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%I #24 Dec 11 2024 00:21:13
%S 257,641,1025,1409,1793,2177,2561,2945,3329,3713,4097,4481,4865,5249,
%T 5633,6017,6401,6785,7169,7553,7937,8321,8705,9089,9473,9857,10241,
%U 10625,11009,11393,11777,12161,12545,12929,13313,13697,14081,14465,14849,15233,15617,16001
%N a(n) = 384*n + 257.
%C Every composite Fermat number has at least two divisors of the form 384*n + 257, n > 0.
%H Arkadiusz Wesolowski, <a href="/A229855/b229855.txt">Table of n, a(n) for n = 0..1000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Fermat_number">Fermat number</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F G.f.: (257 + 127*x)/(1 - x)^2.
%F a(n) = 128*A016789(n) + 1.
%F From _Elmo R. Oliveira_, Dec 08 2024: (Start)
%F E.g.f.: exp(x)*(257 + 384*x).
%F a(n) = 2*a(n-1) - a(n-2) for n > 1. (End)
%p seq(384*n+257, n=0..40);
%t Table[384*n + 257, {n, 0, 40}]
%o (Magma) [384*n+257 : n in [0..40]];
%o (PARI) for(n=0, 40, print1(384*n+257, ", "));
%Y Cf. A000215, A016789, A094358, A229853, A229854, A229856.
%K nonn,easy
%O 0,1
%A _Arkadiusz Wesolowski_, Oct 01 2013