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A124124
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Nonnegative integers n such that 2n^2+2n-3 is square.
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1
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1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| For squares of the form 2n^2-2n-3, the answer is b(n) = 1 + a(n). - Zak Seidov Dec 04 2006.
First differences are apparently in A143608. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 17 2009]
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FORMULA
| It appears that a(n)=2a(n-1)-2a(n-2)+a(n-3) if n is even, a(n)=5a(n-1)-5a(n-2)+a(n-3) if n is odd. Can anyone confirm this?
Yes, these formulae are correct. - Zak Seidov Dec 04 2006
2*a(n)=sqrt[7+2*A077442(n-1)^2]-1. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 03 2006
a(n)=a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 17 2009]
For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - Charlie Marion, Jan 03 2012
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MATHEMATICA
| LinearRecurrence[{1, 6, -6, -1, 1}, {1, 2, 6, 13, 37}, 40] (* From Harvey P. Dale, Nov 05 2011 *)
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CROSSREFS
| Cf. A001108, A046172, A008844.
Sequence in context: A196906 A162057 A026550 * A052450 A001373 A057243
Adjacent sequences: A124121 A124122 A124123 * A124125 A124126 A124127
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KEYWORD
| nonn,easy
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Nov 29 2006
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EXTENSIONS
| More terms from Harvey P. Dale, Feb 07 2011
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