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%I #8 Mar 02 2018 22:44:35
%S 0,1,2,1,4,2,6,1,2,3,10,2,12,4,8,1,16,2,18,3,4,6,22,2,4,7,2,4,28,6,30,
%T 1,9,9,9,2,36,10,5,3,40,4,42,6,8,12,46,2,6,3,10,7,52,2,5,4,6,15,58,6,
%U 60,16,4,1,10,8,66,9,11,13,70,2,72,19,8,10,13,6,78,3,2,21,82,4,24,22,12,6,88,6,11,12,8,24,13,2,96,4,9,3,100,10,102,7,9
%N a(n) = number of steps in simple Euclidean algorithm for gcd(n,k) to reach the termination test n=k when starting with n = n and k = phi(n).
%H Antti Karttunen, <a href="/A300234/b300234.txt">Table of n, a(n) for n = 1..65537</a>
%H Antti Karttunen, <a href="/A286594/a286594.txt">Scheme (Racket) program to compute this sequence</a>
%F a(n) = A285721(n,A000010(n)).
%F a(n) = n - A300238(n).
%e For n = 1, phi(1) = 1, and the arguments for gcd are equal at the start, thus a(1) = 0.
%e For n = 2, eulerphi(2) = 1, gcd(2,1) = gcd(1,1), thus 1 step were required to reach the termination condition, and a(2) = 1.
%e For n = 5, eulerphi(5) = 4, gcd(5,4) = gcd(4,1) = gcd(3,1) = gcd(2,1) = gcd(1,1), four steps required, thus a(5) = 4.
%e For n = 6, eulerphi(6) = 2, gcd(6,2) = gcd(4,2) = gcd(2,2), two steps required, thus a(6) = 2.
%e Here a simple subtracting version of gcd-algorithm is used, where the new versions of two arguments will be the smaller argument and the smaller argument subtracted from the larger, and this is repeated until both are equal.
%o (PARI)
%o A285721(n,k) = if(n==k, 0, 1 + A285721(abs(n-k),min(n,k)));
%o A300234(n) = A285721(n,eulerphi(n));
%Y Cf. A000010, A285721.
%Y Cf. also A286594, A300227, A300228, A300237, A300238.
%K nonn
%O 1,3
%A _Antti Karttunen_, Mar 02 2018
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