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A098703
a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.
5
0, 1, 2, 6, 17, 50, 148, 441, 1318, 3946, 11825, 35454, 106328, 318929, 956698, 2869950, 8609617, 25828474, 77484812, 232453449, 697358750, 2092073666, 6276216817, 18828643686, 56485920112, 169457742625, 508373199218, 1525119551286
OFFSET
0,3
COMMENTS
Sums of antidiagonals of A090888;
Partial sums of A099159;
a(n) = A000045(n) + A094688(n-1);
For n > 2, a(n) = 3a(n-1) - A000045(n-3);
For n > 3, a(n) = 3^2a(n-2) - A000285(n-4);
For n > 4, a(n) = 3^3a(n-3) - A022383(n-5);
Lim_{n -> oo} a(n) / a(n-1) = 3.
a(n) = A101220(1,3,n). - Ross La Haye, Dec 15 2004
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
LINKS
Eric Weisstein, Golden Ratio
Eric Weisstein, Lucas Number
Eric Weisstein, Fibonacci Number
FORMULA
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.
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.
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
Binomial transform of unsigned A084178. Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). - Ross La Haye, Dec 21 2004
G.f.: x*(1-2*x)/((-1+3*x)*(-1+x+x^2)). - Ross La Haye, Aug 09 2005
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
Binomial transform of A052964 beginning 0,1,0,3,1,10,... - Ross La Haye, May 31 2006
EXAMPLE
a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.
MATHEMATICA
f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* Jean-François Alcover, Feb 17 2018 *)
PROG
(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
KEYWORD
nonn
AUTHOR
Ross La Haye, Oct 27 2004
EXTENSIONS
More terms from Robert G. Wilson v, Nov 04 2004
More terms from Ross La Haye, Dec 21 2004
STATUS
approved