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

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, 16, 17, 16, 16, 18, 19, 18, 18, 20, 21, 20, 20, 22, 23, 22, 22, 24, 25, 24, 24, 26, 27, 26, 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

Offset

1,1

Comments

The sum of the first n Fibonacci numbers is sequence A000071.

When we divide the sum by the largest Fibonacci number that divides the sum, we always get a Lucas number.

For n > 3, a(n+4) = a(n) + 2.

Links

Table of n, a(n) for n=1..80.

Tanya Khovanova and the MIT PRIMES STEP senior group, Fibonacci Partial Sums Tricks, arXiv:2409.01296 [math.HO], 2024.

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Formula

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

Example

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.
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.

Mathematica

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 *)

Crossrefs

Cf. A000032, A000045, A000071, A054494, A075850, A372049.

Sequence in context: A308284 A036465 A265339 * A182865 A131469 A309076

Adjacent sequences: A372045 A372046 A372047 * A372049 A372050 A372051

Keyword

nonn,easy

Author

Tanya Khovanova and the MIT PRIMES STEP senior group, Apr 17 2024

Status

approved