OFFSET
1,1
COMMENTS
The series giving the best rational approximations to e is e = 3 - 2/a(1) + 2/a(2) - 2/a(3) + ... The continued fraction for e is [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 [2;1], [2;1,2,1,1], [2;1,2,1,1,4,1, 1] and so forth.
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 A122533, for the best approximations to 1/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]==7, a[2]==497, a[3]==71071}, a, {n, 30}] (* G. C. Greubel, Oct 27 2024 *)
PROG
(Magma) I:=[7, 497, 71071]; [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 = A122523
if n<4: return (0, 7, 497, 71071)[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,changed
AUTHOR
Gene Ward Smith, Sep 17 2006
STATUS
approved