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Sum of non-Fibonacci numbers between successive Fibonacci numbers: a(n) = Sum_{k=F(n)+1..F(n+1)-1} k.
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%I #30 Sep 30 2024 19:01:36

%S 0,0,0,0,4,13,42,119,330,890,2376,6291,16588,43615,114492,300236,

%T 786828,2061233,5398470,14136759,37015990,96917974,253748880,

%U 664346375,1739318904,4553656703,11921726232,31211643384,81713400340,213928875445,560073740226

%N Sum of non-Fibonacci numbers between successive Fibonacci numbers: a(n) = Sum_{k=F(n)+1..F(n+1)-1} k.

%H Colin Barker, <a href="/A109454/b109454.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2104.05593">On the Gap-sum and Gap-product Sequences of Integer Sequences</a>, arXiv:2104.05593 [math.CO], 2021.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-5,-1,1).

%F a(n) = Fibonacci(n+2)*(Fibonacci(n-1)-1)/2, n>1. - _Vladeta Jovovic_, Aug 27 2005

%F a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) + a(n-5) for n>6. - _Colin Barker_, Mar 26 2015

%F G.f.: x^4*(x^2-x-4) / ((x+1)*(x^2-3*x+1)*(x^2+x-1)). - _Colin Barker_, Mar 26 2015

%e F(5) = F(4) + 1 = 4.

%e F(6) = (F(5) + 1) + (F(5) + 2) = 6+7 = 13.

%e F(7) = 9+10+11+12 = 42.

%t CoefficientList[Series[x^4*(x^2 - x - 4)/((x + 1) (x^2 - 3 x + 1) (x^2 + x - 1)), {x, 0, 30}], x] (* _Michael De Vlieger_, Jul 08 2021 *)

%t Total[Range[#[[1]]+1,#[[2]]-1]]&/@Partition[Fibonacci[Range[0,40]],2,1] (* or *) LinearRecurrence[{3,1,-5,-1,1},{0,0,0,0,4,13,42},40] (* _Harvey P. Dale_, Sep 30 2024 *)

%o (PARI) concat([0,0,0,0], Vec(x^4*(x^2-x-4) / ((x+1)*(x^2-3*x+1)*(x^2+x-1)) + O(x^100))) \\ _Colin Barker_, Mar 26 2015

%Y Cf. A000045, A006002.

%K nonn,easy

%O 0,5

%A _Amarnath Murthy_, Aug 27 2005

%E More terms from _Franklin T. Adams-Watters_, Jun 06 2006