A071222 Smallest k such that gcd(n,k) = gcd(n+1,k+1).

1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2

Offset: 0

Author

Benoit Cloitre, Jun 10 2002

Keywords

• easy
• nonn

Comments

a(n) = least m>0 such that gcd(n!+1+m,n-m) = 1. [Clark Kimberling, Jul 21 2012]

From Antti Karttunen, Jan 26 2014: (Start)

a(n-1)+1 = A053669(n) = Smallest k >= 2 coprime to n = Smallest prime not dividing n.

Note that a(n) is equal to A235918(n+1) for the first 209 values of n. The first difference occurs at n=210 and A235921 lists the integers n for which a(n) differs from A235918(n+1).

(End)

Links

• Clark Kimberling & Antti Karttunen, Table of n, a(n) for n = 0..10001 (Terms up to n=1000 from Kimberling)

Crossrefs

One less than A053669(n+1).

Cf. also A007978, A055874, A235918, A235921, A249270.

Programs

Haskell

a071222 n = head [k | k <- [1..], gcd (n + 1) (k + 1) == gcd n k]
-- Reinhard Zumkeller, Oct 01 2014

Mathematica

sgcd[n_]:=Module[{k=1},While[GCD[n,k]!=GCD[n+1,k+1],k++];k]; Array[sgcd,110] (* Harvey P. Dale, Jul 13 2012 *)

PARI

for(n=1,140,s=1; while(gcd(s,n)
				

Scheme

(define (A071222 n) (let loop ((k 1)) (cond ((= (gcd n k) (gcd (+ n 1) (+ k 1))) k) (else (loop (+ 1 k)))))) ;; Antti Karttunen, Jan 26 2014

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A249270 - 1. - Amiram Eldar, Jul 26 2022

Extensions

Added a(0)=1. - N. J. A. Sloane, Jan 19 2014