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

0,1

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).

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.

Mathematica

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

Prog

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

Crossrefs

Cf. A010883, A010873, A130482

Sequence in context: A212176 A070671 A119281 * A173525 A070772 A094937

Adjacent sequences: A328940 A328941 A328942 * A328944 A328945 A328946

Keyword

nonn

Author

David Nacin, Oct 31 2019

Status

approved