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a(1) = 1, then a(n) is the least integer > a(n-1) such that n is the maximum element in the continued fraction for 1/a(1) + 1/a(2) + ... + 1/a(n).
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%I #11 Apr 29 2023 20:47:00

%S 1,2,4,13,38,51,97,124,247,295,348,398,421,494,615,881,1105,1252,1616,

%T 1634,1637,2222,2410,2478,2583,92598,115781,124161,132776,141565,

%U 151072,160466,169593,180127,190982,200776,211896,223261,236300,245322,259018

%N a(1) = 1, then a(n) is the least integer > a(n-1) such that n is the maximum element in the continued fraction for 1/a(1) + 1/a(2) + ... + 1/a(n).

%C Conjectures:

%C The sequence is well defined for any n.

%C Limit_{n->oo} a(n)^(1/n) = 1.

%e The continued fraction for 1/a(1) + 1/a(2) = 1 + 1/2 is [1,2].

%e The c.f. for 1 + 1/2 + 1/3 is [1,1,5].

%e The c.f. for 1 + 1/2 + 1/4 is [1,1,3], whose maximum element is 3, hence a(3)=3.

%o (PARI) s=1;t=1;for(n=2,50,s=s+1/t;t=t+1; while(abs(n-vecmax(contfrac(s+1/t)))>0, t++); print1(t,","))

%K nonn

%O 1,2

%A _Benoit Cloitre_, Mar 23 2010