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a(n) = 18*n + 5.
2

%I #18 Jun 28 2026 15:48:09

%S 5,23,41,59,77,95,113,131,149,167,185,203,221,239,257,275,293,311,329,

%T 347,365,383,401,419,437,455,473,491,509,527,545,563,581,599,617,635,

%U 653,671,689,707,725,743,761,779,797,815,833,851,869,887,905,923,941,959

%N a(n) = 18*n + 5.

%C Conjecture: there are infinitely many composite Fermat numbers such that no one of them has a divisor that belongs to this sequence.

%H Wikipedia, <a href="https://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.: (5 + 13*x)/(1 - x)^2.

%F From _Elmo R. Oliveira_, Dec 08 2024: (Start)

%F E.g.f.: exp(x)*(5 + 18*x).

%F a(n) = 2*a(n-1) - a(n-2) for n > 1. (End)

%p seq(18*n+5, n=0..53);

%t Table[18*n + 5, {n, 0, 53}]

%t LinearRecurrence[{2,-1},{5,23},60] (* _Harvey P. Dale_, Aug 25 2017 *)

%o (Magma) [18*n+5: n in [0..53]];

%o (PARI) for(n=0, 53, print1(18*n+5, ", "));

%Y Supersequence of A061240.

%Y Cf. A229855.

%K nonn,easy

%O 0,1

%A _Arkadiusz Wesolowski_, May 07 2014