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%I #5 Nov 14 2016 10:32:33
%S 5,2,5,2,5,7,2,12,2,12,2,12,7,4,14,14,10,14,7,14,6,26,10,4,20,33,2,33,
%T 2,33,2,33,2,38,2,38,2,38,7,4,40,40,10,40,7,14,6,78,10,4,46,85,2,85,2,
%U 85,2,85,2,85,2,85,2,85,2,85,2,85,2,85,2,85,2,85,2,85,2
%N Relative of Hofstadter Q-sequence: a(1) = 5, a(2) = 2; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).
%C In calculating terms of this sequence, use the convention that a(n)=0 for n<=0.
%C Most terms in this sequence alternate between 2 and a term of A275163. These runs are separated by 18 other terms, and each run is approximately twice as long as the previous.
%H Nathan Fox, <a href="/A278066/b278066.txt">Table of n, a(n) for n = 1..10000</a>
%H N. Fox, <a href="https://vimeo.com/191094180">Hofstadter-like Sequences over Nonstandard Integers"</a>, Talk given at the Rutgers Experimental Mathematics Seminar, November 10 2016.
%F a(1) = 5, a(2) = 2, a(3) = 5, a(4) = 2, a(5) = 5, a(6) = 7, a(7) = 2; thereafter, for k>=0,
%F a(A275163(k)+1)=A275163(k)+5
%F a(A275163(k)+2)=2
%F a(A275163(k)+3)=A275163(k)+5
%F a(A275163(k)+4)=2
%F a(A275163(k)+5)=A275163(k)+5
%F a(A275163(k)+6)=7
%F a(A275163(k)+7)=4
%F a(A275163(k)+8)=A275163(k)+7
%F a(A275163(k)+9)=A275163(k)+7
%F a(A275163(k)+10)=10
%F a(A275163(k)+11)=A275163(k)+7
%F a(A275163(k)+12)=7
%F a(A275163(k)+13)=14
%F a(A275163(k)+14)=6
%F a(A275163(k)+15)=2*A275163(k)+12
%F a(A275163(k)+16)=10
%F a(A275163(k)+17)=4
%F a(A275163(k)+18)=A275163(k)+13
%F a(A275163(k)+i)=A275163(k+1), i odd, 19<=i<A275163(k+1)-A275163(k)
%F a(A275163(k)+i)=2, i odd, 20<=i<=A275163(k+1)-A275163(k).
%Y Cf. A005185, A275163, A278067, A278068.
%K nonn
%O 1,1
%A _Nathan Fox_, Nov 13 2016
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