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.
%I A093387 #62 Dec 23 2024 14:53:42 %S A093387 0,0,1,2,6,12,29,58,130,260,562,1124,2380,4760,9949,19898,41226,82452, %T A093387 169766,339532,695860,1391720,2842226,5684452,11576916,23153832, %U A093387 47050564,94101128,190876696,381753392,773201629,1546403258,3128164186,6256328372,12642301534 %N A093387 a(n) = 2^(n-1) - binomial(n, floor(n/2)). %C A093387 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}. %C A093387 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. %C A093387 For each set of indistinguishable vectors we choose one vector, which is called the representative. a(n) is the number of representatives. %C A093387 Hankel transform is A127365. - _Paul Barry_, Jan 11 2007 %C A093387 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 %C A093387 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 (<s, v> = 0). - _Isaac Grosof_, Jan 16 2023 %H A093387 Vincenzo Librandi, <a href="/A093387/b093387.txt">Table of n, a(n) for n = 1..1000</a> %H A093387 Matthijs Coster, <a href="http://www.coster.demon.nl/sequences/index.html">Sequences</a> %H A093387 Matthijs Coster, <a href="/A093387/a093387.pdf">Statements and Representatives</a>, 2004. %H A093387 Vladimir Shevelev, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2017-November/018107.html">A Mathar's conjecture</a>, Seqfan, Nov 17 2017. %F A093387 a(n) = A000079(n-1) - A001405(n). %F A093387 a(n+1) = Sum_{k=2..n} binomial(n, floor((n-k)/2)). - _Paul Barry_, Jan 11 2007 %F A093387 a(2n) = 2*a(2n-1). - _Emeric Deutsch_, May 30 2011 %F A093387 a(n+1) = Sum_{k>=0} k*A191310(n,k). - _Emeric Deutsch_, May 30 2011 %F A093387 G.f.: (1-sqrt(1-4*z^2))^2/(4*z*(1-2*z)). - _Emeric Deutsch_, May 30 2011 %F A093387 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 %F A093387 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 %e A093387 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 %e A093387 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 %p A093387 A093387:=n->2^(n-1)-binomial(n, floor(n/2)); seq(A093387(n), n=1..50); # _Wesley Ivan Hurt_, Dec 01 2013 %t A093387 Table[2^(n - 1) - Binomial[n, Floor[n/2]], {n, 50}] (* _Wesley Ivan Hurt_, Dec 01 2013 *) %o A093387 (PARI) a(n) = 2^(n-1) - binomial(n, n\2); \\ _Michel Marcus_, Aug 13 2013 %Y A093387 Cf. A000079, A001405, A000108, A127365, A191310. %K A093387 nonn %O A093387 1,4 %A A093387 _Matthijs Coster_, Apr 29 2004 %E A093387 Offset corrected by _R. J. Mathar_, Jun 04 2011