A371561 Numbers with multiplicative digital root of 5 that are free of 1s and have their digits in ascending order.

5, 35, 57, 355, 359, 557, 579, 3335, 3357, 5579, 5777, 33557, 35559, 333555, 357799, 557779, 3335779, 3355777, 33333577

Offset: 1

Author

Sergio Pimentel, Mar 27 2024

Keywords

• nonn
• more
• base

Comments

Conjectured to be complete.

If it exists, a(20) > 10^500. - Michael S. Branicky, Apr 18 2024

Crossrefs

Cf. A034052, A263473, A263479, A031347.

Programs

Mathematica

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 *)

Python

from math import prod
from itertools import count, islice, combinations_with_replacement as mc
def A031347(n):
    while n > 9: n = prod(map(int, str(n)))
    return n
def bgen(): yield from (m for d in count(1) for m in mc((3,5,7,9), d))
def agen(): yield from (int("".join(map(str, t))) for t in bgen() if A031347(prod(t)) == 5)
print(list(islice(agen(), 19))) # Michael S. Branicky, Apr 17 2024, edited Apr 18 2024 after Chai Wah Wu

Python

from math import prod
from itertools import count, islice
def A371561_gen(): # generator of terms
    for l in count(1):
        for a in range(l,-1,-1):
            a3 = 3**a
            for b in range(l-a,-1,-1):
                b3 = a3*5**b
                for c in range(l-a-b,-1,-1):
                    d = l-a-b-c
                    d3 = b3*7**c*9**d
                    while d3 > 9:
                        d3 = prod(int(x) for x in str(d3))
                    if d3==5:
                        yield (10**(a+b+c+d)-1)//3+(10**d*(10**c*(10**b+1)+1)-3)*2//9
A371561_list = list(islice(A371561_gen(),19)) # Chai Wah Wu, Apr 17 2024