A301809 Group the natural numbers such that the first group is (1) then (2),(3),(4,5),(6,7,8),... with the n-th group containing F(n) sequential terms where F(n) is the n-th Fibonacci number (A000045(n)). Sequence gives the sum of terms in the n-th group.
1, 2, 3, 9, 21, 55, 140, 364, 945, 2465, 6435, 16821, 43992, 115102, 301223, 788425, 2063817, 5402651, 14143524, 37026936, 96935685, 253777537, 664392743, 1739393929, 4553778096, 11921922650, 31211961195, 81713914569, 213929707485, 560075086495, 1466295355580, 3838810662436, 10050136117497
Offset: 1
Examples
a(7) = 14 + 15 + 16 + ... + 21 = (F(9)+1)*F(6)/2 = 140.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Wikipedia, Hofstadter sequence.
- Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1, 1).
Programs
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Magma
[1] cat [(Fibonacci(n+2)+1)*Fibonacci(n-1) div 2 : n in [2..35] ]; // Vincenzo Librandi, Apr 18 2018
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Mathematica
a[n_] := If[n==1, 1, (Fibonacci[n+2]+1)Fibonacci[n-1]/2]; Array[a, 50] Join[{1}, LinearRecurrence[{3, 1, -5, -1, 1}, {2, 3, 9, 21, 55}, 40]] (* Vincenzo Librandi, Apr 18 2018 *) -
PARI
a(n) = if (n==1, 1, (fibonacci(n+2)+1)*fibonacci(n-1)/2); \\ Michel Marcus, Apr 21 2018
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PARI
Vec(x*(1 - x)*(1 - 4*x^2 - x^3 + x^4) / ((1 + x)*(1 - 3*x + x^2)*(1 - x - x^2)) + O(x^60)) \\ Colin Barker, May 11 2018
Formula
a(1) = 1 and for n > 1, a(n) = (F(n+2)+1)*F(n-1)/2, where F(n) is the n-th Fibonacci number (A000045).
From Colin Barker, Mar 27 2018: (Start)
G.f.: x*(1 - x)*(1 - 4*x^2 - x^3 + x^4) / ((1 + x)*(1 - 3*x + x^2)*(1 - x - x^2)).
a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) + a(n-5) for n>6. (End)
Comments