A084222 a(n) = -2*a(n-1) + 3*a(n-2), with a(0)=1, a(1)=2.
1, 2, -1, 8, -19, 62, -181, 548, -1639, 4922, -14761, 44288, -132859, 398582, -1195741, 3587228, -10761679, 32285042, -96855121, 290565368, -871696099, 2615088302, -7845264901, 23535794708, -70607384119, 211822152362, -635466457081, 1906399371248
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Alexandru Ciolan and Pieter Moree, Browkin's discriminator conjecture, arXiv:1707.02183 [math.NT], 2017.
- Pieter Moree and Ana Zumalacárregui, Salajan's conjecture on discriminating terms in an exponential sequence, arXiv:1504.05718 [math.NT], 2015; Journal of Number Theory 160 (2016), pp. 646-665.
- Index entries for linear recurrences with constant coefficients, signature (-2,3).
Crossrefs
Cf. A211866.
Programs
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Magma
[(5-(-3)^n)/4: n in [0..40]]; // Vincenzo Librandi, Feb 26 2014
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Mathematica
CoefficientList[Series[(1 + 4 x)/((1 - x) (1 + 3 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 26 2014 *) LinearRecurrence[{-2, 3}, {1, 2}, 28] (* Jean-François Alcover, Sep 27 2017 *) -
PARI
a(n) = (5-(-3)^n)/4; \\ Joerg Arndt, Jul 14 2013
Formula
Binomial transform is A084221.
a(n) = (5-(-3)^n)/4.
G.f.: (1+4*x)/((1-x)*(1+3*x)).
E.g.f.: (5*exp(x)-exp(-3*x))/4.
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
a(n) = 9*a(n-2) - 10 with a(0) = 1 and a(1) = 2. - Philippe Deléham, Feb 24 2014
a(2n) = -A211866(n), n>0. - Philippe Deléham, Feb 24 2014