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%I #71 Apr 27 2023 15:10:35
%S 0,1,1,1,1,2,2,2,3,4,4,5,7,8,9,12,15,17,21,27,32,38,48,59,70,86,107,
%T 129,156,193,236,285,349,429,521,634,778,950,1155,1412,1728,2105,2567,
%U 3140,3833,4672,5707,6973,8505,10379,12680,15478,18884,23059,28158,34362
%N a(0)=0, a(1)=1, a(2)=1, a(3)=1, a(n) = a(n-3) + a(n-4) for n > 3.
%C P(0)=P(1)=P(2)=P(3)=1, for m > 3: P(m) = P(m-3) + P(m-4) is the 3rd sequence in the series: Fibonacci sequence, Padovan sequence, ... The Padovan sequence (whose ratio of successive terms approaches the plastic constant) is similar to the Perrin sequence. - _Jonathan Vos Post_, Jan 23 2005
%C Binomial transform yields A079398 without the initial (0,1,1,1). - _R. J. Mathar_, Apr 09 2008
%C a(n+1) corresponds to the diagonal sums of "triangle": 1; 1; 1; 1,1; 1,1; 1,1; 1,2,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated three times. - _Philippe Deléham_, Dec 13 2008
%C a(n) is the number of pairs of rabbits living at month n with the following rules: a pair of rabbits born in month n begins to procreate in month n + 3, procreates again in month n + 4, and dies at the end of this month (each pair therefore gives birth to 2 pairs); warning! The first pair is born in month 2. - _Robert FERREOL_, Oct 24 2017
%H Vincenzo Librandi, <a href="/A079398/b079398.txt">Table of n, a(n) for n = 0..1000</a>
%H Vedran Krcadinac, <a href="http://www.fq.math.ca/Papers1/44-4/quartkrcadinac04_2006.pdf">A new generalization of the golden ratio</a>, Fibonacci Quart. 44 (2006), no. 4, 335-340.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PadovanSequence.html">Padovan Sequence</a>.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,1).
%F a(0)=0, a(1)=1, a(2)=1, a(3)=1, a(n) = a(n-3) + a(n-4) for n > 3. - _Colin Barker_, Sep 18 2013
%F From _Paul Barry_, Jul 06 2004: (Start)
%F a(n) = Sum_{k=0..floor((n-1)/2)} binomial(floor((n-k-1)/3), k)} (offset 0).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(floor((n-k-1)/3), k)}-0^n (offset 0). (End)
%F For n > 1, a(n) = P(n-2) where P(n) is defined by: P(0)=P(1)=P(2)=P(3)=1, for m > 3: P(m) = P(m-3) + P(m-4). - _Jonathan Vos Post_, Jan 23 2005
%F The same sequence may be constructed as follows: Let M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {1, 1, 0, 0}}; v[1] = {1, 1, 1, 1}; v[n] = M.v[n - 1]. Then a(n) = v[n][[1]]. - _Roger L. Bagula_, Sep 16 2006
%F O.g.f.: -x^2*(1+x+x^2)/(-1+x^3+x^4). a(n) = A017817(n-1) + A017817(n-2) + A017817(n-3). - _R. J. Mathar_, Apr 09 2008
%t CoefficientList[Series[x (1 + x + x^2)/(1 - x^3 - x^4), {x, 0, 60}], x] (* _Vincenzo Librandi_, Mar 16 2014 *)
%t LinearRecurrence[{0, 0, 1, 1}, {0, 1, 1, 1}, 60] (* _Jean-François Alcover_, Dec 05 2017 *)
%t nxt[{a_,b_,c_,d_}]:={b,c,d,a+b}; NestList[nxt,{0,1,1,1},60][[;;,1]] (* _Harvey P. Dale_, Apr 27 2023 *)
%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,0,0]^n*[0;1;1;1])[1,1] \\ _Charles R Greathouse IV_, Oct 03 2016
%o (PARI) x='x+O('x^50); concat([0], Vec(x*(1+x+x^2)/(1-x^3-x^4))) \\ _G. C. Greubel_, Apr 30 2017
%Y Cf. A000931.
%K nonn,easy
%O 0,6
%A _Benoit Cloitre_, Feb 16 2003
%E Recurrence corrected by _Colin Barker_, Sep 18 2013
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