OFFSET
1,1
COMMENTS
The series giving the best rational approximations to 1/e is 1/e = 1/3 + 2/a(1) - 2/a(2) + 2/a(3) - ... The continued fraction for 1/e is [0;2,1, 2,1,1,4,1,1,6,1,1,8...] and the above best approximations give every third convergent, the convergents deriving from [0;2,1], [0;2,1,2, 1,1], [0;2,1,2,1,1,4,1,1] and so forth are the partial sums of the above infinite series.
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..200
FORMULA
a(n+3) = (16*n^2 + 96*n + 141) * a(n+2) + (2*n+7)*(16*n^2 + 64*n + 61)/(2*n+3) * a(n+1) - (2*n+7)/(2*n+3) * a(n). This recurrence relationship is identical to A122523, for the best approximations to e.
MATHEMATICA
RecurrenceTable[{a[n]== ((2*n-3)*(16*n^2 -3)*a[n-1] +(2*n+1)*(16*(n-1)^2 - 3)*a[n-2] -(2*n+1)*a[n-3])/(2*n-3), a[1]==57, a[2]==3667, a[3]==525153}, a, {n, 30}] (* G. C. Greubel, Oct 27 2024 *)
PROG
(Magma) I:=[57, 3667, 525153]; [n le 3 select I[n] else ((2*n-3)*(16*n^2 - 3)*Self(n-1) + (2*n+1)*(16*(n-1)^2 -3)*Self(n-2) - (2*n+1)*Self(n-3))/(2*n-3): n in [1..30]]; // G. C. Greubel, Oct 27 2024
(SageMath)
@CachedFunction
def a(n): # a = A122533
if n<4: return (0, 57, 3667, 525153)[n]
else: return ((2*n-3)*(16*n^2 -3)*a(n-1) +(2*n+1)*(16*(n-1)^2 -3)*a(n-2) -(2*n+1)*a(n-3))/(2*n-3)
[a(n) for n in range(1, 31)] # G. C. Greubel, Oct 27 2024
CROSSREFS
KEYWORD
frac,nonn
AUTHOR
Gene Ward Smith, Sep 17 2006
STATUS
approved