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%I #36 Apr 17 2022 01:14:41
%S 1,6,102,543,37008
%N Numbers k such that the sum of the distinct prime divisors of the product of all legal permutations of the digits of k is equal to k-1.
%C 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.
%C From _David A. Corneth_, Feb 15 2019: (Start)
%C Let S(m) be the sum of the distinct prime divisors of the product of all legal permutations of the digits of m.
%C 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.
%C All terms with at most k digits can be found by iterating only over terms in A179239 with at most k digits.
%C 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.
%C 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.
%C 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).
%C 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.
%C 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.
%C 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)
%e 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.
%e 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.
%Y Cf. A179239.
%K nonn,more,base
%O 1,2
%A _Christopher Hohl_, Feb 13 2019
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