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A111047
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Product of continued fraction terms of H(n) = Sum_{k=1..n} 1/k.
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0
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1, 2, 5, 24, 48, 32, 100, 140, 840, 1872, 54000, 12960, 51840, 533871, 322371, 31104, 709632, 1921500, 4147200, 3701376, 124416, 262080, 2488320, 21811680, 403107840, 146966400, 2538086400, 1074954240, 14370048000, 415704960000
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OFFSET
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1,2
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COMMENTS
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The last term of each continued fraction is considered to be >= 2, for n >= 2.
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LINKS
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EXAMPLE
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1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)), so the 6th term of the sequence is 2*2*4*2 = 32.
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PROG
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(PARI) for(n=1, 30, v=contfrac(sum(k=1, n, 1/k)); print1(prod(j=1, length(v), v[j]), ", "))
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CROSSREFS
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KEYWORD
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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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