A208537 Number of 7-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

0, 0, 18, 312, 2340, 11160, 39990, 117648, 299592, 683280, 1428570, 2783880, 5118828, 8964072, 15059070, 24408480, 38347920, 58619808, 87460002, 127695960, 182857140, 257298360, 356336838, 486403632, 655210200, 871930800, 1147401450

Offset: 1

Author

R. H. Hardin, Feb 27 2012

Keywords

• nonn
• easy

Comments

This sequence would be better defined as a(n) = (n^7-n)/7 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023

Row 7 of A208535.

Also, row 7 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017

Examples

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

References

• J. Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.

Links

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

Crossrefs

Cf. A208535.

Programs

Mathematica

A208537[n_]:=((n-1)^7-(n-1))/7;Array[A208537,50] (* Paolo Xausa, Nov 14 2023 *)

PARI

Vec(6*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) = (1/7)*n^7 - 1*n^6 + 3*n^5 - 5*n^4 + 5*n^3 - 3*n^2 + (6/7)*n.

Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).

a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 7. - Danny Rorabaugh, Nov 10 2017

From Colin Barker, Nov 11 2017: (Start)

G.f.: 6*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)

a(n) = ((n-1)^7 - (n-1))/7. (inspired by Hassler's formula in A208536) - Eric M. Schmidt, Dec 08 2017