A270994 - OEIS

%I #46 Sep 11 2022 12:17:17

%S 9454129,20638939,31823749,43008559,54193369,65378179,76562989,

%T 87747799,98932609,110117419,121302229,132487039,143671849,154856659,

%U 166041469,177226279,188411089,199595899,210780709,221965519,233150329,244335139,255519949,266704759,277889569,289074379,300259189

%N a(n) = 9454129 + 11184810*n.

%C See A270971 for the motivation.

%C These are all Sierpiński numbers.

%C 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 Also 9454129 + 28 = 9454157 is a term of A244561. So, with the same proof, a(n) + 28 is a Sierpiński number too.

%C Are a(n) and a(n) + 28 always consecutive Sierpiński numbers?

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

%F G.f.: (9454129 + 1730681*x)/(1 - x)^2.

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

%e a(1) = 9454129 + 11184810*1 = 20638939.

%p A270994:=n->9454129 + 11184810*n: seq(A270994(n), n=0..40); # _Wesley Ivan Hurt_, Apr 02 2016

%t Table[9454129 + 11184810*n, {n, 0, 100}] (* _G. C. Greubel_, Mar 28 2016 *)

%o (PARI) a(n) = 9454129 + 11184810*n;

%o (PARI) x='x+O('x^99); Vec((9454129+1730681*x)/(1-x)^2)

%o (Magma) [9454129 + 11184810*n: n in [0..30]]; // _Vincenzo Librandi_, Mar 29 2016

%o (Python) for n in range(0,100):print(9454129+11184810*n) # _Soumil Mandal_, Apr 03 2016

%Y Cf. A076336, A244561, A270971, A270993.

%K nonn,easy

%O 0,1

%A _Altug Alkan_, Mar 28 2016