Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
A371561
Numbers with multiplicative digital root of 5 that are free of 1s and have their digits in ascending order.
1
5, 35, 57, 355, 359, 557, 579, 3335, 3357, 5579, 5777, 33557, 35559, 333555, 357799, 557779, 3335779, 3355777, 33333577
OFFSET
1,1
COMMENTS
Conjectured to be complete.
If it exists, a(20) > 10^500. - Michael S. Branicky, Apr 18 2024
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,more,base
AUTHOR
Sergio Pimentel, Mar 27 2024
STATUS
approved