mathoverflow

deckard @slimer48484

2026-08-03 — saved image

deckard @slimer48484
Fable resolved a MathOverflow question as a side effect of making artwork about it

[attached generative artwork, dark background with dot-grid and glowing diagonal/staircase lines, titled 'THE LEDGER OF HALVES', annotated:]
every n < 2^20 hangs at (log2 odd part, how many times 2 divides n); each row is exactly half the light of the row below.
MO 513837, resolved: Σ_k (2 − 2^(1−k)) B_k = H_(2^N − 1)  exactly —
the dyadic weights are the harmonic series regrouped by odd part;
the weight 2 − 2^(1−k) is the chain of halvings the frame can hold,
and γ = lim ( H_(2^N−1) − N ln 2 ) is the classical limit in disguise.
the shoreline j = N − log2 m: beyond it, the ghost halvings — their shortfall per column is exactly the missing 2^(k−N).
m = 1: the powers of two
odd numbers — the shore every integer hangs from
digits of γ per layer: −log10|S_N − γ| = 0.301·N + 0.549... (error = ψ(2^N) − N ln 2 = −2^(−N−1) − 4^(−N)/12 − ...)
Richardson twin 2S_N − S_(N−1): slope doubles
10:07 PM · Aug 2, 2026 · 1,237 Views
1 reply, 3 reposts, 6 likes, 1 bookmark

[below, a second post from deckard @slimer48484 · 11h, partially visible, titled 'THE HALF-STEP':]
gold record d = 99,890,389 : period 28,965 — the smallest x has 14,869 digits
all 60,792,693 squarefree d ≤ 10^8 · x = log10 d · height = log10( R / ln 2√d ), R = log ε_d, ε_d = fundamental solution of x² − d y²
above the horizon: continued-fraction period ODD — ε has norm −1, the ladder takes a half-step and x² − d y² = −1 is solve[d]
below, mirrored: period EVEN — cyan was allowed −1 (no prime ≡ 3 mod 4) and refused; violet was forbidden from the sta[rt] [cut off]

fablegenerative artmathematicsmathoverflowtwitternumber theory