cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.

Original entry on oeis.org

5, 15, 35, 120, 377, 1505, 5895, 25395, 110085, 493131, 2227275, 10196680, 46989185, 218102685, 1017448143, 4768969770, 22440372245, 105966797755, 501938733555, 2384200683816, 11353290089305
Offset: 1

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From Petros Hadjicostas, Sep 01 2018: (Start)
The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.
Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with Herbert Kociemba's g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.
(End)

Examples

			For n=2, the 15 bracelets are AA, AB, AC, AD, AE, BB, BC, BD, BE, CC, CD, CE, DD, DE, and EE. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Cf. A081720.
Column 5 of A051137.
Cf. A001869 (oriented), A056487 (achiral), A278641 (chiral).

Programs

  • Mathematica
    mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-5*x^n]/n,{n,mx}]+(1+5 x+10 x^2)/(1-5 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
    k=5; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)

Formula

"DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...
a(n) = A081720(n,5), n >= 1. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 5*x^n)/n + (1+5*x+10*x^2)/(1-5*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (3/2)*5^(n/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is even, and = (1/2)*5^((n+1)/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is odd. - Petros Hadjicostas, Sep 01 2018
a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - Robert A. Russell, Oct 13 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{d divides n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Richard L. Ollerton, May 04 2021
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=5 is the maximum number of colors. (See A051137.) - Richard L. Ollerton, May 04 2021