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

1,2

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)

David A. Corneth, a(n) = [product of digits of a(n)] * [sum of digits of a(n)] * [some factor]

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

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

Sequence in context: A037714 A037616 A048532 * A075231 A085773 A066042

Adjacent sequences: A049099 A049100 A049101 * A049103 A049104 A049105

Keyword

nonn,base

Author

Olivier Gérard

Status

approved