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A083698
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Partial quotients of the continued fraction which has convergents with the least possible prime denominators (A072999).
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6
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0, 2, 1, 1, 2, 2, 4, 6, 8, 4, 6, 38, 10, 14, 16, 6, 2, 12, 24, 100, 36, 74, 46, 44, 52, 18, 8, 46, 114, 20, 70, 6, 38, 190, 44, 76, 14, 118, 218, 34, 14, 82, 32, 28, 110, 76, 126, 230, 46, 578, 138, 192, 306, 424, 38, 148, 468, 218, 210, 174, 300, 244, 60, 744, 482, 190, 344
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OFFSET
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0,2
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COMMENTS
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Equivalently: a(1) = 2 and for n >= 2, a(n) is the least integer such that the numerator of the continued fraction [a(1),a(2),...,a(n)] is prime.
These prime numerators, listed in A072999, are the same as the prime denominators in the definition of this sequence A083698. The equivalence comes from the fact that 1/[a1, ..., aN] = [0, a1, ..., aN] for any continued fraction [a1, ..., aN] with a1 != 0. That is, numerators and denominators of the convergents are exchanged when considering the continued fraction with/without integer part, which amounts to inserting or deleting a leading 0.
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LINKS
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FORMULA
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EXAMPLE
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The partial quotients of the continued fraction 2 + 1/(1 + 1/(1 + 1/(2 + 1/(2 + 1/(4 + ...))))) are by definition the coefficients [2, 1, 1, 2, 2, 4, ...]. The convergents of this continued fraction are:
2 = 3/1, 2 + 1/1 = 3 = 3/1, 2 + 1/(1 + 1/1) = 2 + 1/2 = 5/2, ...
Here the primes listed in A072999 appear as numerators (cf. equivalent definition in comments). These primes appear as denominators if the terms [2, 1, 1, 2, 2, 4, ...] are considered as coefficients that appear in the pure fraction 1/(a(1) + 1/(a(2) + 1/...))), with convergents: 1/2, 1/(2 + 1/1) = 1/3, 1/(5/2) = 2/5, etc.
This amounts to including the initial a(0) = 0 for the integer part, which "shifts down into the denominator" the coefficients (2, 1, 1, ...) of the earlier mentioned continued fraction 2 + 1/(...).
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MATHEMATICA
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Nest[Append[#, Block[{k = 1}, While[! PrimeQ@ Denominator@ FromContinuedFraction@ Append[#, k], k++]; k]] &, {2}, 64] (* Michael De Vlieger, Dec 22 2019 *)
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PROG
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(PARI) l=1; h=2; print1(h, ", "); while(l<2^512, t=l+h; while(!isprime(t), t+=h); print1(floor(t/h), ", "); l=h; h=t)
(PARI) v=[2]; for(k=1, 70, m=1; while(isprime(contfracpnqn(concat(v, [m]))[1, 1])==0, m++); v=concat(v, [m])); a(n)=if(n<2, 2, v[n]); \\ Benoit Cloitre, Jan 15 2013.
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CROSSREFS
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KEYWORD
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cofr,nonn
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AUTHOR
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EXTENSIONS
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Edited by M. F. Hasler, Dec 29 2019, merging information from the duplicate A209270, following an observation by Hans Havermann on the SeqFan list.
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STATUS
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approved
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