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A353299 a(n) is the length of the continued fraction for the sum of the reciprocals of the first n primes.

%I A353299 #10 Feb 16 2025 08:34:03
%S A353299 2,3,2,5,9,10,11,16,13,20,27,27,31,43,37,41,43,47,50,58,53,57,65,83,
%T A353299 69,62,80,84,88,93,88,110,119,117,104,111,116,126,114,140,130,164,166,
%U A353299 132,158,154,166,168,178,178,146,176,192,188,190,203,213,191,224,236,234,238,236,236,251
%N A353299 a(n) is the length of the continued fraction for the sum of the reciprocals of the first n primes.
%H A353299 Ilya Gutkovskiy, <a href="/A353299/a353299.jpg">Scatterplot of a(n)/(n*log(n)) up to n=10000</a>
%H A353299 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>
%H A353299 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/HarmonicSeriesofPrimes.html">Harmonic Series of Primes</a>
%e A353299 Sum_{k=1..2} 1/prime(k) = 1/2 + 1/3 = 5/6 = 0 + 1/(1 + 1/5), so a(2) = 3.
%e A353299 Sum_{k=1..4} 1/prime(k) = 1/2 + 1/3 + 1/5 + 1/7 = 247/210 = 1 + 1/(5 + 1/(1 + 1/(2 + 1/12))), so a(4) = 5.
%t A353299 Table[Length[ContinuedFraction[Sum[1/Prime[k], {k, 1, n}]]], {n, 1, 65}]
%o A353299 (PARI) a(n) = #contfrac(sum(k=1, n, 1/prime(k))); \\ _Michel Marcus_, Apr 10 2022
%Y A353299 Row lengths of A260615.
%Y A353299 Cf. A002110, A024451, A055573.
%K A353299 nonn
%O A353299 1,1
%A A353299 _Ilya Gutkovskiy_, Apr 09 2022