A028356 Simple periodic sequence underlying clock sequence A028354.

1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From Klaus Brockhaus, May 15 2010: (Start)

Continued fraction expansion of (28+sqrt(2730))/56.

Decimal expansion of 1112/9009.

Partial sums of 1 followed by A130151.

First differences of A028357. (End)

References

• Zdeněk Horský, "Pražský orloj" ("The Astronomical Clock of Prague", in Czech), Panorama, Prague, 1988, pp. 76-78.

Links

• Michal Křížek, Alena Šolcová and Lawrence Somer, Construction of Šindel sequences, Comment. Math. Univ. Carolin., 48 (2007), 373-388.
• N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
• Index entries for linear recurrences with constant coefficients, signature (1,0,-1,1).

Crossrefs

Cf. A000034, A028354, A068073, A118382, A118383.

Cf. A177924 (decimal expansion of (28+sqrt(2730))/56), A130151 (repeat 1, 1, 1, -1, -1, -1), A028357 (partial sums of A028356). - Klaus Brockhaus, May 15 2010

Programs

Magma

&cat [[1, 2, 3, 4, 3, 2]^^20]; // Klaus Brockhaus, May 15 2010

Maple

A028356:=n->[1, 2, 3, 4, 3, 2][(n mod 6)+1]: seq(A028356(n), n=0..100); # Wesley Ivan Hurt, Jun 23 2016

Mathematica

CoefficientList[ Series[(1 + 2x + 3x^2 + 4x^3 + 3x^4 + 2x^5)/(1 - x^6), {x, 0, 85}], x]
LinearRecurrence[{1,0,-1,1},{1,2,3,4},120] (* or *) PadRight[{},120,{1,2,3,4,3,2}] (* Harvey P. Dale, Apr 15 2016 *)

Python

def A028356(n): return (1,2,3,4,3,2)[n%6] # Chai Wah Wu, Apr 18 2024

Sage

def A():
    a, b, c, d = 1, 2, 3, 4
    while True:
        yield a
        a, b, c, d = b, c, d, a + (d - b)
A028356 = A(); [next(A028356) for n in range(106)] # Peter Luschny, Jul 26 2014

Formula

Sum of any six successive terms is 15.

G.f.: (1 + 2*x + 3*x^2 + 4*x^3 + 3*x^4 + 2*x^5)/(1 - x^6).

From Wesley Ivan Hurt, Jun 23 2016: (Start)

a(n) = a(n-1) - a(n-3) + a(n-4) for n>3.

a(n) = (15 - cos(n*Pi) - 8*cos(n*Pi/3))/6. (End)

E.g.f.: (15*exp(x) - exp(-x) - 8*cos(sqrt(3)*x/2)*(sinh(x/2) + cosh(x/2)))/6. - Ilya Gutkovskiy, Jun 23 2016

a(n) = abs(((n+3) mod 6)-3) + 1. - Daniel Jiménez, Jan 14 2023

Extensions

Additional comments from Robert G. Wilson v, Mar 01 2002