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A343744 Zuckerman numbers which divided by the product of their digits give integers which are also divisible by the product of their digits, and so on, until result is 1.

%I A343744 #48 Mar 31 2023 09:17:53
%S A343744 1,2,3,4,5,6,7,8,9,12,15,24,36,128,135,144,175,384,672,735,1296,1575,
%T A343744 82944,139968,1492992,27869184
%N A343744 Zuckerman numbers which divided by the product of their digits give integers which are also divisible by the product of their digits, and so on, until result is 1.
%C A343744 Repunits >= 11 (A002275) are not in the sequence because, as they are fixed points of this map, they don't fit the definition.
%C A343744 Question: is this sequence finite as the similar sequence with Niven numbers (A114440) that has 15095 terms?
%C A343744 No other terms up to 2*10^9. - _Michel Marcus_, Apr 27 2021
%C A343744 From _David A. Corneth_, Apr 27 2021: (Start)
%C A343744 Terms are 7-smooth. Any prime factor > 7 will not be divided away by dividing by product of digits.
%C A343744 Any number k > a(26)*10^163 with product of digits vp > 0 has k/vp > a(26) so it suffices to check all candidates <= a(26)*10^163. Doing so gives no more terms so this sequence is finite and full. (End)
%C A343744 The number of steps needed to reach 1, has a maximum of 3, which occurs for n = 21, 23..26. - _A.H.M. Smeets_, Apr 29 2021
%H A343744 Giovanni Resta, <a href="https://www.numbersaplenty.com/set/Zuckerman_number/">Zuckerman numbers</a>, Numbers Aplenty.
%e A343744 The integer 1296 is divisible by the product of its digits as 1296/(1*2*9*6) = 12, then 12/(1*2) = 6 and 6/6 = 1; hence, 1296 is a term of this sequence.
%t A343744 f[n_] := If[(prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod], n/prod, 0]; Select[Range[10^5], FixedPointList[f, #][[-1]] == 1 &] (* _Amiram Eldar_, Apr 27 2021 *)
%o A343744 (PARI) isz(n) = my(p=vecprod(digits(n))); p && !(n % p); \\ A007602
%o A343744 isok(n) = if (n==1, return(1)); my(m=n); until(m==1, if (isz(m), my(nm = m/vecprod(digits(m))); if (nm==m, return (0), m = nm), return(0))); return(1); \\ _Michel Marcus_, Apr 27 2021
%o A343744 (Python)
%o A343744 def proddigit(n):
%o A343744     p = 1
%o A343744     while n > 0:
%o A343744         n, p = n//10, p*(n%10)
%o A343744     return p
%o A343744 n, a = 1, 1
%o A343744 while n > 0:
%o A343744     aa, pa = a, proddigit(a)
%o A343744     while pa > 1 and aa%pa == 0 and aa > 1:
%o A343744         aa = aa//pa
%o A343744         pa = proddigit(aa)
%o A343744     if aa == 1:
%o A343744         print(n,a)
%o A343744         n = n+1
%o A343744     a = a+1 # _A.H.M. Smeets_, Apr 29 2021
%Y A343744 Cf. A007602, A288069.
%Y A343744 Cf. A114440 (similar for Harshad numbers).
%Y A343744 Subsequence of A002473 and of A343681.
%K A343744 nonn,base,fini,full
%O A343744 1,2
%A A343744 _Bernard Schott_, Apr 27 2021
%E A343744 a(26) from _Michel Marcus_, Apr 27 2021