OFFSET
1,2
COMMENTS
Repunits >= 11 (A002275) are not in the sequence because, as they are fixed points of this map, they don't fit the definition.
Question: is this sequence finite as the similar sequence with Niven numbers (A114440) that has 15095 terms?
No other terms up to 2*10^9. - Michel Marcus, Apr 27 2021
From David A. Corneth, Apr 27 2021: (Start)
Terms are 7-smooth. Any prime factor > 7 will not be divided away by dividing by product of digits.
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)
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
LINKS
Giovanni Resta, Zuckerman numbers, Numbers Aplenty.
EXAMPLE
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.
MATHEMATICA
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 *)
PROG
(PARI) isz(n) = my(p=vecprod(digits(n))); p && !(n % p); \\ A007602
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
(Python)
def proddigit(n):
p = 1
while n > 0:
n, p = n//10, p*(n%10)
return p
n, a = 1, 1
while n > 0:
aa, pa = a, proddigit(a)
while pa > 1 and aa%pa == 0 and aa > 1:
aa = aa//pa
pa = proddigit(aa)
if aa == 1:
print(n, a)
n = n+1
a = a+1 # A.H.M. Smeets, Apr 29 2021
CROSSREFS
KEYWORD
nonn,base,fini,full
AUTHOR
Bernard Schott, Apr 27 2021
EXTENSIONS
a(26) from Michel Marcus, Apr 27 2021
STATUS
approved