cp's OEIS Frontend

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

%I A049102 #36 Feb 16 2019 01:10:26
%S A049102 1,12,111,112,135,144,216,432,2112,11112,11115,11232,12312,13824,
%T A049102 14112,21112,23112,27216,31212,41112,81216,93312,111132,122112,124416,
%U A049102 131112,132192,186624,212112,221112,221184,222912,239112,248832,311472,316224
%N A049102 Positive numbers n such that n is a multiple of (product of digits of n) * (sum of digits of n).
%C A049102 Empirically, it looks as if every number of the form (10^3^n-1)/9 has this property. - _David W. Wilson_, Dec 12 2001
%C A049102 From _David A. Corneth_, Jan 23 2019: (Start)
%C A049102 Indeed, (10^3^n-1)/9 is in the sequence. It has digital sum times product of digits equal to 3^n.
%C A049102 Proof: (10^3^0-1)/9 = (10^1-1)/9 = 1 is in the sequence.
%C A049102 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)
%H A049102 David A. Corneth, <a href="/A049102/b049102.txt">Table of n, a(n) for n = 1..19637</a> (first 100 terms from Vincenzo Librandi, terms < 10^14)
%H A049102 David A. Corneth, <a href="/A049102/a049102.gp.txt">a(n) = [product of digits of a(n)] * [sum of digits of a(n)] * [some factor]</a>
%e A049102 432 is a term because: 4*3*2=24, 4+3+2=9, 24*9=216 and 432/216 = 2.
%t A049102 Select[ Range[10^6], IntegerQ[ # /(Apply[ Times, IntegerDigits[ # ]] * Apply[ Plus, IntegerDigits[ # ]] ) ] & ]
%o A049102 (PARI) isok(n) = my(d=digits(n)); vecprod(d) && (n % (vecsum(d)*vecprod(d)) == 0); \\ _Michel Marcus_, Jan 23 2019
%Y A049102 Cf. A038369, A049101, A049105, A049106, A052382.
%K A049102 nonn,base
%O A049102 1,2
%A A049102 _Olivier Gérard_