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A056053
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a(n) = smallest odd number 2m+1 such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.
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10
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3, 15, 113, 837, 6183, 45691, 337607, 2494595, 18432707, 136200301, 1006391657, 7436284415, 54947122715, 406007372211, 3000011249847, 22167251422541, 163795064320249, 1210290918990281, 8942907496445513, 66079645178783351
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OFFSET
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1,1
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COMMENTS
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a(2) = 15 and a(3) = 113 are related to the Borwein integrals. Concretely, a(2) = 15 is the smallest odd m such that the integral Integral_{x=-oo..oo} Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi, and a(3) = 113 is the smallest odd m such that the integral Integral_{x=-oo..oo} cos(x) * Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi/2. See the Wikipedia link and the 3Blue1Brown video link below. - Jianing Song, Dec 10 2022
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REFERENCES
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Calvin C. Clawson, "Mathematical Mysteries, The Beauty and Magic of Numbers," Plenum Press, NY and London, 1996, page 64.
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LINKS
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FORMULA
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The next term is approximately the previous term * e^2.
a(n) = A092315(n)*2 + 1 = floor(exp(n*2-Euler)/4+1/8)*2+1 for all n (conjectured). - M. F. Hasler, Jan 24 2017
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MATHEMATICA
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s = 0; k = 1; Do[ While[s = N[s + 1/k, 24]; s <= n, k += 2]; Print[k]; k += 2, {n, 1, 11}]
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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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