OFFSET
1,2
COMMENTS
Empirically, it looks as if every number of the form (10^3^n-1)/9 has this property. - David W. Wilson, Dec 12 2001
From David A. Corneth, Jan 23 2019: (Start)
Indeed, (10^3^n-1)/9 is in the sequence. It has digital sum times product of digits equal to 3^n.
Proof: (10^3^0-1)/9 = (10^1-1)/9 = 1 is in the sequence.
If (10^3^k-1)/9 is in the sequence then (10^3^(k + 1)-1)/9 = ((10^3^k-1)/9) * (10^(2*3^k) + 10^(3^k) + 1) = 3 * m * ((10^3^k-1)/9) for some m. This number is divisible by 3 * 3^k = 3^(k + 1) so (10^3^(k+1) - 1)/9 is in the sequence and so (10^3^n - 1) / 9 is in the sequence from which it follows that the sequence is infinite. (End)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..19637 (first 100 terms from Vincenzo Librandi, terms < 10^14)
EXAMPLE
432 is a term because: 4*3*2=24, 4+3+2=9, 24*9=216 and 432/216 = 2.
MATHEMATICA
Select[ Range[10^6], IntegerQ[ # /(Apply[ Times, IntegerDigits[ # ]] * Apply[ Plus, IntegerDigits[ # ]] ) ] & ]
PROG
(PARI) isok(n) = my(d=digits(n)); vecprod(d) && (n % (vecsum(d)*vecprod(d)) == 0); \\ Michel Marcus, Jan 23 2019
CROSSREFS
KEYWORD
nonn,base
AUTHOR
STATUS
approved