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A071293
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a(0)=1, a(n) is the smallest integer > a(n-1) such that the continued fraction for 1/a(0)+1/a(1)+1/a(2)+...+1/a(n) contains exactly 2^n elements.
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0
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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 continued fraction for 1/a(0)+1/a(1)+1/5 = 1+1/2+1/5 is {1, 1, 2, 3} which contains 2^2 elements. 5 is the smallest integer > 2 with this property, hence a(2)=5.
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PROG
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(PARI) s=1; t=1; for(n=1, 5, 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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hard,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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