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 A106387 Numbers j such that 6j^2 + 6j + 1 = 11k. 3

%I

%S 4,6,15,17,26,28,37,39,48,50,59,61,70,72,81,83,92,94,103,105,114,116,

%T 125,127,136,138,147,149,158,160,169,171,180,182,191,193,202,204,213,

%U 215,224,226,235,237,246,248,257,259,268,270,279,281,290,292,301,303

%N Numbers j such that 6j^2 + 6j + 1 = 11k.

%C k sequence = A106388.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F j(1)=4, j(2)=6 then j(n)=j(n-2)+11.

%F a(n) = 11*n - a(n-1) - 12 (with a(1)=4). - _Vincenzo Librandi_, Nov 13 2010

%F a(2k-1) = 11k - 7, a(2k) = 11k - 5. - _Ralf Stephan_, Nov 15 2010

%F From _Bruno Berselli_, Nov 16 2010: (Start)

%F a(n) = (22*n - 7*(-1)^n - 13)/4.

%F G.f.: x*(4+2*x+5*x^2)/((1+x)*(1-x)^2).

%F a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.

%F a(n) - a(n-2) = 11 for n > 2.

%F a(n) - 2*a(n-1) + a(n-2) = -7*(-1)^n for n > 2. (End)

%t Select[Range[320],Divisible[6#^2+6#+1,11]&] (* _Harvey P. Dale_, Sep 10 2011 *)

%o (PARI) Vec((4+2*x+5*x^2)/(1+x)/(1-x)^2+O(x^99)) \\ _Charles R Greathouse IV_, Dec 28 2011

%Y Cf. A106388, A106389, A106390.

%K nonn,easy

%O 1,1

%A _Pierre CAMI_, May 01 2005

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Last modified August 21 06:54 EDT 2019. Contains 326162 sequences. (Running on oeis4.)