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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A093387 a(n) = 2^(n-1) - binomial(n, floor(n/2)).

Original entry on oeis.org

0, 0, 1, 2, 6, 12, 29, 58, 130, 260, 562, 1124, 2380, 4760, 9949, 19898, 41226, 82452, 169766, 339532, 695860, 1391720, 2842226, 5684452, 11576916, 23153832, 47050564, 94101128, 190876696, 381753392, 773201629, 1546403258, 3128164186, 6256328372, 12642301534
Offset: 1

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Author

Matthijs Coster, Apr 29 2004

Keywords

Comments

Suppose n >= 3. Let e_1,...,e_n be n unit-vectors which generate Euclidean space R_n and let l_n = {x = sum a_i e_i | a_1 >= a_2 >= ... >= a_n >= 0 }. Consider the hypercube H_n with vertices h_1,...,h_{2^n} = {epsilon_1 e_1 + ... + epsilon_n e_n}.
For each element x in l_n we build 2^n "statements" by taking the inner product of x with h_i. We call a statement true if (x,h_i) > 0 and false if (x,h_i) < 0. Two vectors x and y are indistinguishable if all statements produced by x and y are equal.
For each set of indistinguishable vectors we choose one vector, which is called the representative. a(n) is the number of representatives.
Hankel transform is A127365. - Paul Barry, Jan 11 2007
Number of up-steps starting at level 0 in all dispersed Dyck paths of length n-1 (that is, in Motzkin paths of length n-1 with no (1,0)-steps at positive heights). - Emeric Deutsch, May 30 2011
Number of north-east paths of length n on the square lattice, starting with a north step, which ever visit a point strictly east of the main diagonal. Equivalently, the number of vectors v in [1,-1]^n with first coordinate v_1=1 for which there exists a vector s with strictly ordered, strictly positive entries (s_1 > s_2 > ... > s_n > 0) that is orthogonal to v ( = 0). - Isaac Grosof, Jan 16 2023

Examples

			a(5)=6 because, denoting U=(1,1), D=(1,-1), H=(1,0), in HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD we have 0+1+1+1+2+1=6 U steps starting at level 0. - _Emeric Deutsch_, May 30 2011
a(5)=6 because there are 6 north-east paths starting with N which visit a point strictly east of the main diagonal: NNEEE, NENEE, NEENN, NEENE, NEEEN, NEEEE. - _Isaac Grosof_, Jan 16 2023

Links

Crossrefs

Cf. A000079, A001405, A000108, A127365, A191310.

Programs

Formula

a(n) = A000079(n-1) - A001405(n).
a(n+1) = Sum_{k=2..n} binomial(n, floor((n-k)/2)). - Paul Barry, Jan 11 2007
a(2n) = 2*a(2n-1). - Emeric Deutsch, May 30 2011
a(n+1) = Sum_{k>=0} k*A191310(n,k). - Emeric Deutsch, May 30 2011
G.f.: (1-sqrt(1-4*z^2))^2/(4*z*(1-2*z)). - Emeric Deutsch, May 30 2011
Conjecture: (n+1)*a(n) + 2*(-n-1)*a(n-1) + 4*(-n+2)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
a(2*n+1) = 2*a(2*n) + A000108(n). Together with the first formula by Emeric Deutsch, we have a simple system of recursions. Using them, we can prove Mathar's conjecture. For example, let n be odd, n=2*m+1. By the left hand side of Mathar's conjecture, we have (2*m+2)*a(2*m+1) - 2*(2*m+2)*a(2*m) - 4*(2*m-1)*a(2*m-1) + 8(2*m-1)*a(2*m-2) = (2*m+2)*(2*a(2*m) + A000108(m) - 2*a(2*m)) - 4*(2*m-1)*(2*a(2*m-2) + A000108(m-1) - 2*a(2*m-2)) = (2*m+2)*A000108(m) - 4*(2*m-1)*A000108(m-1) = 0, since A000108(m) = binomial(2*m, m)/(m+1). - Vladimir Shevelev, Nov 17 2017

Extensions

Offset corrected by R. J. Mathar, Jun 04 2011
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