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.

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 24, 36, 128, 135, 144, 175, 384, 672, 735, 1296, 1575, 82944, 139968, 1492992, 27869184

Offset: 1

Author

Bernard Schott, Apr 27 2021

Keywords

• nonn
• base
• fini
• full

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

Examples

			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.
		

Links

• Giovanni Resta, Zuckerman numbers, Numbers Aplenty.

Crossrefs

Cf. A007602, A288069.

Cf. A114440 (similar for Harshad numbers).

Subsequence of A002473 and of A343681.

Programs

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 *)

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

Extensions

a(26) from Michel Marcus, Apr 27 2021