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Integers n, divisible by six, for which t = (4^n - 5 * 2^n - 4 * 4^(n^2 - 2n) + 8 * 2^(n^2 - 2n)) / (4 * (n^3 - n)) is an integer
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%I #7 Dec 31 2024 18:09:33

%S 6,12,18,30,36,42,60,72,108,180,192,240,270,312,420,432,462,600,660,

%T 810,882,1092,1152,1290,1296,1302,1320,1620,1722,1872,2028,2112,2268,

%U 2310,2340,2592,2688,2700,2790,2970,3000,3120,3258,3300,3360,3390,3528,3540

%N Integers n, divisible by six, for which t = (4^n - 5 * 2^n - 4 * 4^(n^2 - 2n) + 8 * 2^(n^2 - 2n)) / (4 * (n^3 - n)) is an integer

%C Then n - 1 and n + 1 are almost always a pair of twin primes, and the set of these should be an infinite subset of all twin primes.

%C If n - 1 and n + 1 are simultaneously composite, this occurs very rarely.

%H Harvey P. Dale, <a href="/A162864/b162864.txt">Table of n, a(n) for n = 1..100</a>

%t Select[6*Range[600],IntegerQ[(4^#-5*2^#-4*4^(#^2-2#)+8*2^(#^2-2#))/(4*(#^3-#))]&] (* _Harvey P. Dale_, Dec 31 2024 *)

%Y Cf. A001097

%Y Cf. A001359, A014574, A006512

%K easy,nonn

%O 1,1

%A _Reikku Kulon_, Jul 15 2009