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A231692 Define a sequence of rationals by f(0)=0, thereafter f(n)=f(n-1)-1/n if that is >= 0, otherwise f(n)=f(n-1)+1/n; a(n) = numerator of f(n). 3

%I #37 Oct 19 2017 10:41:58

%S 0,1,1,1,5,13,1,27,19,451,199,4709,2399,3467,29207,5183,55411,221267,

%T 300649,1628251,5508127,259001,762881,6460903,5694791,11476403,

%U 27820203,326206681,5151783667,69088293143,146724611903,2219373406193,8951357840311,4575492601111,328329280711,4454145077671

%N Define a sequence of rationals by f(0)=0, thereafter f(n)=f(n-1)-1/n if that is >= 0, otherwise f(n)=f(n-1)+1/n; a(n) = numerator of f(n).

%C It is conjectured that the terms of the {f(n)} sequence are distinct.

%C If that is true, then the {f(n)} sequence is a fractional analog of Recamán's sequence A005132.

%C The denominators of {f(n)} form A231693 (a non-monotonic sequence).

%C From _Don Reble_, Nov 16 2013: (Start)

%C Here is a proof that the f(n) are distinct: Suppose not. Then the difference between the terms (which is zero) is a number of the form +- 1/(a+1) +- 1/(a+2) +- 1/(a+3) +- ... +- 1/(a+n).

%C Consider any harmonic sum

%C S = +- 1/(a+1) +- 1/(a+2) +- 1/(a+3) +- ... +- 1/(a+n)

%C where one puts any sign on any term, and there is at least one term. Let G be the LCM of the denominator(s). Then for any denominator D, G/D is an integer, and G*S = sum(+- G/D_i) is an integer. Let E be the highest power of two that divides G. Then there is only one multiple of E among the denominators. (If there were two, they would be consecutive multiples of E, and one would be divisible by 2*E.) Call that denominator F. So (+- G/F) is an odd integer, and for all other denominators D, (+- G/D) is an even integer. Therefore G*S is odd, therefore not zero, so S is not zero. (End)

%D David Wilson, Posting to Sequence Fans Mailing List, Nov 14 2013.

%H David W. Wilson and R. Zumkeller, <a href="/A231692/b231692.txt">Table of n, a(n) for n = 0..1000</a>

%e 0, 1, 1/2, 1/6, 5/12, 13/60, 1/20, 27/140, 19/280, 451/2520, 199/2520, 4709/27720, ...

%p f:=proc(n) option remember;

%p if n=0 then 0 elif

%p f(n-1) >= 1/n then f(n-1)-1/n else f(n-1)+1/n; fi; end;

%o (PARI) s=0;vector(30,n,numerator(s-=(-1)^(n*s<1)/n)) \\ - _M. F. Hasler_, Nov 15 2013

%o (Haskell)

%o a231692_list = map numerator $ 0 : wilson 1 0 where

%o wilson x y = y' : wilson (x + 1) y'

%o where y' = y + (if y < 1 % x then 1 else -1) % x

%o -- _Reinhard Zumkeller_, Nov 16 2013

%Y Cf. A231693, A005132.

%K nonn,frac

%O 0,5

%A _N. J. A. Sloane_, Nov 15 2013, Nov 16 2013

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