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A179229 a(n)=number of Abelian groups of order 2n which are not isomorphic to any Galois field GF(k) of that order. 1
0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 1, 1, 1, 0, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,8

LINKS

Table of n, a(n) for n=1..16.

McCarro J., Small Groups

FORMULA

a(n)+A197366(n) = A000688(2n). - R. J. Mathar, Jun 14 2019

EXAMPLE

a(1)=0 because we have 1 Abelian group or order 2*1=2 and we have 3 different k with GF(k) containing of 2 elements these k->and isomorphic Abelian groups are  following: 3 -> 2.1, 4 -> 2.1, 6 -> 2.1 <which means that only one group is represented and no Abelian group of order 2 remains unrepresented by GF(k)>.

a(2)=0 because we have 2 Abelian groups or order 2*2=4: 4.1, 4.2 and we have 4 different k with GF(k) containing of 4 elements these k->and isomorphic Abelian groups are  following: 15 -> 4.1, 8 -> 4.2, 10 -> 4.1, 12 -> 4.2 <which means that both groups are represented and no Abelian group of order 4 is unrepresented by GF(k)>.

a(3)=0 because we have 1 Abelian group or order 2*3=6 and we have 4 different k with GF(k) containing of 6 elements these k->and isomorphic Abelian groups are  following: 7 -> 6.1, 9 -> 6.1, 14 -> 6.1, 18 -> 6.1 <which means that only one group is represented and no Abelian group of order 6 is unrepresented by GF(k)>.

a(4)=1 because we have 3 different Abelian groups or order 2*4=8: 8.1, 8.2, 8.3 and we have 5 different k with GF(k) containing of 8 elements these k->and isomorphic Abelian groups are  following: 15 -> 8.2, 16 -> 8.2, 20 -> 8.2, 24 -> 8.3, 30 -> 8.2 <what mean that 8.1 is not represented by any GF(k)>. Group 8.1 <C8 cyclic group> don't have direct isomorphism with any Galois Field, but have Galois subfield e.g. set {1, 11, 19, 29, 31, 41, 49, 59} from GF(60).

a(5)=0 because we have 1 Abelian group or order 2*5=10: 10.1 and we have 2 different k with GF(k) containing of 10 elements these k->and isomorphic Abelian groups are  following: 11 -> 10.1, 22 -> 10.1 <which means that only one group is represented and no Abelian group of order 22 is unrepresented by GF(k)>.

a(6)=0 because we have 2 Abelian groups or order 2*6=12: 12.1, 12.2 and we have 5 different k with GF(k) containing of 12 elements these k->and isomorphic Abelian groups are  following: 13 -> 12.1, 21 -> 12.2, 26 -> 12.1, 28 -> 12.2, 36 -> 12.2, 42 -> 12.2 <which means that both groups are represented and no Abelian group of order 12 is unrepresented by GF(k)>.

a(7)=1 because we have 1 Abelian group of order 2*7=14: 14.1 and we don't have any GF(k) containing of 14 elements <which means that 14.1 is not represented by any GF(k)>.

a(8)=2 because we have 5 different Abelian groups or order 2*8=16: 16.1, 16.2, 16.3, 16.4, 16.5 and we have 6 different k with GF(k) containing of 16 elements these k->and isomorphic Abelian groups are  following: 17 -> 16.1, 32 -> 16.2, 34 -> 16.1, 40 -> 16.4, 48 -> 16.4, 60 -> 16.4 <which means that 16.3 and 16.5 are not represented by any GF(k)>.

a(9)=1 because we have 2 different Abelian groups or order 2*9=18: 18.1, 18.2 and we have 4 different k with GF(k) containing of 18 elements these k->and isomorphic Abelian groups are  following: 19 -> 18.1, 27 -> 18.1, 38 -> 18.1, 54 -> 18.1 <which means that 18.2 is not represented by any GF(k)>.

a(10)=0 because we have 2 Abelian groups or order 2*10=20: 20.1, 20.2 and we have 5 different k with GF(k) containing of 20 elements these k->and isomorphic Abelian groups are  following: 20.1, 33 -> 20.2, 44 -> 20.2, 50 -> 20.1, 66 -> 20.2 <which means that both groups are represented and no Abelian group of order 20 remains unrepresented by GF(k)>.

a(11)=0 because we have 1 Abelian group or order 2*11=22: 32.1 and we have 2 different k with GF(k) containing of 22 elements these k->and isomorhiic Abelian groups are  following: 23->22.1, 46->22.1 <which means that only one group is represented and no Abelian group of order 22 is unrepresented by GF(k)>.

a(12)=1 because we have 3 different Abelian groups or order 2*12=24: 24.1, 24.2, 24.3 and we have 10 different k with GF(k) containing of 24 elements these k->and isomorphic Abelian groups are  following: 35 -> 24.2, 39 -> 24.2, 45 -> 24.2, 52 -> 24.2, 56 -> 24.3, 70 -> 24.2, 72 -> 24.3, 78 -> 24.2, 84 -> 3, 90 -> 24.2 <which means that 24.1 is not represented by any GF(k)>.

a(13)=1 because we have 1 Abelian group or order 2*13=26: 26.1 but we don't have anyone k with GF(k) containing of 26 elements <which means that 26.1 is not represented by any GF(k)>.

a(14)=1   because we have 2 different Abelian groups or order 2*14=28: 28.1, 28.2 and we have 2 different k with GF(k) containing of 28 elements these k->and isomorphic Abelian groups are  following: 29 -> 28.1, 58 -> 28.1 <which means that 28.2 is not represented by any GF(k)>.

a(15)=0   because we have 1 Abelian group or order 2*15=30: 30.1 and we have 2 different k with GF(k) containing of 30 elements these k->and isomorphic Abelian groups are following: 31->30.1, 62->30.1 <which means that only one group is represented and no Abelian group of order 30 is unrepresented by GF(k)>.

a(16)=3 because we have 7 different Abelian groups or order 2*16=32: 32.1, 32.2, 32.3, 32.4, 32.5, 32.6, 32.7 and we have 7 different k with GF(k) containing of 32 elements these k->and isomorphic Abelian groups are  following:   51 -> 32.3, 64 -> 32.3, 68 -> 32.3, 80 -> 32.4, 96 -> 32.5, 102 -> 32.3, 120 -> 32.6  <which means that 32.1 and 32.2 and 32.7 are not represented by any GF(k)>.

MATHEMATICA

(*procedure to determination of Abelian groups of GF(k)*)

cc = {}; bb = {}; ct = {}; gf = 90; v = {}; Do[If[GCD[p, gf] == 1, AppendTo[v, p]], {p, 1, gf - 1}]; len = Length[v]; Do[aa = {}; Do[AppendTo[aa, Mod[v[[n]] v[[m]], gf]], {m, 1, len}]; AppendTo[ct, aa], {n, 1, len}]; Do[If[ct[[n]][[n]] == 1, AppendTo[cc, ct[[n]][[1]]]]; AppendTo[bb, ct[[n]][[n]]], {n, 1, len}]; bb = Union[bb]; Print[bb]; Print["-1->", cc]; Print[Length[bb]]; ct // MatrixForm

CROSSREFS

Cf. A046660, A197366

Sequence in context: A309047 A255317 A309168 * A117201 A060953 A339367

Adjacent sequences:  A179226 A179227 A179228 * A179230 A179231 A179232

KEYWORD

more,nonn,hard

AUTHOR

Artur Jasinski, Jul 03 2010

STATUS

approved

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Last modified July 27 02:39 EDT 2021. Contains 346302 sequences. (Running on oeis4.)