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A138743
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Let r_1 = 1. Let r_{m+1} = r_1 + 1/(r_2 + 1/(r_3 +...(r_{m-1} + 1/r_m)...)), a continued fraction of rational terms. Then a(n) is the number of (positive integer) terms in the simple continued fraction of r_n.
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2
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1, 1, 1, 3, 6, 6, 11, 26, 48, 82, 201, 379, 836, 1554, 3197, 6420, 12639, 25298, 50675, 101675, 203379, 405946, 811519, 1622692, 3249540, 6494117, 12998399, 25991681
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OFFSET
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1,4
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COMMENTS
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This sequence is the number of terms in the n-th row of irregular array A138742.
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LINKS
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EXAMPLE
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{r_n}: 1, 1, 2, 5/3, 31/18, 1231/720,...
r_5 = 31/18, for instance, equals the simple continued fraction 1+ 1/(1 + 1/(2 + 1/(1 + 1/(1 +1/2)))). There are six integer terms (1,1,2,1,1,2) in this continued fraction, so a(5) = 6.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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