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A106387
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Numbers j such that 6j^2 + 6j + 1 = 11k.
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3
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| k sequence = A106388
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (1,1,-1).
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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) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 13 2010]
a(2k-1)=11k-7, a(2k)=11k-5. - Ralf Stephan, Nov 15 2010
Contribution from Bruno Berselli (berselli.bruno(AT)yahoo.it), 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)
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MATHEMATICA
| Select[Range[320], Divisible[6#^2+6#+1, 11]&] (* From Harvey P. Dale, Sep 10 2011 *)
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PROG
| (PARI) Vec((4+2*x+5*x^2)/(1+x)/(1-x)^2+O(x^99)) \\ Charles R Greathouse IV, Dec 28 2011
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CROSSREFS
| Cf. A106388, A106389, A106390.
Sequence in context: A048753 A055719 A117883 * A034771 A034764 A119034
Adjacent sequences: A106384 A106385 A106386 * A106388 A106389 A106390
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KEYWORD
| nonn,easy
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AUTHOR
| Pierre CAMI (pierre-cami(AT)bbox.fr), May 01 2005
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