cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A372050 The index of the largest Fibonacci number that divides the sum of Fibonacci numbers with indices 1 through A000217(n) (the n-th triangular number).

Original entry on oeis.org

2, 3, 5, 7, 8, 12, 14, 18, 24, 28, 35, 41, 46, 54, 60, 68, 78, 89, 97, 107, 116, 128, 138, 150, 164, 176, 191, 205, 218, 234, 248, 264, 282, 298, 317, 335, 352, 372, 390, 410, 432, 452, 475, 497, 518, 542, 564, 588, 614, 638, 665, 691, 716, 744, 770, 798, 828, 856, 887, 917, 946, 978
Offset: 1

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Author

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

Keywords

Comments

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

Examples

			For example, the sum of the first ten Fibonacci numbers is 143. The largest Fibonacci that divides this sum is 13, the seventh Fibonacci number. Thus, as 10 is the fourth triangular number a(4) = 7. After the division we get 143/13 = 11, the fifth Lucas number.
		

Crossrefs

A372718 Triangular numbers that are 2 mod 4, halved.

Original entry on oeis.org

3, 5, 33, 39, 95, 105, 189, 203, 315, 333, 473, 495, 663, 689, 885, 915, 1139, 1173, 1425, 1463, 1743, 1785, 2093, 2139, 2475, 2525, 2889, 2943, 3335, 3393, 3813, 3875, 4323, 4389, 4865, 4935, 5439, 5513, 6045, 6123, 6683, 6765, 7353, 7439, 8055, 8145, 8789, 8883, 9555, 9653
Offset: 1

Views

Author

Tanya Khovanova and the PRIMES STEP senior group, May 11 2024

Keywords

Comments

The sum of the first 2*a(n) numbers of any Fibonacci-like sequence equals its (a(n)+2)-nd term times the a(n)-th Lucas number.

Examples

			10 is a triangular number that has a remainder of 2 when divided by 4. Therefore, its half, 5, is in this sequence. Moreover, the sum of the first 5*2 Fibonacci numbers is 143 (not counting zero). This sum is a product of 13, which is the (5+2 = 7)-th term of the Fibonacci sequence times 11, which is the fifth Lucas number.
		

Crossrefs

Programs

  • Mathematica
    Select[Table[n (n + 1)/2, {n, 200}], Mod[#, 4] == 2 &]/2

Formula

a(n) = A000217(A047457(n))/2 = A372070(n)/2.
Showing 1-2 of 2 results.