## Tribonacci numbers

Will there be a formula for a ‘n bonacci series’? Like, a ‘tri’ bonacci is 1, 1, 1, 3, 5, 9, 17, 31, 57, … .  So that an n-bonacci is 1, 1, 1, … ntimes…, n, 2n-1,4 n-3, …?

(Proposed by Souradeep/FB/Math)

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For 1,1,1,3,5,9,17,31,57…. we have a recursion formula
a(n) + a(n+1) + a(n+2) = a(n+3). To solve this first find λ from the characteristic equation 1 + λ + λ² = λ³. Then a(n) = A λ₁ⁿ + B λ₂ⁿ + C λ₃ⁿ where A, B, C are determined by the initial conditions a(1) = a(2) = a(3) = 1.

The result is

a(n) = (λ₂ – 1)·(λ₃ – 1)/(λ₁·(λ₁ – λ₂)·(λ₁ – λ₃))·λ₁ⁿ + (1 – λ₃)·(λ₁ – 1)/(λ₂·(λ₁ – λ₂)·(λ₂ – λ₃))·λ₂ⁿ + (λ₁ – 1)·(λ₂ – 1)/(λ₃·(λ₁ – λ₃)·(λ₂ – λ₃))·λ₃ⁿ

where

λ1 = – (19/216 – √33/72)1/3 – (√33/72 + 19/216)1/3 + 1/3 + i·((19·√3/72 + √11/8)1/3 – (19·√3/72 – √11/8)1/3)

λ2 = – (19/216 – √33/72)1/3 – (√33/72 + 19/216)1/3 + 1/3 + i·((19·√3/72 – √11/8)1/3 – (19·√3/72 + √11/8)1/3)

λ3 = (19/27 – √33/9)1/3 + (√33/9 + 19/27)1/3 + 1/3

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The problem is that the solutions to third degree equations are usual complicated. To get something simpler modify the recursion with suitable coefficients a,b,c so that the third degree eauation a + bλ + cλ² = λ³ gets simple solutions. You can start choosing some numbers λ₁, λ₂, λ₃ and then set a = λ₁ λ₂ λ₃, etc.

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