Improve lower bound for C_3a#71
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Summary
This updates the lower bound for the Gyamarti-Hennecart-Ruzsa sum-difference constant from
[
C_{3a} \geq 1.173077
]
to
[
C_{3a} \geq 1.1740744.
]
The bound comes from the finite digit construction
[
A={0,2,3,4,5,6,7,8,9,10},\qquad B=21,\qquad d=80,\qquad T=150,
]
and
[
U=\left{\sum_{i=0}^{79}a_i21^i:\ a_i\in A,\ \sum_{i=0}^{79}a_i\le150\right}.
]
For this set,
[
\max(U)=10\sum_{i=65}^{79}21^i
]
so
[
\max(U)=
2995805288150731620410662946668903948341032736664352641511848666717243994160370658179324073879212562136150.
]
The exact cardinalities are
[
\lvert U+U\rvert =
75448362167176243488362019935078206851619643198150854886920234689186981134888,
]
and
[
\lvert U-U\rvert =
195351744295266763842135520514417052287242446785296742323733058216909095059024572338564089814415.
]
Thus
[
1+\frac{\log(\lvert U-U\rvert/\lvert U+U\rvert)}
{\log(2\max(U)+1)}
1.174074447693521163363531806658755676155543896382679659744422686584815723921237399321826308369811589\ldots,
]
which certifies the stated conservative lower bound.
Certificate
The construction uses the standard finite-set lower-bound lemma already stated on the
3a.mdpage:[
C_{3a}\geq
1+\frac{\log(\lvert U-U\rvert/\lvert U+U\rvert)}
{\log(2\max(U)+1)}.
]
Sums have no carries because every digit sum lies in ([0,20]), while the base is (21).
Differences are digitwise unique. If
[
\sum_{i=0}^{79}c_i21^i=0,\qquad c_i\in[-20,20],
]
then reducing modulo (21) gives (c_0=0), and induction gives every (c_i=0).
For sums, write (Y=A+A) and
[
P_y={a\in A:y-a\in A}.
]
A sum digit vector (y=(y_i)) is feasible exactly when one can choose (a_i\in P_{y_i}) such that
[
\sum_i a_i\le150,\qquad
\sum_i(y_i-a_i)\le150.
]
For differences, write (\Delta=A-A) and
[
q_\delta=\min{b\in A:b+\delta\in A}.
]
A difference digit vector (\delta=(\delta_i)) is feasible exactly when
[
\sum_i q_{\delta_i}\le150,\qquad
\sum_i(q_{\delta_i}+\delta_i)\le150.
]
The exact counting program below verifies the cardinalities and gives a rigorous logarithmic comparison.
Verification
Expected output includes:
Therefore
[
\log\frac{\lvert U-U\rvert}{\lvert U+U\rvert}
0.1740744\log(2\max(U)+1)
0,
]
so
[
C_{3a}\geq 1.1740744.
]
AI assistance disclosure
The construction, patch, PR text, and verification script were prepared with assistance from ChatGPT 5.5 Pro. The submitter rechecked the construction and arithmetic before submission.