A208545 Number of 7-bead necklaces of n colors allowing reversal, with no adjacent beads having the same color.

0, 0, 9, 156, 1170, 5580, 19995, 58824, 149796, 341640, 714285, 1391940, 2559414, 4482036, 7529535, 12204240, 19173960, 29309904, 43730001, 63847980, 91428570, 128649180, 178168419, 243201816, 327605100, 435965400, 573700725, 747168084

Offset: 1

Author

R. H. Hardin, Feb 27 2012

Keywords

• nonn
• easy

Comments

Row 7 of A208544.

Examples

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

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210
• Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Crossrefs

Cf. A208537.

Programs

PARI

Vec(3*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8 + O(x^40)) \\ Colin Barker, Nov 11 2017

Formula

Empirical: a(n) = (1/14)*n^7 - (1/2)*n^6 + (3/2)*n^5 - (5/2)*n^4 + (5/2)*n^3 - (3/2)*n^2 + (3/7)*n.

From Colin Barker, Nov 11 2017: (Start)

G.f.: 3*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8.

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.

(End)