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Numbers n such that n = (product of digits of n) * (sum of digits of n) in some base.
0

%I #5 Oct 03 2014 12:03:37

%S 0,1,6,12,16,20,30,42,48,54,56,72,90,96,110,128,132,135,144,156,160,

%T 162,176,182,210,231,240,250,272,300,306,324,336,342,380,384,420,432,

%U 448,455,462,480,495,504,506,540,552,576,600,624,650,663,686,702,720,750

%N Numbers n such that n = (product of digits of n) * (sum of digits of n) in some base.

%C This sequence is infinite since b^2+b is in the sequence for all b>1: in base b, b^2+b has digits {1,b} and (1*b)*(1+b)=b^2+b.

%e 12 is in the sequence since in base 9, 12 has digits {1,3} and (1*3)*(1+3)=12.

%Y Cf. A038369.

%K base,nonn

%O 1,3

%A _Matthew Conroy_, Feb 02 2006