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%I #35 Sep 08 2022 08:45:15
%S 0,1,2,6,17,50,148,441,1318,3946,11825,35454,106328,318929,956698,
%T 2869950,8609617,25828474,77484812,232453449,697358750,2092073666,
%U 6276216817,18828643686,56485920112,169457742625,508373199218,1525119551286
%N a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.
%C Sums of antidiagonals of A090888;
%C Partial sums of A099159;
%C a(n) = A000045(n) + A094688(n-1);
%C For n > 2, a(n) = 3a(n-1) - A000045(n-3);
%C For n > 3, a(n) = 3^2a(n-2) - A000285(n-4);
%C For n > 4, a(n) = 3^3a(n-3) - A022383(n-5);
%C Lim_{n -> oo} a(n) / a(n-1) = 3.
%C a(n) = A101220(1,3,n). - _Ross La Haye_, Dec 15 2004
%C Form an array with m(0,n) = A000045(n), the Fibonacci numbers, and m(i,j) = Sum_{k<i} m(k,j) + Sum_{k<j} m(i,k), which is the sum of the terms above m(i,j) plus the sum of the terms to the left of m(i,j). The sum of the terms in antidiagonal(n) = a(n+1). - _J. M. Bergot_, May 27 2013
%H Vincenzo Librandi, <a href="/A098703/b098703.txt">Table of n, a(n) for n = 0..1000</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/GoldenRatio.html">Golden Ratio</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/LucasNumber.html">Lucas Number</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-3).
%F a(n) = (((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^n*sqrt(5))) + ((3^n - (((1 + sqrt(5)) / 2)^(n+1) + ((1 - sqrt(5)) / 2)^(n+1))) / 5); a(n) = (3^n + (((1 + sqrt(5)) / 2)^(n-1) + ((1 - sqrt(5)) / 2)^(n-1))) / 5.
%F Let Luc(n) denote the n-th Lucas number [A000032] and Fib(n) denote the n-th Fibonacci number (A000045). Then a(n) = (3^n + Luc(n-1)) / 5; a(n) = Fib(n) + ((3^n - Luc(n+1)) / 5); a(n) = (3^n + Fib(n) + Fib(n-2)) / 5; a(n) = (3^n + 4Fib(n) - Fib(n+2)) / 5; a(n) = Sum_{k=0...n-1} Fib(k)*3^(n-k-1) - Fib(k-2)*2^(n-k-1), ... and so on.
%F a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
%F Binomial transform of unsigned A084178. Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). - _Ross La Haye_, Dec 21 2004
%F G.f.: x*(1-2*x)/((-1+3*x)*(-1+x+x^2)). - _Ross La Haye_, Aug 09 2005
%F a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3^(n-2) for n > 1. - _Ross La Haye_, Aug 20 2005
%F Binomial transform of A052964 beginning 0,1,0,3,1,10,... - _Ross La Haye_, May 31 2006
%e a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.
%t f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* _Robert G. Wilson v_, Nov 04 2004 *)
%t LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* _Jean-François Alcover_, Feb 17 2018 *)
%o (Magma) I:=[0,1,2]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-3*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 18 2018
%Y Cf. A001622, A000032, A000045, A090888, A099159, A094688, A000285, A022383, A000244.
%K nonn
%O 0,3
%A _Ross La Haye_, Oct 27 2004
%E More terms from _Robert G. Wilson v_, Nov 04 2004
%E More terms from _Ross La Haye_, Dec 21 2004
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