A106387 Numbers j such that 6j^2 + 6j + 1 = 11k.
4, 6, 15, 17, 26, 28, 37, 39, 48, 50, 59, 61, 70, 72, 81, 83, 92, 94, 103, 105, 114, 116, 125, 127, 136, 138, 147, 149, 158, 160, 169, 171, 180, 182, 191, 193, 202, 204, 213, 215, 224, 226, 235, 237, 246, 248, 257, 259, 268, 270, 279, 281, 290, 292, 301, 303
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Programs
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Mathematica
Select[Range[320],Divisible[6#^2+6#+1,11]&] (* Harvey P. Dale, Sep 10 2011 *)
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PARI
Vec((4+2*x+5*x^2)/(1+x)/(1-x)^2+O(x^99)) \\ Charles R Greathouse IV, Dec 28 2011
Formula
j(1)=4, j(2)=6 then j(n)=j(n-2)+11.
a(n) = 11*n - a(n-1) - 12 (with a(1)=4). - Vincenzo Librandi, Nov 13 2010
a(2k-1) = 11k - 7, a(2k) = 11k - 5. - Ralf Stephan, Nov 15 2010
From Bruno Berselli, Nov 16 2010: (Start)
a(n) = (22*n - 7*(-1)^n - 13)/4.
G.f.: x*(4+2*x+5*x^2)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) - a(n-2) = 11 for n > 2.
a(n) - 2*a(n-1) + a(n-2) = -7*(-1)^n for n > 2. (End)
Comments