[seqfan] Re: Re counting problem

[Prof. Dr. Alois Heinz]

I've just added more terms.

Alois

%S A151753 
5,7,3,9,5,3,3,9,1,3,7,7,3,7,7,5,1,5,7,9,3,7,3,3,9,9,9,5,3,7,1,7,9,5,7,
%T A151753 
9,3,1,3,9,5,7,5,5,7,7,1,1,3,1,9,5,9,1,9,5,5,7,7,7,7,7,9,7,9,9,7,7,3,3,
%U A151753 
5,9,7,7,1,1,5,9,7,1,5,1,7,5,9,1,7,9,3,5,3,9,1,7,9,9,1,9,9,1,5,3,7,5,9
%E A151753 More terms from Alois P. Heinz (heinz(AT)hs-heilbronn.de), 
Jun 18 2009
%p A151753 a:= proc(n) option remember; local k, l; if n=1 then b(1):= 
5; 5 else a(n-1); l:= b(n-1); for k from 1 to 9 by 2 while (parse 
(cat(k, l)) mod 5^n)<>0 do od; b(n):= parse (cat(k, l)); k fi end: seq 
(a(n), n=1..120); [From Alois P. Heinz (heinz(AT)hs-heilbronn.de), Jun 
18 2009]
 
%S A151752 5,75,375,9375,59375,359375,3359375,93359375,193359375,3193359375,
%T A151752 
73193359375,773193359375,3773193359375,73773193359375,773773193359375,
%U A151752 5773773193359375,15773773193359375,515773773193359375
%E A151752 More terms from Alois P. Heinz (heinz(AT)hs-heilbronn.de), 
Jun 18 2009
%p A151752 a:= proc(n) option remember; local k, l; if n=1 then 5 else 
l:= a(n-1); for k from 1 to 9 by 2 while (parse (cat(k, l)) mod 5^n)<>0 
do od; parse (cat(k, l)) fi end: seq (a(n), n=1..30); [From Alois P. 
Heinz (heinz(AT)hs-heilbronn.de), Jun 18 2009]

N. J. A. Sloane schrieb:

>I'm adding three entries, A151752-A151754, based 
>on this thread!
>Neil
>
>
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