A308047 Sum of subgroup indices of dihedral group, Sum_{H <= D(n)} [D(n):H].

3, 11, 18, 35, 38, 74, 66, 115, 117, 166, 146, 266, 198, 298, 308, 403, 326, 533, 402, 630, 564, 682, 578, 970, 713, 934, 900, 1162, 902, 1444, 1026, 1491, 1316, 1558, 1396, 2093, 1446, 1930, 1812, 2390, 1766, 2692, 1938, 2730, 2522, 2794, 2306, 3658, 2565, 3441

Offset

1,1

Comments

This is a generalization of the sum of divisors to finite groups:

Let G be a finite group. sigma(G) = Sum_{H <= G} [G:H]. For G = D(n) = dihedral group with 2*n elements, we get a(n) = 2*n * Sum_{H <= G} 1/|H|.

Relation to Lagarias inequality: Let H(G) be the harmonic numbers of D(n) relative to a generating set S <= D(n). Then the conjecture (Lagarias inequality) is:

a(n) <= H(D(n)) + exp(H(D(n)))*log(H(D(n))).

Relation to sum of divisors: sigma(n) = sigma(C_n), where C_n = cyclic group.

Links

Table of n, a(n) for n=1..50.

MathStackexchange, Two questions about the dihedral group, Proof of the formula a(n) = sigma_2(n) + 2*sigma_1(n).

MathOverflow, A Group theoretic interpretation of Lagarias inequality, Definition of the Harmonic numbers H(G) for each finite group G.

MathOverflow, Conjectured upper bound to number of subgroups, Definition of sigma(G)=sigma_1 for each finite group G.

Formula

a(n) = 2*n * Sum_{H <= G} 1/|H|, where the sum runs through all subgroups H of G.

a(n) = sigma_2(n) + 2*sigma_1(n).

Example

a(1) = 3 as there are two subgroups of order 1 and 2 so 2*1*(1/1 + 1/2) = 3.

Mathematica

Table[DivisorSigma[2, n] + 2*DivisorSigma[1, n], {n, 50}] (* G. C. Greubel, Jul 15 2019 *)

Prog

(Sage)
[sigma(n, 2)+2*sigma(n) for n in range(1, 51)]
(PARI) a(n) = sigma(n, 2) + 2*sigma(n); \\ Michel Marcus, May 13 2019
(Magma) [DivisorSigma(2, n) + 2*DivisorSigma(1, n): n in [1..50]]; // G. C. Greubel, Jul 15 2019

Crossrefs

Sequence in context: A194128 A127995 A316839 * A119141 A303520 A225144

Adjacent sequences: A308044 A308045 A308046 * A308048 A308049 A308050

Keyword

nonn

Author

Orges Leka, May 10 2019

Status

approved