Families Of Seqs Derived From Rationals

Leroy Quet qq-quet at mindspring.com
Thu Dec 30 20:25:02 CET 2004

>Just a suggestion for those looking for ideas of sequences to submit:
>We can take a sequence of rationals {r(k)}, where the numerators and 
>denominators of {r(k)} might already by in the EIS, and form a new 
>sequence which is the concatenation of the continued fractions of each 
>r(k), as I have done in this sequence I just submitted:
>>%S A000001 1,1,2,1,1,5,2,12,2,3,1,1,8,2,2,4,2
>>%N A000001 Array where nth row (of A055573(n) terms) is the continued 
>>fraction terms of the nth harmonic number, sum{k=1 to n} 1/k.
>>%e A000001 Since the 3rd harmonic number is 11/6 = 1 +1/(1 +1/5), the 3rd 
>>row is 1,1,5.
>>%Y A000001 A055573,A001008,A002805
>>%O A000001 1
>>%K A000001 ,more,nonn,tabl,cofr
>Also we can form the sequence which gives the number of terms in each 
>continued fraction (as is sequence A055573 relating to my example above).
>(It should be noted in the sequence's comment line that it is understood 
>that each CF's final term is not 1 when r(k) is not 1.)
>And we can also form the sequence which gives the SUM of the CF terms for 
>each r(k).
>And don't forget the sequence of final (non-1) terms of each continued 
>fraction, as well as {floor(r(k))} (the initial terms of the {positive} 
>rationals' CFs).
>Leroy Quet

PS: Of course if we have instead a sequence of reals, we can form an 
infinite array of the continued fractions and then read off the CF terms 
by diagonals to get a sequence.

By the way, I am not saying that the ideas given above have not been done 
already before, so check the OEIS to see if any sequence already exists 
in the database, of course! 

Leroy Quet

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