A261301 a(n+1) = abs(a(n) - gcd(a(n), n)), a(1) = 1.

1, 0, 2, 1, 0, 5, 4, 3, 2, 1, 0, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 47, 46, 45, 40, 39, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 79, 78, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60

Offset: 1

Author

M. F. Hasler, Aug 14 2015

Keywords

• nonn

Comments

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

It is conjectured that a(n) = 0 implies that n is prime, see A186253. (This is the sequence {u(n)} mentioned there.)

a(A186253(n)-1) = 1; a(A186253(n)) = 0; a(A186253(n)+1) = A186253(n). - Reinhard Zumkeller, Sep 07 2015

Examples

			a(2) = a(1) - gcd(a(1),1) = 1 - 1 = 0.
a(3) = |a(2) - gcd(a(2),2)| = gcd(0,2) = 2 is prime.
a(3+2) = a(5) = 0, a(6)  = gcd(0,5) = 5 is prime.
a(6+5) = a(11) = 0, a(12) = gcd(0,11) = 11 is prime.
a(12+11) = a(23) = 0, a(24) = 23 is prime.
a(24+23) = a(47) = 0, a(48) = 47 is prime.
a(50) = 45 and gcd(45,50) = 5, thus a(51) = 45 - 5 = 40.
a(52) = 39 and gcd(39,52) = 13, thus a(53) = 39 - 13 = 26. Then, a(53+26) = 0 and 79 = a(80) is prime.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A186253 - A186263, A261302 - A261310, A106108.

Programs

Haskell

a261301 n = a261301_list !! (n-1)
a261301_list = 1 : map abs
   (zipWith (-) a261301_list (zipWith gcd [1..] a261301_list))
-- Reinhard Zumkeller, Sep 07 2015

Mathematica

FoldList[Abs[#1-GCD[#1,#2]]&,1,Range@96] (* Ivan N. Ianakiev, Aug 15 2015 *)

PARI

print1(a=1);m=1;for(n=1,199, print1( ",",a=abs(a-gcd(a,n))))