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%I #18 Jan 12 2024 10:04:54
%S 1,2,3,3,2,2,3,4,4,3,4,2,3,2,3,3,3,5,4,5,3,5,6,5,4,5,5,6,4,5,3,4,5,3,
%T 4,3,3,4,5,3,6,4,5,6,6,3,3,7,8,8,5,5,6,6,4,6,7,4,7,5,7,7,5,4,6,4,5,5,
%U 3,3,4,3,3,4,4,3,3,3,3,4,6,4,4,4,3,7,5,5,6,4,6,5,5,4,4,5,3,4,6,7,5,5,5,4,6
%N Length of continued fraction for prime(n)/n.
%H David A. Corneth, <a href="/A107901/b107901.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>.
%F a(n) = A107898(n) - 1.
%e From _David A. Corneth_, Jan 09 2024: (Start)
%e a(10) = 3 as prime(10)/10 = 29/10 whose continued fraction is [2; 1, 9] having length 3
%e a(18) = 5 as prime(18)/18 = 61/18 whose continued fraction is [3; 2, 1, 1, 3] having length 5. (End)
%t A107901[n_]=Length[ContinuedFraction[Prime[n]/n]
%o (PARI) a(n) = #contfrac(prime(n)/n) \\ _David A. Corneth_, Jan 09 2024
%o (PARI) first(n) = my(res = primes(n)); for(i = 1, n, res[i] = #contfrac(res[i]/i)); res \\ _David A. Corneth_, Jan 09 2024
%Y Cf. A107898, A107899, A107900.
%K nonn
%O 1,2
%A _Zak Seidov_, May 26 2005