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A071013
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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.
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0
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1, 2, 5, 7, 12, 25, 119, 152, 154, 163, 164, 178, 179, 183, 190, 192, 245, 267, 279, 290, 306, 313, 359, 420, 454, 496, 528, 576, 592, 615, 649, 661, 674, 702, 760, 830, 868, 945, 967, 978, 1000, 1167, 1190, 1194, 1209, 1288, 1289, 1325, 1452, 1892, 2084
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OFFSET
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0,2
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LINKS
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EXAMPLE
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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.
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PROG
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(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, ", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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