A319171 Square array, read by antidiagonals, upwards: T(n,k) is the number of groups of order prime(k+1)^n.

1, 1, 1, 2, 1, 1, 5, 2, 1, 1, 14, 5, 2, 1, 1, 51, 15, 5, 2, 1, 1, 267, 67, 15, 5, 2, 1, 1, 2328, 504, 77, 15, 5, 2, 1, 1, 56092, 9310, 684, 83, 15, 5, 2, 1, 1, 10494213, 1396077, 34297, 860, 87, 15, 5, 2, 1, 1, 49487367289, 5937876645

Offset: 0

Author

Franck Maminirina Ramaharo, Sep 12 2018

Keywords

• tabl
• nonn
• hard
• more

Comments

In 1960, Higman conjectured that the function f(n,p) giving the number of groups of prime-power order p^n, for fixed n and varying p, is a "Polynomial in Residue Classes" (PORC), i.e., there exist an integer M and polynomials q_i(x) in Z[x] (i = 1, 2, ..., M) such that if p = i mod M, then f(n,p) = q_i(p). The conjecture is confirmed for n <= 7.

Examples

			Array begins:
  (p = 2) (p = 3) (p = 5) (p = 7) (p = 11) (p = 13) ...
       1       1       1       1        1        1  ...
       1       1       1       1        1        1  ...
       2       2       2       2        2        2  ...
       5       5       5       5        5        5  ...
      14      15      15      15       15       15  ...
      51      67      77      83       87       97  ...
     267     504     684     860     1192     1476  ...
    2328    9310   34297  113147   750735  1600573  ...
     ...
		

Links

• H. U. Besche, B. Eick, and E. A. O'Brien. A Millennium Project: Constructing Small Groups, Internat. J. Algebra and Computation, 12 (2002), 623-644.
• David Burrell, On the number of groups of order 1024, Communications in Algebra, 2021, 1-3.
• David Burrell, The number of p-groups of order 19,683 and new lists of p-groups, Communications in Algebra, Vol. 51 - Issue 6 (2023), 2673-2679.
• Heiko Dietrich, Computational aspects of finite p-groups
• Groupprops, Groups of prime power order
• Groupprops, Higman's PORC conjecture
• Groupprops, PORC function
• Graham Higman, Enumerating p-Groups. I: Inequalities, Proc. London Math. Soc. Vol. 10 (1960), 24-30.
• Graham Higman, Enumerating p-Groups. II: Problem whose solution is PORC, Proc. London Math. Soc. Vol. 10 (1960), 566-582.
• Eamonn O'Brien, Polycyclic groups
• Gordon Royle, Numbers of Small Groups
• Michael Vaughan-Lee, Graham Higman’s PORC Conjecture, Jahresbericht der Deutschen Mathematiker-Vereinigung Vol. 114 (2012), 89-16.
• Michael Vaughan-Lee, Groups of order p^8 and exponent p, International Journal of Group Theory Vol. 4 (2015), 25-42.
• Brett E. Witty, Enumeration of groups of prime-power order, PhD thesis, 2006.
• Index entries for sequences related to groups

Crossrefs

Inspired by A158106.

Cf. A000001, A000679, A090091, A090130, A090140, A128604, A232105, A232106, A232107.

Programs

GAP

# This program computes the first 45 terms, rows 0..8.
P:=Filtered([1..300],IsPrime);;
T1:=List([0..7],n->List([0..15],k->NumberSmallGroups(P[k+1]^n)));;
T2:=[Flat(Concatenation(List([8],n->List([0],k->NumberSmallGroups(P[k+1]^n))),List([1..14],i->0)))];;
T:=Concatenation(T1,T2);;
b:=List([2..10],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Oct 01 2018

Maple

with(GroupTheory): T:=proc(n,k) NumGroups(ithprime(k+1)^n); end proc: seq(seq(T(n-k,k),k=0..n),n=0..10); # Muniru A Asiru, Oct 03 2018

Mathematica

(* This program uses Higman's PORC functions to compute the rows 0 to 7 *)
f[0, p_] := 1; f[1, p_] := 1; f[2, p_] := 2; f[3, p_] := 5;
f[4, p_] := If[p == 2, 14, 15];
f[5, p_] := If[p == 2, 51, If[p == 3, 67, 61 + 2*p + 2*GCD[p - 1, 3] + GCD[p - 1, 4]]];
f[6, p_] := If[p == 2, 267, If[p == 3, 504, 3*p^2 + 39*p + 344 + 24*GCD[p - 1, 3] + 11*GCD[p - 1, 4] + 2*GCD[p - 1, 5]]];
f[7, p_] := If[p == 2, 2328, If[p == 3, 9310, If[p == 5, 34297, 3*p^5 + 12*p^4 + 44*p^3 + 170*p^2 + 707*p + 2455 + (4*p^2 + 44*p + 291)*GCD[p - 1, 3] + (p^2 + 19*p + 135)*GCD[p - 1, 4] + (3*p + 31)*GCD[p - 1, 5] + 4*GCD[p - 1, 7] + 5*GCD[p - 1, 8] + GCD[p - 1, 9]]]];
tabl[kk_] := TableForm[Table[f[n, Prime[k+1]], {n, 0, 7}, {k, 0, kk}]];

Formula

T(n,0) = A000679(n).

T(n,1) = A090091(n).

T(n,2) = A090130(n).

T(n,3) = A090140(n).

T(0,n) = 1, T(1,n) = 1, T(2,n) = 2 and T(3,n) = 5.

T(4,0) = 14 and T(4,n) = 15, n > 0.

T(5,n) = A232105(n+1).

T(6,n) = A232106(n+1).

T(7,n) = A232107(n+1).

Extensions

a(55)=T(10,0) corrected by David Burrell, Jun 07 2022

a(56)=T(9,1) from David Burrell, Sep 01 2023