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%I #27 Nov 20 2020 04:49:01
%S 1,2,3,4,5,6,7,8,9,12,14,15,16,18,21,24,25,27,28,32,35,36,42,45,48,54,
%T 56,63,64,72,75,84,96,112,125,126,128,135,144,147,162,168,175,189,192,
%U 224,225,252,256,288,294,336,375,378,384,441,448,486,512,567,576,588,625,672,675,729
%N Numbers that are a divisor of the product of the digits of its divisors.
%C In the first 130 million numbers there are 117 terms, the last being 234375. It is possible no more terms exist, although this is unknown.
%C The number with the largest ratio of it divisors' digit product to the number itself is 5376, with a ratio of ~2.2*10^43.
%C From _David A. Corneth_, Aug 22 2020: (Start)
%C The product of the digits of divisors must be nonzero. For example, 10 isn't a term as the product of digits of divisors is 0.
%C Terms are 7-smooth.
%C The largest term is 234375. Proof:
%C 2^k1 has a 0 for k1 = 10, 3^k2 has a 0 for k2 = 10, 5^k3 has a 0 for k3 = 8 and 7^k4 has a 0 for k4 = 4.
%C Checking all numbers of the form 2^e1 * 3^e2 * 5^e3 * 7^e4 with ei bounded by the respective ki gives the 117 terms mentioned above with the largest of them being 234375. (End)
%H Scott R. Shannon, <a href="/A337230/b337230.txt">Table of n, a(n) for n = 1..117</a>
%e a(4) = 4 is a term as the divisors of 4 are 1,2,4 and 1*2*4 = 8 which is a divisible by 4.
%e a(10) = 12 is a term as the divisors of 12 are 1,2,3,4,6,12 and 1*2*3*4*6*1*2 = 288 which is divisible by 12.
%e a(19) = 28 is a term as the divisors of 28 are 1,2,4,7,14,28 and 1*2*4*7*1*4*2*8 = 3584 which is divisible by 28.
%t Select[Range[3^6], (prod = Times @@ (Times @@@ IntegerDigits @ Divisors[#])) > 0 && Divisible[prod, #] &] (* _Amiram Eldar_, Aug 22 2020 *)
%o (PARI) is(n) = {if(n < 10, return(n > 0)); f = factor(n); if(f[#f~, 1] > 7, return(0)); my(d = divisors(n), p = 1); for(i = 2, #d, dd = digits(d[i]); for(j = 1, #dd, p *= dd[j]); if(p == 0, return(0))); p % n == 0} \\ _David A. Corneth_, Aug 22 2020
%Y Cf. A002473, A027750, A000005, A007954, A007602, A005349, A038186, A333617.
%K nonn,fini,full,base
%O 1,2
%A _Scott R. Shannon_, Aug 20 2020
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