Is there an algorithm to find the Lengths of period of continued fraction sqrt((an)^2+4a)) whith a,n are odd integer.And can we find the continued fraction of the previous number.
Andrej Dujella
If n=1, then length = 6.
If n is odd and n>1, then length = 10.
The expansion is
[an; ((n-1)/2, 1, 1, (an-1)/2, 2n, (an-1)/2, 1, 1, (n-1)/2, 2an)].
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Andrej Dujella
Indeed, the expansion for a=5 odd is
[-an-1; (1, (n-3)/2, 2, (-an-3)/2, 1, 2n-2, 1, (-an-3)/2, 2, (n-3)/2, 1, -2an-2)]
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