A049102 Positive numbers n such that n is a multiple of (product of digits of n) * (sum of digits of n).

1, 12, 111, 112, 135, 144, 216, 432, 2112, 11112, 11115, 11232, 12312, 13824, 14112, 21112, 23112, 27216, 31212, 41112, 81216, 93312, 111132, 122112, 124416, 131112, 132192, 186624, 212112, 221112, 221184, 222912, 239112, 248832, 311472, 316224

Offset: 1

Author

Olivier Gérard

Keywords

• nonn
• base

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)

Examples

			432 is a term because: 4*3*2=24, 4+3+2=9, 24*9=216 and 432/216 = 2.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..19637 (first 100 terms from Vincenzo Librandi, terms < 10^14)
• David A. Corneth, a(n) = [product of digits of a(n)] * [sum of digits of a(n)] * [some factor]

Crossrefs

Cf. A038369, A049101, A049105, A049106, A052382.

Programs

Mathematica

Select[ Range[10^6], IntegerQ[ # /(Apply[ Times, IntegerDigits[ # ]] * Apply[ Plus, IntegerDigits[ # ]] ) ] & ]

PARI

isok(n) = my(d=digits(n)); vecprod(d) && (n % (vecsum(d)*vecprod(d)) == 0); \\ Michel Marcus, Jan 23 2019