A124697 -id:A124697 - cp's OEIS frontend

A208773 Number of n-bead necklaces labeled with numbers 1..4 not allowing reversal, with no adjacent beads differing by more than 1.

4, 7, 10, 18, 30, 65, 128, 293, 658, 1544, 3622, 8711, 20924, 50889, 124150, 304718, 750334, 1855429, 4600696, 11442853, 28528618, 71294416, 178529670, 447923761, 1125756860, 2833917147, 7144466842, 18036449390, 45591671454, 115381885423, 292329164912, 741411257693, 1882219950046, 4782783122992, 12163730636250

Offset: 1

R. H. Hardin, Mar 01 2012

nonn

			All solutions for n=3:
..2....4....1....2....1....2....3....3....1....3
..2....4....1....2....1....3....3....4....2....3
..3....4....2....2....1....3....3....4....2....4

Andrew Howroyd, Table of n, a(n) for n = 1..100
Arnold Knopfmacher, Toufik Mansour, Augustine Munagi, Helmut Prodinger, Smooth words and Chebyshev polynomials, arXiv:0809.0551v1 [math.CO], 2008.

Column 4 of A208777.

Cf. A215336 (cyclically smooth Lyndon words with 4 colors).

sn[n_, k_] := 1/n*Sum[ Sum[ EulerPhi[j]*(1 + 2*Cos[i*Pi/(k + 1)])^(n/j), {j, Divisors[n]}], {i, 1, k}]; Table[sn[n, 4], {n, 1, 35}] // FullSimplify (* Jean-François Alcover, Oct 31 2017, after Joerg Arndt *)

/* from the Knopfmacher et al. reference */
default(realprecision,99); /* using floats */
sn(n,k)=1/n*sum(i=1,k,sumdiv(n,j,eulerphi(j)*(1+2*cos(i*Pi/(k+1)))^(n/j)));
vector(66,n, round(sn(n,4)) )
/* Joerg Arndt, Aug 09 2012 */

a(n) = Sum_{ d | n } A215336(d). - Joerg Arndt, Aug 13 2012

a(n) = (1/n) * Sum_{d | n} totient(n/d) * A124697(n). - Andrew Howroyd, Mar 18 2017

A215500 a(n) = ((sqrt(5) + 3)^n + (-sqrt(5) -1)^n + (-sqrt(5) + 3)^n + (sqrt(5) - 1)^n) / 2^n.

Original entry on oeis.org

4, 2, 10, 14, 54, 112, 340, 814, 2254, 5702, 15250, 39404, 104004, 270922, 711490, 1859134, 4873054, 12748472, 33391060, 87394454, 228841254, 599050102, 1568437210, 4106054164, 10750060804, 28143585362, 73681573690, 192899714414, 505019869254, 1322156172352

Offset: 0

Peter Luschny, Aug 13 2012

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 02 2014

			G.f. = 4 + 2*x + 10*x^2 + 14*x^3 + 54*x^4 + 112*x^5 + 340*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-4,1).

Cf. A051927, A215502, A215503.

Cf. A124697.

I:=[4,2,10,14]; [n le 4 select I[n] else 2*Self(n-1) + 3*Self(n-2) - 4*Self(n-3) + Self(n-4): n in [1..30]]; // G. C. Greubel, Apr 23 2018

A215500 := x -> ((sqrt(5)+3)^x+(-sqrt(5)-1)^x+(-sqrt(5)+3)^x+(sqrt(5)-1)^x)/2^x;
seq(simplify(A215500(i)),i=0..29);

a[n_] := ((Sqrt[5] + 3)^n + (-Sqrt[5] - 1)^n + (-Sqrt[5] + 3)^n + (Sqrt[5] - 1)^n)/2^n; Table[a[n] // Simplify, {n, 0, 29}] (* Jean-François Alcover, Jul 02 2013 *)
LinearRecurrence[{2,3,-4,1}, {4, 2, 10, 14}, 50] (* G. C. Greubel, Apr 23 2018 *)

{a(n) = polsym( (1 + (-1)^(n>0)*x - x^2) * (1 - 3*x + x^2), abs(n))[1 + abs(n)]}; /* Michael Somos, Jun 02 2014 */

def A215500(x) :
    return ((sqrt(5)+3)^x+(-sqrt(5)-1)^x+(-sqrt(5)+3)^x+(sqrt(5)-1)^x)/2^x
[A215500(i).round() for i in (0..29)]

G.f.: 2*(2-x)*(1+x)*(1-2*x)/((1-3*x+x^2)*(1+x-x^2)). - Colin Barker, Aug 19 2012
a(n) = 2*a(n-1)+3*a(n-2)-4*a(n-3)+a(n-4). - Colin Barker, Aug 20 2012
a(-n) = A124697(n) if n>0. - Michael Somos, Jun 02 2014

cp's OEIS Frontend

A208773 Number of n-bead necklaces labeled with numbers 1..4 not allowing reversal, with no adjacent beads differing by more than 1.

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