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Search: id:A105051
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| A105051 |
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Define a(1)=0, a(2)=0, a(3)=15, a(4)=111 then a(n)=254*a(n-2)+126-a(n-4) also sequence such that 7*(a(n)^2) + 7*a(n) + 1 = a square. |
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| 0, 0, 15, 111, 3936, 28320, 999855, 7193295, 253959360, 1827068736, 64504677711, 464068265775, 16383934179360, 117871512438240, 4161454776879855, 29938900091047311, 1056993129393303936, 7604362751613578880
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OFFSET
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1,3
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COMMENT
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This sequence is such that 7*(a(n)^2) + 7*a(n) + 1 = j^2 = a square integer positive root of a(n)^2 + a(n) - (j^2-1)/7 = 0 is such that 2*a(n)+1=sqrt((4*j^2+3)/7), ((4*j^2+3)/7) need to be a square so j^2 = (7*k^2-3)/4, put k=2*a(n)+1 you found that (7*k^2-3)/4 = 7*(a(n)^2) + 7*a(n) + 1
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FORMULA
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a(n)=A105040(n-2). G.f.: 3x^2(5+32x+5x^2)/((1-x)(1+16x+x^2)(1-16x+x^2)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 28 2008]
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CROSSREFS
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Sequence in context: A034184 A092646 A105040 this_sequence A110822 A001849 A115150
Adjacent sequences: A105048 A105049 A105050 this_sequence A105052 A105053 A105054
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KEYWORD
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nonn
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AUTHOR
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Pierre CAMI (pierrecami(AT)tele2.fr), Apr 04 2005
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EXTENSIONS
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Extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 28 2008
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