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a(n+1) = abs(a(n) - gcd(a(n), 9n+8)), u(1) = 1.
2

%I #5 Aug 22 2015 05:07:26

%S 1,0,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,

%T 2,1,0,269,268,267,266,265,264,263,262,261,260,259,258,257,256,255,

%U 254,253,252,251,250,249,248,247,246,245,244,243,242,241,240,239,238,237,236,235,234,233,232,231

%N a(n+1) = abs(a(n) - gcd(a(n), 9n+8)), u(1) = 1.

%C It is conjectured that for all n > 2, u(n) = 0 implies that u(n+1) = 9n+8 is prime, cf. A186261. (This is the sequence {u(n)} mentioned there.)

%e a(2) = a(1) - gcd(a(1),9+8) = 1 - 1 = 0.

%e a(3) = |a(2) - gcd(a(2),9*2+8)| = gcd(0,26) = 26.

%e a(3+26) = a(29) = 0 and a(29+1) = gcd(0,9*29+8) = 269 is prime.

%o (PARI) print1(a=1);for(n=1,99,print1(",",a=abs(a-gcd(a,9*n+8))))

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

%K nonn

%O 1,3

%A _M. F. Hasler_, Aug 14 2015