%I #36 Apr 18 2024 18:06:10
%S 5,35,57,355,359,557,579,3335,3357,5579,5777,33557,35559,333555,
%T 357799,557779,3335779,3355777,33333577
%N Numbers with multiplicative digital root of 5 that are free of 1s and have their digits in ascending order.
%C Conjectured to be complete.
%C If it exists, a(20) > 10^500. - _Michael S. Branicky_, Apr 18 2024
%t A031347 = Table[NestWhile[Times @@ IntegerDigits[#] &, n, # > 9 &], {n, 1, 100000}]; Select[Range[100000], A031347[[#]] == 5 && DigitCount[#, 10, 1] == 0 && Sort[IntegerDigits[#]] == IntegerDigits[#] &] (* _Vaclav Kotesovec_, Apr 17 2024 *)
%o (Python)
%o from math import prod
%o from itertools import count, islice, combinations_with_replacement as mc
%o def A031347(n):
%o while n > 9: n = prod(map(int, str(n)))
%o return n
%o def bgen(): yield from (m for d in count(1) for m in mc((3,5,7,9), d))
%o def agen(): yield from (int("".join(map(str, t))) for t in bgen() if A031347(prod(t)) == 5)
%o print(list(islice(agen(), 19))) # _Michael S. Branicky_, Apr 17 2024, edited Apr 18 2024 after _Chai Wah Wu_
%o (Python)
%o from math import prod
%o from itertools import count, islice
%o def A371561_gen(): # generator of terms
%o for l in count(1):
%o for a in range(l,-1,-1):
%o a3 = 3**a
%o for b in range(l-a,-1,-1):
%o b3 = a3*5**b
%o for c in range(l-a-b,-1,-1):
%o d = l-a-b-c
%o d3 = b3*7**c*9**d
%o while d3 > 9:
%o d3 = prod(int(x) for x in str(d3))
%o if d3==5:
%o yield (10**(a+b+c+d)-1)//3+(10**d*(10**c*(10**b+1)+1)-3)*2//9
%o A371561_list = list(islice(A371561_gen(),19)) # _Chai Wah Wu_, Apr 17 2024
%Y Cf. A034052, A263473, A263479, A031347.
%K nonn,more,base
%O 1,1
%A _Sergio Pimentel_, Mar 27 2024