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a(n+1) = abs(a(n) - gcd(a(n), 4n+3)), a(1) = 1.
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%I #6 Aug 22 2015 05:23:50

%S 1,0,11,10,9,8,7,6,5,4,3,2,1,0,59,58,57,56,55,54,53,52,39,38,37,36,35,

%T 34,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,

%U 11,10,9,8,7,6,5,4,3,2,1,0,251,250,249,248,247,246,245,244,243,242,241,240,239,238,237,236

%N a(n+1) = abs(a(n) - gcd(a(n), 4n+3)), a(1) = 1.

%C The absolute value is relevant only when a(n) = 0, in which case a(n+1) = gcd(a(n), 4n+3) = 4n+3.

%C It is conjectured that a(n) = 0 implies that 4n+3 = a(n+1) is prime, cf. A186256. (This is the sequence {u(n)} mentioned there.)

%o (PARI) print1(a=1);for(n=1,199,print1(",",a=abs(a-gcd(a,4*n+3))))

%Y Cf. A261301 - A261310, A186253 - A186263, A106108.

%K nonn

%O 1,3

%A _M. F. Hasler_, Aug 14 2015