OFFSET
0,2
COMMENTS
For n > 0, a(n) = v(n+1)/v(n), where v = A203305 is the Vandermonde determinant of the first n of the numbers -2^j - 1; see the Mathematica section. - Clark Kimberling, Jan 01 2012
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
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.
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)
MAPLE
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 *)
PROG
(PARI) a(n)=if(n<0, 0, prod(k=1, n, 2^k-1)*2^((n^2+n)/2)) /* Michael Somos, May 09 2005 */
(Magma) [1] cat [(&*[2^(n+1) - 2^(j+1): j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Aug 31 2023
(SageMath) [product(2^(n+1) - 2^(k+1) for k in range(n)) for n in range(21)] # G. C. Greubel, Aug 31 2023
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
STATUS
approved