A300234 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).

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, 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, 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

Offset: 1

Author

Antti Karttunen, Mar 02 2018

Keywords

• nonn

Examples

			For n = 1, phi(1) = 1, and the arguments for gcd are equal at the start, thus a(1) = 0.
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.
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.
For n = 6, eulerphi(6) = 2, gcd(6,2) = gcd(4,2) = gcd(2,2), two steps required, thus a(6) = 2.
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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Antti Karttunen, Scheme (Racket) program to compute this sequence

Crossrefs

Cf. A000010, A285721.

Cf. also A286594, A300227, A300228, A300237, A300238.

Programs

PARI

A285721(n,k) = if(n==k, 0, 1 + A285721(abs(n-k),min(n,k)));
A300234(n) = A285721(n,eulerphi(n));

Formula

a(n) = A285721(n,A000010(n)).

a(n) = n - A300238(n).