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A372048 The index of the largest Fibonacci number that divides the sum of Fibonacci numbers with indices 1 through n.

%I A372048 #30 Jul 29 2025 00:20:27
%S A372048 2,3,3,2,4,5,4,4,6,7,6,6,8,9,8,8,10,11,10,10,12,13,12,12,14,15,14,14,
%T A372048 16,17,16,16,18,19,18,18,20,21,20,20,22,23,22,22,24,25,24,24,26,27,26,
%U A372048 26,28,29,28,28,30,31,30,30,32,33,32,32,34,35,34,34,36,37,36,36,38,39,38,38,40,41,40,40
%N A372048 The index of the largest Fibonacci number that divides the sum of Fibonacci numbers with indices 1 through n.
%C A372048 The sum of the first n Fibonacci numbers is sequence A000071.
%C A372048 When we divide the sum by the largest Fibonacci number that divides the sum, we always get a Lucas number.
%C A372048 For n > 3, a(n+4) = a(n) + 2.
%H A372048 Tanya Khovanova and the MIT PRIMES STEP senior group, <a href="https://arxiv.org/abs/2409.01296">Fibonacci Partial Sums Tricks</a>, arXiv:2409.01296 [math.HO], 2024.
%H A372048 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-1).
%F A372048 G.f.: x*(x^6-2*x^5+2*x^4-2*x^3+x^2-x+2)/((x^2+1)*(x-1)^2). - _Alois P. Heinz_, Jul 25 2025
%e A372048 The sum of the first three Fibonacci numbers is 1+1+2=4. The largest Fibonacci that divides this sum is 2, the third Fibonacci number. Thus, a(3) = 2. After the division, we get 4/2 = 2, the zeroth Lucas number.
%e A372048 The sum of the first ten Fibonacci numbers is 143. The largest Fibonacci that divides this sum is 13, the seventh Fibonacci number. Thus, a(10) = 7. After the division, we get 143/13 = 11, the fifth Lucas number.
%t A372048 LinearRecurrence[{2, -2, 2, -1}, {2, 3, 3, 2, 4, 5, 4}, 80] (* _James C. McMahon_, Apr 30 2024, updated by _Sean A. Irvine_, Jul 29 2025 *)
%Y A372048 Cf. A000032, A000045, A000071, A054494, A075850, A372049.
%K A372048 nonn,easy
%O A372048 1,1
%A A372048 _Tanya Khovanova_ and the MIT PRIMES STEP senior group, Apr 17 2024