A028365 Order of general affine group over GF(2), AGL(n,2).
1, 2, 24, 1344, 322560, 319979520, 1290157424640, 20972799094947840, 1369104324918194995200, 358201502736997192984166400, 375234700595146883504949480652800, 1573079924978208093254925489963584716800
Offset: 0
References
- J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 54 (1.64).
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..57
- Abdalla G. M. Ahmed, Matt Pharr, Victor Ostromoukhov, and Hui Huang, SZ Sequences: Binary-Based (0, 2^q)-Sequences, arXiv:2505.20434 [cs.GR], 2025. See p. 7.
- Marcus Brinkmann, Extended Affine and CCZ Equivalence up to Dimension 4, Ruhr University Bochum (2019).
- Putnam Competition 1999, Question A6, Amer. Math. Monthly 107 (Oct 2000), 721-732; see p. 725.
- I. Strazdins, Universal affine classification of Boolean functions, Acta Applic. Math. 46 (1997), 147-167.
Programs
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Magma
[1] cat [(&*[2^(n+1) - 2^(j+1): j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Aug 31 2023
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Maple
A028365 := n->2^n*product(2^n-2^'i','i'=0..n-1); # version 1 A028365 := n->product(2^'j'-1,'j'=1..n)*2^binomial(n+1,2); # version 2
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Mathematica
RecurrenceTable[{a[1]==1, a[2]==2, a[3]==24, a[n]==(6a[n-1]^2 a[n-3] - 8a[n-1] a[n-2]^2)/(a[n-2] a[n-3])}, a[n], {n,20}] (* Harvey P. Dale, Aug 03 2011 *) (* Next, the connection with Vandermonde determinants *) f[j_]:= 2^j - 1; z = 15; v[n_]:= Product[Product[f[k] - f[j], {j,k-1}], {k,2,n}] Table[v[n], {n,z}] (* A203303 *) Table[v[n+1]/v[n], {n,z}] (* A028365 *) Table[v[n]*v[n+2]/(2*v[n+1])^2, {n,z}] (* A171499 *) (* Clark Kimberling, Jan 01 2011 *) Table[(-1)^n*2^Binomial[n+1,2]*QPochhammer[2,2,n], {n,0,20}] (* G. C. Greubel, Aug 31 2023 *)
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PARI
a(n)=if(n<0,0,prod(k=1,n,2^k-1)*2^((n^2+n)/2)) /* Michael Somos, May 09 2005 */
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SageMath
[product(2^(n+1) - 2^(k+1) for k in range(n)) for n in range(21)] # G. C. Greubel, Aug 31 2023
Formula
a(n) is asymptotic to C*2^(n*(n+1)) where C = 0.288788095086602421278899721... = prod(k>=1, 1-1/2^k) (cf. A048651). - Benoit Cloitre, Apr 11 2003
a(n) = (6*a(n-1)^2*a(n-3) - 8*a(n-1)*a(n-2)^2) / (a(n-2)*a(n-3)). [From Putman Exam]. - Max Alekseyev, May 18 2007
a(n) = 2*A203305(n), n > 0. - Clark Kimberling, Jan 01 2012
From Max Alekseyev, Jun 09 2015: (Start)
a(n) = 2^n * A002884(n).
a(n) = 2^n * n! * A053601(n). (End)
From G. C. Greubel, Aug 31 2023: (Start)
a(n) = Product_{j=0..n-1} (2^(n+1) - 2^(j+1)).
a(n) = (-1)^n * 2^binomial(n+1,2) * QPochhammer(2,2,n). (End)
Comments