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A056053
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.
11
1, 3, 15, 113, 837, 6183, 45691, 337607, 2494595, 18432707, 136200301, 1006391657, 7436284415, 54947122715, 406007372211, 3000011249847, 22167251422541, 163795064320249, 1210290918990281, 8942907496445513, 66079645178783351, 488266205223462461, 3607826381608149807
OFFSET
0,2
COMMENTS
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
REFERENCES
Calvin C. Clawson, "Mathematical Mysteries, The Beauty and Magic of Numbers," Plenum Press, NY and London, 1996, page 64.
LINKS
Grant Sanderson, Researchers thought this was a bug (Borwein integrals), 3Blue1Brown video (2022).
Wikipedia, Borwein integral
FORMULA
a(n) ~ floor((1/2)*A002387(2n)).
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
a(n) ~ exp(2*n - A350763) = (1/2)*exp(2*n - gamma), gamma = A001620. - A.H.M. Smeets, Apr 15 2022
MATHEMATICA
s = 0; k = 1; Do[ While[s = N[s + 1/k, 24]; s <= n, k += 2]; Print[k]; k += 2, {n, 1, 11}]
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jul 25 2000 and Jan 11 2004
EXTENSIONS
Corrected by N. J. A. Sloane, Feb 16 2004
More terms from Robert G. Wilson v, Apr 17 2004
a(17) corrected - see correction in A092315. - Gerhard Kirchner, Jul 25 2020
a(0) prepended by Robert G. Wilson v, Oct 23 2024
STATUS
approved