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Search: id:A001571
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| A001571 |
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a(0) = 0, a(1) = 2, a(n) = 4a(n-1) - a(n-2) + 1.
(Formerly M1928 N0762)
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+0
9
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| 0, 2, 9, 35, 132, 494, 1845, 6887, 25704, 95930, 358017, 1336139, 4986540, 18610022, 69453549, 259204175, 967363152, 3610248434, 13473630585, 50284273907, 187663465044, 700369586270, 2613814880037, 9754889933879, 36405744855480
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Second member of the Diophantine pair (m,k) that solves 3(m^2+m)=k^2+k: a(n)=k. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
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REFERENCES
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S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
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LINKS
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S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
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FORMULA
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a(n)=(A001834(n)-1)/2.
a(n)=-(1/2)-(1/4)*sqrt(3)*[2-sqrt(3)]^n+(1/4)*sqrt(3)*[2+sqrt(3)]^n+(1/4)*[2-sqrt(3)]^n+(1/4) *[2+sqrt(3)]^n, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Jul 31 2008]
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MAPLE
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A001571:=z*(-2+z)/(-1+z)/(z**2-4*z+1); [Conjectured by S. Plouffe in his 1992 dissertation.]
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MATHEMATICA
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a[0] = 0; a[1] = 2; a[n_] := a[n] = 4a[n - 1] - a[n - 2] + 1; Table[ a[n], {n, 0, 24}] (from Robert G. Wilson v Apr 24 2004)
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CROSSREFS
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Adjacent sequences: A001568 A001569 A001570 this_sequence A001572 A001573 A001574
Sequence in context: A140217 A032601 A083141 this_sequence A092431 A147762 A077837
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KEYWORD
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nonn
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AUTHOR
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njas
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EXTENSIONS
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Better description from Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
More terms and new description from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 24 2004
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