A286594 - OEIS

A286594

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 = A000203(n).

0, 2, 3, 4, 5, 1, 7, 8, 6, 5, 11, 4, 13, 5, 4, 16, 17, 7, 19, 11, 12, 6, 23, 3, 10, 6, 15, 1, 29, 5, 31, 32, 7, 7, 9, 13, 37, 7, 9, 5, 41, 6, 43, 11, 7, 8, 47, 7, 14, 14, 7, 11, 53, 7, 11, 8, 15, 9, 59, 6, 61, 9, 12, 64, 10, 8, 67, 11, 8, 20, 71, 9, 73, 10, 13, 9, 23, 9, 79, 17, 42, 11, 83, 4, 11, 11, 8, 23, 89, 5, 7, 9, 16, 12, 8, 6, 97, 17, 9, 16, 101, 11

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OFFSET

1,2

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10000

Antti Karttunen, Scheme (Racket) program to compute this sequence

FORMULA

a(n) = A285721(n, A000203(n)) = A285721(A000203(n), n).

a(n) = n - A300237(n). - Antti Karttunen, Mar 02 2018

EXAMPLE

For n = 1, sigma(1) = 1, and the arguments for gcd are equal at the start, thus a(1) = 0.

For n = 2, sigma(2) = 3, gcd(3,2) = gcd(2,1) = gcd(1,1), thus 2 steps were required to reach the termination condition, and a(2) = 2.

For n = 6, sigma(6) = 12, gcd(12,6) = gcd(6,6), thus a(6) = 1.

For n = 9, sigma(9) = 13, gcd(13,9) = gcd(9,4) = gcd(5,4) = gcd(4,1) = gcd(3,1) = gcd(2,1) = gcd(1,1), thus a(9) = 6.

Here the simple subtracting version of gcd-algorithm is used, where the new arguments will be the smaller argument and the smaller argument subtracted from the larger, and this is repeated until both are equal.

PROG

(Scheme) (define (A286594 n) (A285721bi n (A000203 n))) ;; Requires also code from A000203 and A285721.

(Python)

from sympy import divisor_sigma

def A(n, k): return 0 if n==k else 1 + A(abs(n - k), min(n, k))

def a(n): return A(n, divisor_sigma(n)) # Indranil Ghosh, May 22 2017

(PARI)

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

A286594(n) = A285721(n, sigma(n)); \\ Antti Karttunen, Mar 02 2018

CROSSREFS

Cf. A000203, A009194, A285721.

Cf. A000396 (positions of 1's).