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%I #4 Mar 31 2012 21:08:40
%S 97,1,2,2,316,2,3,3,3,4,12,4,4,12,11,11,316,11,316,316,6,316,316,316,
%T 316,97,316,316,13,316,13,13,13,13,8,13,13,12,13,13,13,13,13,13,14,14,
%U 316,14,316,316,316,97,9,97,97,13,10,10,11,10,14,11,12,12,97,12,97,132
%N Given n, define a(n) as follows: let a sequence b(k) be defined by b(k+1)=b(k)+b(k)mod k; b(1)=2n-1. (Here b(k)mod k denotes the least nonnegative residue of b(k) modulo k). Let a(n) be the common value of b(k+1)-b(k) for all large k if such exists; otherwise let a(n) be 0.
%C Putting b(1)=2n gives essentially the same sequence as putting b(1)=2n-1. It is a plausible conjecture or at least an interesting open problem that a(n) is never zero; that is all the sequences b(k) are arithmetic progressions from some point on. Sequence A073117 is the sequence b(k) with b(1)=1. Do the values a(n) include all positive numbers?
%e n=4: b(1)=7 and the sequence b(k) continues 7,8,10,12,14...with b(k+1)-b(k)=2 for all k>3, so a(4)=2.
%Y Cf. A073117.
%K nonn
%O 1,1
%A _Alex Abercrombie_, Mar 22 2007
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