A328943 a(n) = 2 + (n mod 4).

2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3

Offset: 0

Author

David Nacin, Oct 31 2019

Keywords

• nonn

Comments

Terms of the simple continued fraction of (36+sqrt(1806))/34.

2345/9999=0.234523452345...

Partial sums are given by A130482(n) + 2*n + 2.

Example of a sequence where the largest of any four consecutive terms equals the sum of the two smallest.

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• David Nacin, Van der Laan Sequences and a Conjecture on Padovan Numbers, J. Int. Seq., Vol. 26 (2023), Article 23.1.2.
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1).

Crossrefs

Cf. A010883, A010873, A130482

Programs

Mathematica

PadRight[{}, 120, {2, 3, 4, 5}]

Python

def a(n):
   return n%4+2

Formula

a(n) = 2 + (n mod 4).

G.f.: (5x^3 + 4x^2 + 3x + 2)/(1 - x^4).

a(n) = A010873(n) + 2 = A010883(n) + 1.

a(n) = 14 - a(n-1) - a(n-2) - a(n-3) for n > 2.