login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A319171
Square array, read by antidiagonals, upwards: T(n,k) is the number of groups of order prime(k+1)^n.
0
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
0,4
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.
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.
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
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.
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).
EXAMPLE
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 ...
...
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}]];
PROG
(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
CROSSREFS
KEYWORD
tabl,nonn,hard,more
AUTHOR
EXTENSIONS
a(55)=T(10,0) corrected by David Burrell, Jun 07 2022
a(56)=T(9,1) from David Burrell, Sep 01 2023
STATUS
approved