Continued Fractions With Square Numerators

[Hans Havermann]

> a(1) = 1. a(n) is the lowest positive integer such that the  
> continued fraction [a(1);a(2),a(3),...,a(n)] is equal to a rational  
> with a square numerator.
> The sequence begins: 1, 3, 2, 5,...

Jack Brennen:

> The next few terms of Leroy's sequence are:
> 1,3,2,5, 43, 522, 1104509, ...


... and here's my attempt at a(8)-a(12):

60248974744,
2075863890266492169136,
10942918579397694712648387271683911959312808,
304366130052353180971554734771542912191750299192365265003301401044158903 
63628017565032,
783937267308839841757961409143613666063084358984783689500149071683605360 
504355756300748745937652713062842348773667173541821027267890390773994217 
98511614768052156175104


> The continued fraction [1,3,2,5,43,522,1104509] corresponds to:
>   1220041748025/946166591401
> Where the numerator is 1104555 squared.
>
> I'm sure it's not a coincidence that the square root of the numerator
> is numerically very close to the last term of the continued fraction


My take on this is that it's not a coincidence in the sense that  
there exists a term close to the square root of the resulting  
convergent's numerator that will satisfy the condition of the  
numerator being a square; however, i think it *is* a coincidence with  
the added stipulation (as is the case with Leroy's formulation) that  
the last term of the continued fraction be the *smallest* positive  
integer that satisfies that condition.


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