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Least m>0 such that 2^n-m and n-m are relatively prime.
2

%I #6 Feb 13 2021 18:26:20

%S 1,1,1,3,1,1,1,1,1,3,1,1,1,1,2,5,1,1,1,1,1,3,1,1,1,1,1,3,1,1,1,1,1,3,

%T 1,3,1,1,1,3,1,1,1,1,1,3,1,1,1,1,1,3,1,1,1,3,3,5,1,1,1,1,1,3,1,1,1,1,

%U 1,3,1,1,1,1,1,3,1,3,1,1,1,3,1,1,1,1,1,3,1,1,1,1,1,3,1,3,1,1,2

%N Least m>0 such that 2^n-m and n-m are relatively prime.

%H Clark Kimberling, <a href="/A214071/b214071.txt">Table of n, a(n) for n = 1..1000</a>

%e gcd(15,3) = 3, gcd(14,2) = 2, gcd(13,1) = 1, so that a(4) = 3.

%t b[n_] := 2^n; c[n_] := n;

%t Table[m = 1; While[GCD[b[n] - m, c[n] - m] != 1, m++]; m, {n, 1, 140}]

%t rp[n_]:=Module[{m=1,n2=2^n},While[!CoprimeQ[n2-m,n-m],m++];m]; Array[ rp,100] (* _Harvey P. Dale_, Feb 13 2021 *)

%Y Cf. A214056.

%K nonn,easy

%O 1,4

%A _Clark Kimberling_, Jul 26 2012