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A106328
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Numbers j such that 8*(j^2) + 9 = k^2 for some positive number k.
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11
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0, 3, 18, 105, 612, 3567, 20790, 121173, 706248, 4116315, 23991642, 139833537, 815009580, 4750223943, 27686334078, 161367780525, 940520349072, 5481754313907, 31950005534370, 186218278892313, 1085359667819508, 6325939728024735, 36870278700328902, 214895732473948677
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OFFSET
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1,2
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COMMENTS
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The ratio k(n) /(2*j(n)) tends to sqrt(2) as n increases.
The squares of the numbers in this sequence are one less than a triangular number: a(n)^2 = A164080(n). For example, 18^2 is 324, and 325 is a triangular number. a(n)^2 + 1 = A164055(n). a(n)^2 = A072221(n)(A072221(n)+1)/2 - 1. - Tanya Khovanova & Alexey Radul, Aug 09 2009
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
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FORMULA
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a(1)=0, a(2)=3 then a(n) = 6*a(n-1) - a(n-2).
a(n) = ((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1))*3/4/sqrt(2). - Max Alekseyev, Jan 11 2007
a(n) = 3*A001109(n). - M. F. Hasler, R. J. Mathar, Jun 03 2009
a(n) = (3/4)*A005319(n-1).
G.f.: 3x^2/(1-6x+x^2). - Philippe Deléham, Nov 17 2008
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MATHEMATICA
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s=0; lst={}; Do[s+=n; If[Sqrt[s-1]==Floor[Sqrt[s-1]], AppendTo[lst, Sqrt[s-1]]], {n, 8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
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PROG
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(Haskell)
a106328 n = a106328_list !! (n-1)
a106328_list = 0 : 3 : zipWith (-) (map (* 6) (tail a106328_list)) a106328_list
-- Reinhard Zumkeller, Jan 10 2012
(PARI) concat(0, Vec(3*x^2/(1-6*x+x^2) + O(x^40))) \\ Michel Marcus, Sep 07 2016
(PARI) a(n)=([0, 1; -1, 6]^n*[-3; 0])[1, 1] \\ Charles R Greathouse IV, Sep 07 2016
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CROSSREFS
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Cf. A103328, A164080, A164055, A072221.
Sequence in context: A009021 A124408 A136779 * A007277 A025595 A151331
Adjacent sequences: A106325 A106326 A106327 * A106329 A106330 A106331
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KEYWORD
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nonn,easy
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AUTHOR
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Pierre CAMI, Apr 29 2005
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EXTENSIONS
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More terms from Max Alekseyev, Jan 11 2007
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STATUS
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approved
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