A168670 - OEIS

A168670

Numbers that are congruent to {1, 8} mod 11.

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

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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]

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

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

Adjacent sequences: A168667 A168668 A168669 * A168671 A168672 A168673