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A094584
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Dot product of (1,2,...,n) and first n distinct Fibonacci numbers.
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6
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1, 5, 14, 34, 74, 152, 299, 571, 1066, 1956, 3540, 6336, 11237, 19777, 34582, 60134, 104062, 179320, 307855, 526775, 898706, 1529160, 2595624, 4396224, 7431049, 12537917, 21118814, 35517226, 59646386, 100034456, 167562035, 280348531, 468543802, 782277612
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OFFSET
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1,2
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COMMENTS
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a(n) is the cost of all non-leaf nodes in the Fibonacci tree of order n+2. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node. In a Fibonacci tree the cost of a left (right) edge is defined to be 1 (2). The cost of a node of a Fibonacci tree is defined to be the sum of the costs of the edges that form the path from the root to this node. - Emeric Deutsch, Jun 14 2010
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 14.
D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417. [From Emeric Deutsch, Jun 14 2010]
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LINKS
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FORMULA
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a(n) = F(2) + 2*F(3) + 3*F(4) + ... + n*F(n+1) = (n+1)*F(n+3) - F(n+5) + 3.
G.f.: x*(1+2*x)/((1-x)*(1-x-x^2)^2). - Colin Barker, Nov 11 2012
a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + a(n-4) + a(n-5).
a(n) = Sum_{i=1..n+2} (n-i+1) * F(n-i+2).
a(n) = (30*(-1-sqrt(5))^n + (-15+7*sqrt(5))*2^n - (15+7*sqrt(5))*(-3-sqrt(5))^n + 2n*((5-2*sqrt(5))*2^n + (5+2*sqrt(5))*(-3-sqrt(5))^n)) / (10*(-1-sqrt(5))^n). (End)
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EXAMPLE
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a(4) = (1,2,3,4)*(1,2,3,5) = 1+4+9+20 = 34.
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MAPLE
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MATHEMATICA
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Table[Range[n].Fibonacci[Range[2, n+1]], {n, 40}] (* Harvey P. Dale, Aug 21 2011 *)
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PROG
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(Magma) I:=[1, 5, 14, 34, 74]; [n le 5 select I[n] else 3*Self(n-1)-Self(n-2)-3*Self(n-3)+Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Mar 11 2015
(Magma) [n*Fibonacci(n+3)-Fibonacci(n+4)+3: n in [1..40]]; // G. C. Greubel, Apr 28 2019
(GAP) List([1..40], n->(n+1)*Fibonacci(n+3)-Fibonacci(n+5)+3); # Muniru A Asiru, Apr 27 2019
(PARI) {a(n) = n*fibonacci(n+3) - fibonacci(n+4) +3}; \\ G. C. Greubel, Apr 28 2019
(Sage) [n*fibonacci(n+3) - fibonacci(n+4) +3 for n in (1..40)] # G. C. Greubel, Apr 28 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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