A070899 - OEIS

A070899

a(1)=1, a(n) is the smallest integer > a(n-1) such that the largest element in the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals 3n.

%I #6 Mar 30 2012 18:38:59

%S 1,6,13,16,49,71,124,188,298,326,333,354,440,797,832,954,1006,1040,

%T 1280,1319,1414,2038,2113,2231,2291,2924,2973,3107,3983,3984,4331,

%U 4605,4763,5756,6314,6325,6402,7501,7967,8073,8143,8895,9567,9900,10333

%N a(1)=1, a(n) is the smallest integer > a(n-1) such that the largest element in the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals 3n.

%C sum(k=>1,1/a(k))=C=1.3...

%e The continued fraction for S(7)=1+1/6+1/13+1/16+1/49+1/71+1/124 is [1, 2, 1, 6, 1, 1, 2, 3, 3, 1, 21, 1, 1, 3, 1, 3, 18, 3] where the largest element is 21=3*7 and 124 is the smallest integer > a(6)=71 with this property, hence a(7)=124.

%o (PARI) s=1; t=1; for(n=1,60,s=s+1/t; while(abs(3*n-vecmax(contfrac(s+1/t)))>0,t++); print1(t,","))

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, May 19 2002