Mathematica Asked on June 5, 2021
I tried to use Mathematica to find $a_{n}$. But unfortunately, it returned a equation that I can’t understand. Here is the code below.
eqn = RSolve[
a[n + 1] == a[n] + a[n - 2] && a[2] == 1 && a[3] == 2 && a[4] == 3,
a, n];
Table[a[n] /. First[eqn], {n, 1, 13}]
Try:
eqn = RSolve[
a[n + 1] == a[n] + a[n - 2] && a[2] == 1 && a[3] == 2 && a[4] == 3,
a, n];
Table[a[n] /. First[eqn], {n, 1, 13}] // Re // N // Rationalize
(*{1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88}*)
You can try:
RSolve[a[n + 1] == a[n] + a[n - 2] && a[2] == 1 && a[3] == 2 &&
a[4] == 3, a[n], n] // ToRadicals
(*{{a[n] -> (1/3 - 1/6 (1 - I Sqrt[3]) (29/2 - (3 Sqrt[93])/2)^(1/3) -
1/6 (1 + I Sqrt[3]) (1/2 (29 + 3 Sqrt[93]))^(1/3))^
n (1/3 -
1/186 (1 + I Sqrt[3]) (3844 - 372 Sqrt[93])^(
1/3) - ((1 - I Sqrt[3]) (1/2 (31 + 3 Sqrt[93]))^(1/3))/(
3 31^(2/3))) + (1/3 -
1/6 (1 + I Sqrt[3]) (29/2 - (3 Sqrt[93])/2)^(1/3) -
1/6 (1 - I Sqrt[3]) (1/2 (29 + 3 Sqrt[93]))^(1/3))^
n (1/3 -
1/186 (1 - I Sqrt[3]) (3844 - 372 Sqrt[93])^(
1/3) - ((1 + I Sqrt[3]) (1/2 (31 + 3 Sqrt[93]))^(1/3))/(
3 31^(2/3))) +
1/31 3^(-1 -
n) (1 + (29/2 - (3 Sqrt[93])/2)^(
1/3) + (1/2 (29 + 3 Sqrt[93]))^(1/3))^
n (31 + (3844 - 372 Sqrt[93])^(1/3) +
2^(2/3) (31 (31 + 3 Sqrt[93]))^(1/3))}}*)
For more info execute: ?Root
and ?ToRadicals
Answered by Mariusz Iwaniuk on June 5, 2021
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