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A261303 a(n+1) = abs(a(n) - gcd(a(n), 3n+2)), a(1) = 1. 1

%I #10 Aug 22 2015 05:28:40

%S 1,0,8,7,0,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0,71,70,63,62,61,

%T 60,59,58,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,

%U 37,36,35,34,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9

%N a(n+1) = abs(a(n) - gcd(a(n), 3n+2)), a(1) = 1.

%C The absolute value is relevant only when a(n) = 0, in which case a(n+1) = gcd(a(n), 3n+2) = 3n+2.

%C It is conjectured that a(n) = 0 implies that 3n+2 = a(n+1) is prime, for all n > 2, cf. A186255. (This is the sequence {u(n)} mentioned there.)

%o (PARI) print1(a=1);for(n=1,199,print1(",",a=abs(a-gcd(a,3*n+2))))

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

%K nonn

%O 1,3

%A _M. F. Hasler_, Aug 14 2015

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Last modified August 29 23:34 EDT 2024. Contains 375520 sequences. (Running on oeis4.)