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A168670
Numbers that are congruent to {1, 8} mod 11.
1
1, 8, 12, 19, 23, 30, 34, 41, 45, 52, 56, 63, 67, 74, 78, 85, 89, 96, 100, 107, 111, 118, 122, 129, 133, 140, 144, 151, 155, 162, 166, 173, 177, 184, 188, 195, 199, 206, 210, 217, 221, 228, 232, 239, 243, 250, 254, 261, 265, 272, 276, 283, 287, 294, 298, 305, 309, 316
OFFSET
1,2
COMMENTS
Conjecture: For no term n>1 in the sequence 36*n^2+72*n+35 is equal to p*(p+2), where p, p+2 are twin primes.
This conjecture is evident: in fact, it is sufficient to observe that a(2k) = 11*k-3 and a(2k+1) = 11*k+1, therefore 6*a(2k)+7 = 11*(6*k-1) and 6*a(2k+1)+5 = 11*(6*k+1). [Bruno Berselli, Jan 07 2013, modified Jul 07 2015]
FORMULA
a(n) = (-15+3*(-1)^n+22*n)/4. a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: x*(1+7*x+3*x^2)/((1-x)^2*(1+x)). [Colin Barker, May 15 2012, modified Jul 07 2015]
MATHEMATICA
Select[Range[316], MemberQ[{1, 8}, Mod[#, 11]]&] (* Ray Chandler, Jul 07 2015 *)
LinearRecurrence[{1, 1, -1}, {1, 8, 12}, 58] (* Ray Chandler, Jul 07 2015 *)
Rest[CoefficientList[Series[x*(1+7*x+3*x^2)/((1-x)^2*(1+x)), {x, 0, 58}], x]] (* Ray Chandler, Jul 07 2015 *)
CROSSREFS
Sequence in context: A364776 A033477 A105936 * A291758 A232867 A358574
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Dec 02 2009
EXTENSIONS
3 leading terms added. Conjecture clarified. - R. J. Mathar, Jul 07 2015
STATUS
approved