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%I #6 Mar 30 2012 18:38:59
%S 1,2,5,7,12,25,119,152,154,163,164,178,179,183,190,192,245,267,279,
%T 290,306,313,359,420,454,496,528,576,592,615,649,661,674,702,760,830,
%U 868,945,967,978,1000,1167,1190,1194,1209,1288,1289,1325,1452,1892,2084
%N a(0)=1, a(n) is the smallest number > a(n-1) such that the simple continued fraction for S(n)=1/a(0)+1/a(1)+...+1/a(n) contains exactly 2n elements.
%e The simple continued fraction for S(4)=1+1/2+1/5+1/7+1/12 is [1, 1, 12, 1, 1, 4, 1, 2] which contains exactly 8 elements. The simple continued fraction for 1+1/2+1/5+1/7+1/12+1/25 is [1, 1, 28, 1, 1, 2, 1, 2, 1, 2] which contains exactly 10=2*5 elements and 25 is the smallest integer>12 with this property, hence a(5)=25.
%o (PARI) s=1; t=1; for(n=1,78,s=s+1/t; while(abs(2*n-length(contfrac(s+1/t)))>0,t++); print1(t,","))
%K easy,nonn
%O 0,2
%A _Benoit Cloitre_, May 19 2002
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