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%I #4 Aug 22 2015 05:27:31
%S 1,0,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0,89,88,77,76,75,72,71,70,69,68,
%T 67,66,65,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48,47,46,45,
%U 44,43,42,41,40,39,38,37,36,35,34,33,32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14
%N a(n+1) = abs(a(n) - gcd(a(n), 5*n+4)), u(1) = 1.
%C It is conjectured that a(n) = 0 implies that 5n+4 = a(n+1) is prime for n > 2, cf. A186256. (This is the sequence {u(n)} mentioned there.)
%e a(2) = a(1) - gcd(a(1),5+4) = 1 - 1 = 0.
%e a(3) = |a(2) - gcd(a(2),5*2+4)| = 14.
%e a(19) = 88, thus a(20) = 88 - gcd(88, 5*19+4) = 88 - 11 = 77.
%o (PARI) print1(a=1);for(n=1,199,print1(",",a=abs(a-gcd(a,5*n+4))))
%Y Cf. A261301 - A261310, A186253 - A186263, A106108.
%K nonn
%O 1,3
%A _M. F. Hasler_, Aug 14 2015
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