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Numbers n such that 2n-1 and 2n+1 divide 2^n-1.
1

%I #22 Dec 19 2014 12:04:44

%S 8,128,228,648,3240,5976,13160,23760,23940,24840,32768,37224,78540,

%T 82800,139248,166716,238368,278520,280368,288360,516528,633420,664668,

%U 731808,734448,1145520,1211100,1377240,1425816,1484568,1627640,2055060,2131080,2292780

%N Numbers n such that 2n-1 and 2n+1 divide 2^n-1.

%C Intersection of A081856 and A081858.

%C Numbers n such that 4n^2-1 divides 2^n-1. - _Charles R Greathouse IV_, Dec 04 2013

%H Charles R Greathouse IV, <a href="/A233089/b233089.txt">Table of n, a(n) for n = 1..5000</a>

%e 2^8-1=255 is divisible by 2*8-1=15 and by 2*8+1=17.

%t Select[Range[23*10^5],PowerMod[2,#,2#+{1,-1}]=={1,1}&] (* _Harvey P. Dale_, Dec 19 2014 *)

%o (PARI) isok(n) = !((2^n-1) % (2*n-1)) && !((2^n-1) % (2*n+1));

%o (PARI) is(n)=Mod(2,4*n^2-1)^n==1 \\ _Charles R Greathouse IV_, Dec 04 2013

%K nonn

%O 1,1

%A _Michel Marcus_, Dec 04 2013