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%I #42 Sep 08 2022 08:45:11
%S 1,2,-1,8,-19,62,-181,548,-1639,4922,-14761,44288,-132859,398582,
%T -1195741,3587228,-10761679,32285042,-96855121,290565368,-871696099,
%U 2615088302,-7845264901,23535794708,-70607384119,211822152362,-635466457081,1906399371248
%N a(n) = -2*a(n-1) + 3*a(n-2), with a(0)=1, a(1)=2.
%H Vincenzo Librandi, <a href="/A084222/b084222.txt">Table of n, a(n) for n = 0..200</a>
%H Alexandru Ciolan and Pieter Moree, <a href="https://arxiv.org/abs/1707.02183">Browkin's discriminator conjecture</a>, arXiv:1707.02183 [math.NT], 2017.
%H Pieter Moree and Ana Zumalacárregui, <a href="http://arxiv.org/abs/1504.05718">Salajan's conjecture on discriminating terms in an exponential sequence</a>, arXiv:1504.05718 [math.NT], 2015; Journal of Number Theory 160 (2016), pp. 646-665.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-2,3).
%F Binomial transform is A084221.
%F a(n) = (5-(-3)^n)/4.
%F G.f.: (1+4*x)/((1-x)*(1+3*x)).
%F E.g.f.: (5*exp(x)-exp(-3*x))/4.
%F For n > 1, abs(a(n) - a(n+1)) = 3^n. - _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jul 15 2003; corrected by _Philippe Deléham_, Dec 16 2007
%F a(n) = 9*a(n-2) - 10 with a(0) = 1 and a(1) = 2. - _Philippe Deléham_, Feb 24 2014
%F a(2n) = -A211866(n), n>0. - _Philippe Deléham_, Feb 24 2014
%t CoefficientList[Series[(1 + 4 x)/((1 - x) (1 + 3 x)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Feb 26 2014 *)
%t LinearRecurrence[{-2, 3}, {1, 2}, 28] (* _Jean-François Alcover_, Sep 27 2017 *)
%o (PARI) a(n) = (5-(-3)^n)/4; \\ _Joerg Arndt_, Jul 14 2013
%o (Magma) [(5-(-3)^n)/4: n in [0..40]]; // _Vincenzo Librandi_, Feb 26 2014
%Y Cf. A211866.
%K easy,sign
%O 0,2
%A _Paul Barry_, May 21 2003
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