[seqfan] Re: Pseudo-arithmetic progressions

Vladimir Shevelev shevelev at bgu.ac.il
Fri Jul 16 20:59:55 CEST 2010

As a continuation, I have just submitted also the following two sequences:

%I A179480
%S A179480 1,1,2,1,3,3,2,1,5,2,6,5,5,7,2,1,6,9,6,3,3,6,12
%N A179480 A dual sequence to A179382 
%C A179480 Let m>k>0 be odd numbers. Denote m<->k=A000265(m-k). Then the sequence m<->k, m<->(m<->k), m<->(m<->(m<->k)),... is periodic. In this sequence, a(n) is the smallest period in case of m=2*n-1,k=1. 
%e A179480 If n=14, then m=27 and we have 27<->1=13, 27<->13=7, 27<->7=5, 27<->5=11, 27<->11=1. Thus a(14)=5. 
%Y A179480 A179382, A179383, A000265 
%K A179480 nonn
%O A179480 2,3

%I A179481
%S A179481 3,7,11,19,23,29,37,47
%N A179481 a(n) = 2*t(n)-1 where t(n) is the sequence of records positions of A179480. 
%C A179481 Question. Whether every term of this sequence is prime? 
%Y A179481 A179480 A179460 A179382, A179383 
%K A179481 nonn
%O A179481 2,1

I call A179480 a dual to A179382, since if to replace all <-> by <+> , then, in view of the commutativity of binary operation <+>, we obtain the corresponding pseudo-arithmetic progression.
Now, of course, to every (0,1)-sequence, one can correspond a sequence of this type by the rule that <+> corresponds to 1 and <-> corresponds to 0. It is easy to see that, if a given (0,1)-sequence is (eventually) periodic, then the corresponding sequence of the considered type will be (eventually) periodic as well.


 Shevelev Vladimir‎

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