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COMMENTS
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It appears that these constitute all of the 'non-degenerate' cases for k less than 10^8. That is, k is not allowed to have leading zeros, but all 'legal' permutations of k, where k has length m, must be of length m as well. Therefore leading zeros are allowed in the construction of the total permutation product.
Let S(m) be the sum of the distinct prime divisors of the product of all legal permutations of the digits of m.
Let Z(m) be a number where a zero is inserted after the first digit of m (m > 0). For example, Z(1) = 10, Z(19) = 109.
All terms with at most k digits can be found by iterating only over terms in A179239 with at most k digits.
For example, 345 is in A179239. S(345) = S(543), namely 543. As 543 is a permutation of 345, s = 543 is in the sequence.
Similarily, 445 is in A179239 and S(445) = 341, 445 doesn't produce a term. As S(445) = S(454) = S(544), all these number don't produce a term and do not have to be checked.
We have S(Z(m)) >= S(m). Proof: The permutations of Z(m) give the same distinct prime factors as m does, and maybe more. Therefore, S(Z(m)) >= S(m).
This can be applied to eliminate candidates. For example, S(10378) = 1447642. The largest possible value a number with digits of Z(10378) = 100378 can have is 873100. But 1447642 > 873100. So 100378 can't produce a term and doesn't have to be checked.
To perhaps quickly eliminate a candidate without checking all permutations one might let the last digit d of a permutation be such that gcd(d, 10) = 1 to hopefully get large prime factors (if there is such d). For example, when checking if 1378 gives a candidate, start with the 12 permutations ending in 1 or 3.
To find S(m) where m has digits zero, one might use a known value of S(m') where m' has a digit 0 removed from m and proceed finding S(m) with permutations having leading nonzero digits only. (End)
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EXAMPLE
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6 is a term because it is the product of its legal permutations. The distinct prime factors of 6 are 3 and 2, so 3+2=5 and 6-1=5.
102 is a term because the distinct prime factors of the product of its legal permutations, i.e., 102*120*210*201*21*12 = 130195900800, are 2,3,5,7,17, and 67. So, 2+3+5+7+17+67=101, and 102-1=101.
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