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 A175409 Successive numbers of consecutive positive terms to add when rearranging the alternating harmonic series to sum to log[7/3]. 0
 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Let s = log(7/3). Add a(n) positive terms 1 + 1/3 + 1/5 + ... + 1/(2a(n)-1)until their sum is greater than s, then subtract negative terms 1/2, 1/4, ... until the sum drops below s. Continue alternating in this way, adding a(2) consecutive positive terms, subtracting negative terms, and so on. The numbers a(n) are the terms of the sequence. Since x = (1/4)exp(2s) = 49/36 is rational, a result in the reference shows that this sequence is eventually periodic. REFERENCES Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185. LINKS Table of n, a(n) for n=1..80. Francisco J. Freniche, On Riemann's Rearrangement Theorem for the Alternating Harmonic Series, Amer. Math. Monthly 117(2010), 442-448. EXAMPLE s = log(7/3) = 0.847298. The first term, 1, of the alternating harmonic series already exceeds s, so a(1)=1. Subtracting negative terms, we get 1-1/2 = 1/2, which is less than s. Then, adding 1/3 gives 0.833333, which is less than s, so we also add a second term, 1/5, to get 1.03333 which exceeds s. Thus a(2)=2. CROSSREFS Sequence in context: A161113 A161048 A152830 * A161073 A161112 A161047 Adjacent sequences: A175406 A175407 A175408 * A175410 A175411 A175412 KEYWORD nonn AUTHOR John W. Layman, May 04 2010 STATUS approved

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Last modified August 11 08:13 EDT 2024. Contains 375059 sequences. (Running on oeis4.)