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a(n) is the least k > n such that gcd(k,n) > 1 and gcd(k+1,n+1) > 1.
2

%I #16 Mar 08 2018 16:16:46

%S 8,9,14,15,20,21,14,15,32,33,38,39,20,21,50,51,56,57,26,27,68,69,34,

%T 35,32,33,86,87,92,93,38,39,44,45,110,111,44,45,122,123,128,129,50,51,

%U 140,141,62,63,56,57,158,159,64,65,62,63,176,177,182,183,68,69

%N a(n) is the least k > n such that gcd(k,n) > 1 and gcd(k+1,n+1) > 1.

%H Robert Israel, <a href="/A299143/b299143.txt">Table of n, a(n) for n = 2..10000</a>

%F From _Rémy Sigrist_, Feb 04 2018: (Start)

%F a(p) = 3 * p for any odd prime p.

%F a(2*k + 1) = a(2*k) + 1 for any k > 0.

%F a(n) = n + 2*A172170(n + 1) for any n > 1.

%F (End)

%e 8 is the least k>2 such that gcd(8,2)>1 and gcd(9,3)>1. So a(2)=8.

%e 15 is the least k>9 such that gcd(15,9)>1 and gcd(16,10)>1. Therefore a(9)=15.

%p f:= proc(n) local k;

%p for k from n+1 do if igcd(k,n)>1 and igcd(k+1,n+1)>1 then return k fi od

%p end proc:

%p map(f, [$2..100]); # _Robert Israel_, Mar 08 2018

%t Array[Block[{k = # + 1}, While[Or[CoprimeQ[#, k], CoprimeQ[# + 1, k + 1]], k++]; k] &, 62, 2] (* _Michael De Vlieger_, Feb 03 2018 *)

%o (PARI) a(n) = for (k=n+1, oo, if (gcd(n,k)>1 && gcd(n+1, k+1)>1, return (k))) \\ _Rémy Sigrist_, Feb 04 2018

%Y Cf. A172170.

%Y Cf. A061228 or A159475 (when simply gcd(k,n) > 1).

%K nonn

%O 2,1

%A _Alex Ratushnyak_, Feb 03 2018