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A270994 a(n) = 9454129 + 11184810*n.

%I A270994 #46 Sep 11 2022 12:17:17
%S A270994 9454129,20638939,31823749,43008559,54193369,65378179,76562989,
%T A270994 87747799,98932609,110117419,121302229,132487039,143671849,154856659,
%U A270994 166041469,177226279,188411089,199595899,210780709,221965519,233150329,244335139,255519949,266704759,277889569,289074379,300259189
%N A270994 a(n) = 9454129 + 11184810*n.
%C A270994 See A270971 for the motivation.
%C A270994 These are all Sierpiński numbers.
%C A270994 Since 9454129 is a term of A244561, for every integer k > 0, 9454129*2^k + 1 has a divisor in the set {3, 5, 7, 13, 17, 241}. And because 11184810 = 2*3*5*7*13*17*241, a(n)*2^k + 1 = 9454129*2^k + 1 + 11184810*n*2^k + 1 always has a divisor in the set {3, 5, 7, 13, 17, 241}. Since a(n) is always odd because of its definition, a(n) is a Sierpiński number.
%C A270994 Also 9454129 + 28 = 9454157 is a term of A244561. So, with the same proof, a(n) + 28 is a Sierpiński number too.
%C A270994 Are a(n) and a(n) + 28 always consecutive Sierpiński numbers?
%H A270994 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F A270994 G.f.: (9454129 + 1730681*x)/(1 - x)^2.
%F A270994 a(n) = 2*a(n-1) - a(n-2) for n > 1.
%e A270994 a(1) = 9454129 + 11184810*1 = 20638939.
%p A270994 A270994:=n->9454129 + 11184810*n: seq(A270994(n), n=0..40); # _Wesley Ivan Hurt_, Apr 02 2016
%t A270994 Table[9454129 + 11184810*n, {n, 0, 100}] (* _G. C. Greubel_, Mar 28 2016 *)
%o A270994 (PARI) a(n) = 9454129 + 11184810*n;
%o A270994 (PARI) x='x+O('x^99); Vec((9454129+1730681*x)/(1-x)^2)
%o A270994 (Magma) [9454129 + 11184810*n: n in [0..30]]; // _Vincenzo Librandi_, Mar 29 2016
%o A270994 (Python) for n in range(0,100):print(9454129+11184810*n) # _Soumil Mandal_, Apr 03 2016
%Y A270994 Cf. A076336, A244561, A270971, A270993.
%K A270994 nonn,easy
%O A270994 0,1
%A A270994 _Altug Alkan_, Mar 28 2016