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A337832 Porous numbers: Numbers k which are not multiples of 10 such that every m with sum of digits = k and k a divisor of both m and rev(m) has a zero in its digits.

%I A337832 #42 Apr 07 2021 05:03:12
%S A337832 11,37,74,101,121
%N A337832 Porous numbers: Numbers k which are not multiples of 10 such that every m with sum of digits = k and k a divisor of both m and rev(m) has a zero in its digits.
%C A337832 It is straightforward to prove that a number is not porous by finding an m that does not contain a zero, but satisfies the three requirements. For example, 12 is not a term because 48 is a number with sum of digits equal to 12 and both 48 and 84 are multiples of 12. Many numbers of A333666 can be taken directly to prove that a given value is not a term.
%C A337832 So far only for the five numbers of this sequence no m could be found by computer programs to demonstrate that they are not porous. For a good reason, because for these 5 numbers it can be proven that no such m exists (the proofs are provided in the link below).
%C A337832 Up to k = 1000 no further porous numbers exist (see the linked file with the corresponding m where k | m and k | rev(m)). The steadily increasing number of possibilities to construct an m for a given k suggests that the 5 terms might be the only porous numbers, but a mathematical proof for this conjecture seems a big challenge.
%H A337832 Ruediger Jehn, <a href="/A337832/a337832.txt">Numbers m that disprove k is porous</a>
%H A337832 Ruediger Jehn, <a href="/A337832/a337832_2.pdf">Proof for the first 5 terms</a>
%H A337832 Rüdiger Jehn, <a href="https://arxiv.org/abs/2104.02482">Porous Numbers</a>, arXiv:2104.02482 [math.GM], 2021.
%e A337832 If we search for an m to demonstrate that k = 11 is not porous, we loop through all m where sum of digits = 11 and 11 a divisor of both m and rev(m). We find 209, 308, 407, ... 902, 2090, 3080, ... All m contain at least one zero. If it can be proven that this holds for all m, then 11 is a porous number.
%e A337832 For k = 11 this proof actually is quite easy:
%e A337832 Let "m_s ... m_3 m_2 m_1 m_0" be a number m with digits m_i and the sum of the digits is 11. We define:
%e A337832 A = m_0 + m_2 + m_4 + ... and B = m_1 + m_3 + m_5 + ...
%e A337832 A divisibility rule for 11 requires that the alternating sum of the digits must be a multiple of 11. Hence:
%e A337832    A - B = j * 11
%e A337832 Since the sum of the digits is 11, we have
%e A337832    A + B = 11
%e A337832 Adding the two equations yields
%e A337832    2 * A = (j + 1) * 11
%e A337832 Therefore A must be 0 or 11. If A is 11, then B is 0. This means either A or B must be zero and m must contain a zero. Hence 11 is a porous number.
%Y A337832 Cf. A333666.
%K A337832 nonn,base,more
%O A337832 1,1
%A A337832 _Ruediger Jehn_, Sep 24 2020