sum*product POSTSCRIPT

[Richard Guy]

A small glitch in the following, which shd have 10^k - 1
instead of 10^(k-1).  This enables one to see that k < 60.
Still a bit big for a brute force search!
Note that if 5 occurs, then all digits are odd.
Also that, except possibly for one factor of size
< 9k, the number  n  is 7-smooth.       R.

%I A038369
%S A038369 0,1,135,144
%N A038369 n = (product of digits of n) * (sum of digits of n).
%C A038369 To prove finiteness, let k be the number of digits and consider 9^k*9*k < 10^(k-1) for every k > 100 - Ulrich Schimke (UlrSchimke at aol.com).
%H A038369 E. W. Weisstein, <a href="http://mathworld.wolfram.com/Sum-ProductNumber.html">Link to a section of The World of Mathematics.</a>
%H A038369 E. W. Weisstein, <a href="http://mathworld.wolfram.com/Digit.html">Link to a section of The World of Mathematics.</a>
%e A038369 144 belongs to the sequence because 1*4*4=16, 1+4+4=9 -> 16*9=144
%Y A038369 n = A007953(n) * A007954(n).
%K A038369 nice,nonn,fini,base,full,bref
%O A038369 1,3
%A A038369 Felice Russo (felice.russo at katamail.com)

On Tue, 11 Dec 2001, N. J. A. Sloane wrote:

> i should have looked harder 
> 
> this is A038369, which is labeled as "fini,full"