A121760/1: two (interesting?) sequences

[zak seidov]

Seqfans,
I've just send two (interesting?) sequences,
two b-files
and two relative (nice?) graphs with 1000 points.
Hope they are of some interest,
Thanks, Zak
 

%I A121760
%S A121760
1,2,3,4,5,6,7,8,9,1,11,21,31,41,51,61,71,81,91,20,21,22,23,24,25,26,27,28,29,3,13,23,33,43,53,63,73,83,93
%N A121760 In decimal number system, take negative
power of 10 at odd digits of n.
Sequence gives numerators of result.
%C A121760 See accompanying  sequence A121761 In
decimal number system, take negative power of 10 at 
even digits of n.
%F A121760 If n = sum(d(i)*10^(i-1)), then
a(n)=sum(d(i)*10^((-1)^(1+d(i))*(i-1))).
%e A121760 a(12)=21 because 12=1*10^1+2*10^0 and 
a(12)=numerator[1*10^((-1)^(1)*1)+2*10^((-1)^(0)*0)=1/10+2=21/10]=21.
%A A121760 Zak Seidov (zakseidov at yahoo.com), Aug 20
2006


%I A121761
%S A121761 
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,1,6,11,16,21,26,31,36,41,46,30,31,32,33,34,35,36,37,38,39,2
,7,12
%N A121761 In decimal number system, take negative
power of 10 at even digits of n.
Sequence gives numerators of result.
%C A121761 See accompanying  sequence A121760 In
decimal number system, take negative power of 10 at 
odd digits of n.
%F A121761 If n=sum(d(i)*10^(i-1)), then
a(n)=sum(d(i)*10^((-1)^(1+d(i))*(i-1)))
%e A121761 a(23)=16 because 23=2*10^1+3*10^0 and 
a(23)=numerator[2*10^((-1)^(1+2)*1)+3*10^((-1)^(1+3)*0)=2/10+3=32/10=16/5]=16.
%A A121761 Zak Seidov (zakseidov at yahoo.com), Aug 20
2006


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