A028361 Number of totally isotropic spaces of index n in orthogonal geometry of dimension 2n.
1, 2, 6, 30, 270, 4590, 151470, 9845550, 1270075950, 326409519150, 167448083323950, 171634285407048750, 351678650799042888750, 1440827432323678715208750, 11804699153027899713705288750, 193419995622362136809061156168750, 6338179836549184861096125026493768750
Offset: 0
References
- W. Cary Huffman and Vera Pless, Fundamentals of Error Correcting Codes, Cambridge University Press, 2003, Page 366. - Nathan J. Russell, Mar 04 2016
Links
- T. D. Noe, Table of n, a(n) for n=0..50
- C. Bachoc and P. Gaborit, On extremal additive F_4 codes of length 10 to 18, J. Théorie Nombres Bordeaux, 12 (2000), 255-271.
- Julien Clément and Antoine Genitrini, Binary Decision Diagrams: from Tree Compaction to Sampling, arXiv:1907.06743 [cs.DS], 2019.
- John P. D'Angelo, Counting Tournament Brackets, J. Int. Seq., Vol. 25 (2022), Article 22.6.8.
- Duane W. DeTemple, Carlyle circles and the Lemoine simplicity of polygon constructions, The American Mathematical Monthly 98.2 (1991): 97-108.
- Davide Gaiotto and Justin Kulp, Orbifold groupoids, arXiv:2008.05960 [hep-th], 2020, page 42.
Programs
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Magma
[1] cat [ (&*[1+2^j: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Jun 06 2020
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Maple
seq( mul((1+2^j), j=0..n-1), n = 0..20); # G. C. Greubel, Jun 06 2020
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Mathematica
Table[QPochhammer[-1, 2, n], {n, 0, 15}] (* Arkadiusz Wesolowski, Oct 29 2012 *) Table[Product[2^i + 1, {i, 0, n/2 - 2}], {n, 2, 32, 2}] (* Nathan J. Russell, Mar 04 2016 *) Table[Product[2^i + 1, {i, 0, n - 1}], {n, 0, 15}] (* Nathan J. Russell, Mar 04 2016 *) FoldList[Times,1,2^Range[0,20]+1] (* Harvey P. Dale, Apr 11 2016 *)
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PARI
{a(n) = prod(k=0, n-1, 2^k + 1)};
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PARI
{a(n)=polcoeff(sum(m=0,n,2^(m*(m-1)/2)*x^m/prod(k=0,m,1-2^k*x+x*O(x^n))),n)} /* Paul D. Hanna, May 02 2012 */
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Python
for n in range(2,50,2): product = 1 for i in range(0,n//2-2 + 1): product *= (2**i+1) print(product) # Nathan J. Russell, Mar 01 2016
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Sage
[product( 1+2^j for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Jun 06 2020
Formula
a(n) = Product_{i=0..n-1} ( 2^i + 1 ).
Asymptotic to C*2^(n*(n-1)/2) where C = A081845 = 4.76846205806274344829979857... = Product_{k>=0} (1 + 1/2^k). - Benoit Cloitre, Apr 09 2003
It appears that a(n) = 2^((1/2)*(n - 1)*n) * Product_{k>=0} (1 + 1/(2^k)) / Product_{k>=0} (1 + 1/(2^(n + k))). - Peter Moxey (pmoxey(AT)live.com), Mar 21 2010
G.f.: Sum_{n>=0} 2^(n*(n-1)/2) * x^n / Product_{k=0..n} (1 - 2^k*x). - Paul D. Hanna, May 02 2012
a(n) = (a(n-2)^3 + a(n-1) * a(n-3) * (a(n-1) - 2 * a(n-2))) * a(n-1) / (a(n-2)^2 * (a(n-2) - a(n-3))) if n>2. - Michael Somos, Aug 21 2012
0 = a(n)*(+a(n+1) + a(n+2)) + a(n+1)*(-2*a(n+1)) for all n>=0. - Michael Somos, Oct 10 2014
Sum_{k=0..n} 2^k/a(k) = 3-2/a(n) and Sum_{k=0..n} 4^k/a(k) = 9-(4*(1+2^n))/a(n) for n >= 0. - Werner Schulte, Dec 25 2016
G.f. A(x) satisfies: A(x) = (1 + x * A(2*x)) / (1 - x). - Ilya Gutkovskiy, Jun 06 2020
a(n) = Sum_{k=0..n} q_binomial(n, k, q=2) * 2^(k*(k-1)/2). - Jingjie Yang, Jul 30 2025
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