A175409 - OEIS

%I #11 Sep 29 2017 02:52:48

%S 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,

%T 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,

%U 2,1,1,2,1,2,1,1,2,1,1,2

%N Successive numbers of consecutive positive terms to add when rearranging the alternating harmonic series to sum to log[7/3].

%C 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.

%C Since x = (1/4)exp(2s) = 49/36 is rational, a result in the reference shows that this sequence is eventually periodic.

%D 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.

%H Francisco J. Freniche, <a href="http://www.jstor.org/stable/10.4169/000298910x485969">On Riemann's Rearrangement Theorem for the Alternating Harmonic Series</a>, Amer. Math. Monthly 117(2010), 442-448.

%e 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.

%K nonn

%O 1,2

%A _John W. Layman_, May 04 2010