OFFSET
0,2
COMMENTS
Second binomial transform of A032443 with interpolated zeros.
a(n) is the total number of lattice points, taken over all Dyck n-paths (A000108), that (i) lie on or above ground level and (ii) lie on or directly below a peak. For example with n = 2, UUDD has 1 peak contributing 3 lattice points--(2, 0), (2, 1) and (2, 2) when the path starts at the origin--and UDUD has 2 peaks, each contributing 2 lattice points and so a(2) = 3 + 4 = 7. - David Callan, Jul 14 2006
Hankel transform is binomial(n + 2, 2). - Paul Barry, Dec 04 2007
Image of (-1)^n under the Riordan array ((1/2)(1/(1 - 4x) + 1/sqrt(1 - 4x)), c(x) - 1), c(x) the g.f. of A000108. - Paul Barry, Jun 15 2008
From Gus Wiseman, Jun 21 2021: (Start)
Also the even bisection of A116406 = number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(3) = 26 compositions are:
(6) (33) (114) (1122) (11112) (111111)
(42) (123) (1131) (11121)
(51) (132) (1221) (11211)
(213) (2112) (12111)
(222) (2121) (21111)
(231) (2211)
(312) (3111)
(321)
(411)
(End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
G.-S. Cheon, H. Kim, L. W. Shapiro, Mutation effects in ordered trees, arXiv:1410.1249 [math.CO], 2014
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
FORMULA
a(n) = Sum_{k=0..n} C(n, k)*2^(n-k-2)*(2^k + C(k, k/2))*(1 + (-1)^k).
From Paul Barry, Jun 15 2008: (Start)
G.f.: (1 - 4*x + (1 - 2*x)*sqrt(1 - 4*x))/(2*(1 - 4*x)^(3/2)).
a(n) = Sum_{k=0..n} ( Sum_{j=0..n} C(2*n, n-k-j)*(-1)^j ). (End)
a(n) = Sum_{k=0..n} C(2*n, n-k)*(1 + (-1)^k)/2. - Paul Barry, Aug 06 2009
From Paul Barry, Sep 07 2009: (Start)
a(n) = C(2*n-1, n-1) + (4^n + 3*0^n)/4.
Integral representation a(n) = (1/(2*pi))*(Integral_{x=0..4} x^n/sqrt(x(4 - x))) + (4^n + 0^n)/4. (End)
a(n) = Sum_{k=0..floor(n/2)} C(2*n, 2*k + (n mod 2)). - Mircea Merca, Jun 21 2011
Conjecture: n*a(n) + 2*(3 - 4*n)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 07 2012
Conjecture verified using the differential equation (16*x^2-8*x+1)*g'(x) + (8*x-2)*g(x)-2*x=0 satisfied by the G.f. - Robert Israel, Jul 27 2020
a(n) = Sum_{i=0..n} (sum_{j=0..n} binomial(n, i+j)*binomial(n, j-i)). - Yalcin Aktar, Jan 07 2013.
G.f.: (1 + (1 - 4*x)^(-1/2))^2 / 4. Convolution square of A088218. - Michael Somos, Dec 31 2013
0 = (1 + 2*n)*b(n) - (5 + 4*n)*b(n+1) + (4 + 2*n)*b(n+2) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
0 = b(n+3) * (2*b(n+2) - 7*b(n+1) + 5*b(n)) + b(n+2) * (-b(n+2) + 7*b(n+1) - 7*b(n)) + b(n+1) * (-b(n+1) + 2*b(n)) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
For n > 0, a(n) = 2^(2n-1) - A008549(n). - Gus Wiseman, Jun 27 2021
a(n) = [x^n] 1/((1-2*x) * (1-x)^(n-1)). - Seiichi Manyama, Apr 10 2024
EXAMPLE
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 99*x^4 + 382*x^5 + 1486*x^6 + 5812*x^7 + ...
MAPLE
seq(sum(binomial(2*n, 2*k+irem(n, 2)), k=0..floor((1/2)*n)), n=0..20)
seq(binomial(2*n-1, n)+4^(n-1)-(1/4)*0^n, n=0..20)
MATHEMATICA
a[ n_] := SeriesCoefficient[((1 + 1/Sqrt[1 - 4 x])/2)^2, {x, 0, n}] (* Michael Somos, Dec 31 2013 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]], {i, Length[y]}]; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], ats[#]>=0&]], {n, 0, 15, 2}] (* Gus Wiseman, Jun 21 2021 *)
CROSSREFS
The case of alternating sum = 0 is A001700.
The case of alternating sum < 0 is A008549.
This is the even bisection of A116406.
The restriction to reversed partitions is A344611.
A124754 gives the alternating sum of standard compositions.
A316524 is the alternating sum of the prime indices of n.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Feb 13 2006
STATUS
approved