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Search: id:A114040
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| A114040 |
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a(0) = 1, a(1) = 9, a(n) = 6*a(n-1) - a(n-2) - 4. |
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+0
1
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| 1, 9, 49, 281, 1633, 9513, 55441, 323129, 1883329, 10976841, 63977713, 372889433, 2173358881, 12667263849, 73830224209, 430314081401, 2508054264193, 14618011503753, 85200014758321, 496582077046169, 2894292447518689, 16869172608065961, 98320743200877073
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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The most straightforward test for "triangularity" is istriangle(n) <===> issquare(8*n+1). If this sequence could be proved to be free of squares beyond the first three terms, that would lead directly to a proof that 0, 1 and 6 are the only triangular numbers whose squares are triangular numbers.
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FORMULA
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a(n)=1-sqrt(2)*[3-2*sqrt(2)]^n+[3+2*sqrt(2)]^n*sqrt(2), with n>=0 - Paolo P. Lava (ppl(AT)spl.at), Jun 25 2008
G.f.: (1+2x-7x^2)/((1-x)(1-6x+x^2)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 09 2008]
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CROSSREFS
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Equals 8*A001109(n)+1. It is also A081554(n)+1.
Adjacent sequences: A114037 A114038 A114039 this_sequence A114041 A114042 A114043
Sequence in context: A055428 A012231 A123270 this_sequence A090390 A069665 A066558
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KEYWORD
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nonn
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AUTHOR
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njas, based on email from Jack Brennen, Feb 01 2006
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