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Relative of Hofstadter Q-sequence: a(n) = 0 if n <= 0, a(1) = a(2) = a(3) = 1; a(n) = a(n-a(n-1)) + a(n-a(n-2)) + a(n-a(n-3)) for n > 3.
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%I #22 Feb 23 2016 13:34:30

%S 1,1,1,3,7,8,3,8,10,4,5,12,19,11,4,19,27,11,8,22,23,11,13,25,23,21,12,

%T 29,35,12,9,35,48,23,12,44,34,24,11,62,39,10,51,30,16,31,55,62,11,11,

%U 124,78,10,30,39,79,23,20,55,103,47,7,98

%N Relative of Hofstadter Q-sequence: a(n) = 0 if n <= 0, a(1) = a(2) = a(3) = 1; a(n) = a(n-a(n-1)) + a(n-a(n-2)) + a(n-a(n-3)) for n > 3.

%C This sequence has exactly 416 terms. a(416) = 0, so an attempt to calculate a(417) would refer to itself.

%C Without the convention that a(n) = 0 for n <= 0, this sequence would have exactly five terms, since computing a(6) refers to a(-1).

%H Nathan Fox, <a href="/A267501/b267501.txt">Table of n, a(n) for n = 1..416</a> (full sequence)

%e a(10) = 4 because a(9) is 10, a(8) is 8, and a(7) is 3, so we take a(10 - 10) + a(10 - 8) + a(10 - 3) = a(0) + a(2) + a(7) = 0 + 1 + 3 = 4.

%Y Cf. A005185

%K nonn,fini,full

%O 1,4

%A _Nathan Fox_, Feb 23 2016