

A093387


a(n) = 2^(n1)  binomial(n, floor(n/2)).


8



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
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OFFSET

1,4


COMMENTS

Suppose n >= 3. Let e_1,...,e_n be n unitvectors 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 chose one vector, which is called the representative. The sequence gives the number of representatives.
Hankel transform is A127365.  Paul Barry, Jan 11 2007
Number of upsteps starting at level 0 in all dispersed Dyck paths of length n1 (that is, in Motzkin paths of length n1 with no (1,0)steps at positive heights).  Emeric Deutsch, May 30 2011


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Matthijs Coster, Sequences
Matthijs Coster, Statements and Representatives, 2004.
Vladimir Shevelev, A Mathar's conjecture, Seqfan, Nov 17 2017.


FORMULA

a(n) = A000079(n1)  A001405(n).
a(n+1) = Sum_{k=2..n} binomial(n, floor((nk)/2)).  Paul Barry, Jan 11 2007
a(2n) = 2*a(2n1).  Emeric Deutsch, May 30 2011
a(n+1) = Sum_{k>=0} k*A191310(n,k).  Emeric Deutsch, May 30 2011
G.f.: (1sqrt(14*z^2))^2/(4*z*(12*z)).  Emeric Deutsch, May 30 2011
Conjecture: (n+1)*a(n) + 2*(n1)*a(n1) + 4*(n+2)*a(n2) + 8*(n2)*a(n3) = 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*m1)*a(2*m1) + 8(2*m1)*a(2*m2) = (2*m+2)*(2*a(2*m) + A000108(m)  2*a(2*m))  4*(2*m1)*(2*a(2*m2) + A000108(m1)  2*a(2*m2)) = (2*m+2)*A000108(m)  4*(2*m1)*A000108(m1) = 0, since A000108(m) = binomial(2*m, m)/(m+1).  Vladimir Shevelev, Nov 17 2017


EXAMPLE

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


MAPLE

A093387:=n>2^(n1)binomial(n, floor(n/2)); seq(A093387(n), n=1..50); # Wesley Ivan Hurt, Dec 01 2013


MATHEMATICA

Table[2^(n  1)  Binomial[n, Floor[n/2]], {n, 50}] (* Wesley Ivan Hurt, Dec 01 2013 *)


PROG

(PARI) a(n) = 2^(n1)  binomial(n, n\2); \\ Michel Marcus, Aug 13 2013


CROSSREFS

Cf. A000079, A001405, A000108, A127365, A191310.
Sequence in context: A183467 A057582 A094779 * A324408 A229487 A195166
Adjacent sequences: A093384 A093385 A093386 * A093388 A093389 A093390


KEYWORD

nonn


AUTHOR

Matthijs Coster, Apr 29 2004


EXTENSIONS

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


STATUS

approved



