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%I #18 Feb 08 2022 22:42:15
%S 1,1,2,1,3,3,2,1,5,2,6,5,5,7,2,1,6,9,6,3,3,6,12,10,4,13,10,3,15,15,2,
%T 1,17,10,18,2,10,14,20,13,21,2,14,4,6,4,18,11,9,25,26,4,27,9,18,5,22,
%U 4,12,27,10,25,2,1,33,6,18,15,35,22,30,3,22,37,6,12,10,13,26
%N Let m>k>0 be odd numbers and denote by the symbol "m<->k" the value A000265(m-k). Then the sequence m<->k, m<->(m<->k), m<->(m<->(m<->k)), ... is periodic; a(n) is the smallest period in the case m=2*n-1, k=1.
%C A dual sequence to A179382.
%C Let b = (2*n-1) and k = A003558(n-1). If a(n) is odd, b divides (2^k + 1); but if a(n) is even, b divides (2^k - 1). Examples: a(14) = 5, odd; with b = 27 and A003558(13) = 9. Then 27 divides (2^9 + 1) or 513 = 27 * 19. a(18) = 6, even. b = 35, with k= A003558(17) = 12. Then 35 divides (2^12 - 1). - _Gary W. Adamson_, Aug 20 2012.
%C Iff a(n) = n/2 or (n-1)/2, then 2*n - 1 is a prime with one coach and is in A216371. Examples: a(19) = 9, so 37 is in A216371. a(12) = 6, so 23 is in A216371. - _Gary W. Adamson, Sep 08 2012.
%e 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.
%p Contribution from _R. J. Mathar_, Nov 04 2010: (Start)
%p A179480aux := proc(x,y) local xtrack,xitr,xpos ; xtrack := [y] ; while true do xitr := A000265(x-op(-1,xtrack)) ; if not member(xitr, xtrack,'xpos') then xtrack := [op(xtrack),xitr] ; else return 1+nops(xtrack)-xpos ; end if; end do: end proc:
%p A179480 := proc(n) A179480aux(2*n-1,1) ; end proc: seq(A179480(n),n=2..80) ; (End)
%t oddres[n_] := n/2^IntegerExponent[n, 2];
%t b[x_, y_] := Module[{xtrack = {y}, xitr}, While[True, xitr = oddres[x - Last@ xtrack]; If[FreeQ[xtrack, xitr], AppendTo[xtrack, xitr], Return[ Length[xtrack]]]]];
%t a[n_] := b[2n-1, 1];
%t a /@ Range[2, 80] (* _Jean-François Alcover_, Apr 13 2020, after _R. J. Mathar_ *)
%Y Cf. A179382, A179383, A000265.
%Y Cf. A003558.
%Y Cf. A216371.
%K nonn
%O 2,3
%A _Vladimir Shevelev_, Jul 16 2010
%E Edited by _N. J. A. Sloane_, Jul 18 2010
%E More terms from _R. J. Mathar_, Nov 04 2010
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