A078690 Continued fraction expansion of e^(2/5).
1, 2, 30, 12, 1, 1, 17, 90, 27, 1, 1, 32, 150, 42, 1, 1, 47, 210, 57, 1, 1, 62, 270, 72, 1, 1, 77, 330, 87, 1, 1, 92, 390, 102, 1, 1, 107, 450, 117, 1, 1, 122, 510, 132, 1, 1, 137, 570, 147, 1, 1, 152, 630, 162, 1, 1, 167, 690, 177, 1, 1, 182, 750, 192, 1, 1, 197, 810, 207
Offset: 0
Links
- G. Xiao, Contfrac
- K. Matthews, Finding the continued fraction of e^(l/m)
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).
Crossrefs
Cf. A069951.
Programs
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Mathematica
Block[{$MaxExtraPrecision=1000},ContinuedFraction[E^(2/5),70]] (* Harvey P. Dale, Sep 04 2011 *)
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PARI
contfrac(exp(2/5))
Formula
For k>=0, a(5k+1)=15k+2 a(5k+2)=60k+30 a(5k+3)=15k+12 a(5k)=a(5k+4)=1.
G.f.: -(x^9-3*x^8-30*x^7-13*x^6+x^5-x^4-12*x^3-30*x^2-2*x-1) / ((x-1)^2*(x^4+x^3+x^2+x+1)^2). - Colin Barker, Jun 24 2013