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a(n) = 64^n + 1.
12

%I #34 Sep 08 2022 08:46:05

%S 2,65,4097,262145,16777217,1073741825,68719476737,4398046511105,

%T 281474976710657,18014398509481985,1152921504606846977,

%U 73786976294838206465,4722366482869645213697,302231454903657293676545,19342813113834066795298817,1237940039285380274899124225

%N a(n) = 64^n + 1.

%C These numbers can be written as the sum of two relatively prime squares and also as the sum of two relatively prime cubes (i.e., 2^(6*n) + 1 = (2^(3*n))^2 + 1^2 = (2^(2*n))^3 + 1^3).

%H Arkadiusz Wesolowski, <a href="/A228081/b228081.txt">Table of n, a(n) for n = 0..100</a>

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (65,-64).

%F a(n) = 64*a(n-1) - 63.

%F a(n) = A089357(n) + 1 = 2^A008588(n) + 1.

%F G.f.: (2 - 65*x)/((1 - x)*(1 - 64*x)).

%F E.g.f.: e^x + e^(64*x).

%e a(2) = 64^2 + 1 = 4097.

%t Table[64^n + 1, {n, 0, 15}]

%t LinearRecurrence[{65,-64},{2,65},20] (* _Harvey P. Dale_, Jul 17 2020 *)

%o (Magma) [64^n+1 : n in [0..15]]

%o (PARI) for(n=0, 15, print1(64^n+1, ", "))

%Y Cf. A000051 (2^n + 1), A052539 (4^n + 1), A062395 (8^n + 1).

%K easy,nonn

%O 0,1

%A _Arkadiusz Wesolowski_, Aug 09 2013