the test bound, not additional assumptions about the series. The original quadratic bounds satisfy the lower-limit condition, so the remaining problem is still the existence of the segments. Equivalence alone says nothing about the relative difficulty of the formulations; a useful arithmetic proof must exploit further information about the numerators $m_a$.

A fixed segment can have a lower bound greater than $1$. Write $W$ and $F$ for its product and accumulated numerator. Every prime factor of $W$ lies in $\{2,3,5\}$, so an eligible multiplier $B$ is a unit modulo $W$. Congruence then gives

For $(W,F)=(6,4)$, the possible residues are $2$ and $4$, attained at $B=1$ and $11$; thus $2$ is a lower bound for this segment, and the residues are not units. For $(W,F)=(60,47)$, however, ${\mathrm{lpr}}_{60}(-37\cdot 47)=1$, with $37$ coprime to $30$. Thus no lower bound greater than $1$ works for all segments and eligible multipliers, although a particular segment may have one.