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a(1) = 1; a(n) = prime(n) - prime(n-1)* a(n-1) if n > 1.
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%I #3 Feb 11 2014 19:05:41

%S 1,1,2,-3,32,-339,4424,-75189,1428614,-32858093,952884728,

%T -29539426531,1092958781688,-44811310049165,1926886332114142,

%U -90563657609364621,4799873853296324972,-283192557344483173287,17274745998013473570574,-1157407981866902729228387

%N a(1) = 1; a(n) = prime(n) - prime(n-1)* a(n-1) if n > 1.

%C 1. Let s(n) be a sequence such that lim s(n)/s(n+1) = K different from -1. The "oscillator sequence" (or simply "oscillator") of s(n) is the sequence s'(n) defined by the rules: s'(1) = 1; s'(n) = 1 - (s(n-1)/s(n)) s'(n-1). 2. It is an open problem whether the oscillator (prime)' converges to 1/2 or diverges. 3. For s(n) = prime(n), one has s'(n) = 1 - (p(n-1)/p(n)) s'(n-1) = [p(n) - p(n-1) s'(n-1)]/p(n). The numerator is the expression p(n) - p(n-1) s'(n-1), which motivates the definition of the above sequence a(n). a(n) is called the "integral oscillator" of prime(n). In general the integral oscillator of s(n) can be defined similarly.

%t t = {1}; gt = 1; For[i = 2, i <= 24, i++, gt = Prime[i] - Prime[i - 1] gt; t = Append[t, gt]]; t ListPlot[t, PlotJoined -> True]

%Y Cf. A069942.

%K sign

%O 1,3

%A _Joseph L. Pe_, Feb 20 2003