OFFSET
0,2
COMMENTS
a(n) is the number of peaks at odd height in all Motzkin paths of length n+2. Example: a(2)=5 counts the peaks shown between parentheses in the 9 Motzkin paths of length 4: HHHH, HH(UD), H(UD)H, HUHD, (UD)HH, (UD)(UD), UHDH, UHHD and UUDD.
Binomial transform of 1,1,2,2,6,6,20,20,70,70...... (A000984 doubled). It would appear that the Hankel transform of this sequence is a signed version of A128055, with sign pattern given by s(n)=(2/3-sqrt(3)/3)cos(5*Pi*n/6)-sin(5*Pi*n/6)/3+(sqrt(3)/3+2/3)*cos(Pi*n/6)-sin(Pi*n/6)/3-cos(Pi*n/2)/3+sin(Pi*n/2)/3. - Paul Barry, Jan 03 2008
The subsequence of primes in this sequence of partial sums begins: 2, 5, 31, 223, 1723, no more through a(26). - Jonathan Vos Post, May 12 2010
Designate a triangle to have T(n,1) = T(n,n) = 1 and T(r,c) = T(r,c-1) + T(r-1,c-1) + T(r-2,c-1). The sum of the terms in row(n) = a(n+1). - J. M. Bergot, Apr 30 2013
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Moa Apagodu and Doron Zeilberger, Using the "Freshman's Dream" to Prove Combinatorial Congruences, arXiv:1606.03351 [math.CO], 2016. Also Amer. Math. Monthly. 124 (2017), 597-608.
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
Karl Dilcher and Larry Ericksen, Polynomials and algebraic curves related to certain binary and b-ary overpartitions, arXiv:2405.12024 [math.CO], 2024. See p. 14.
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
FORMULA
G.f.: 1/((1-z)*sqrt(1-2*z-3*z^2)).
a(n) = Sum_{0<=j<=i<=n} C(i, i-j)*C(j, i-j). - Benoit Cloitre, Oct 23 2004
a(n) = sum_{k=0..n} Sum_{j=0..n-k} C(k,j)C(n-k,j)C(2j,j). - Paul Barry, Jan 03 2008
Logarithm g.f. atan(x*M(x)), M(x) - o.g.f. for Motzkin numbers (A001006). - Vladimir Kruchinin_, Aug 11 2010
Conjecture: -n*a(n) +(3*n-1)*a(n-1) +(n-2)*a(n-2) +3*(1-n)*a(n-3)=0. - R. J. Mathar, Nov 09 2012 [Since A002426(n) = a(n) - a(n-1), this third-order recurrence follows easily from the second-order recurrence given in A002426. - Peter Bala, Oct 28 2015]
G.f.: G(0)/(1-x), where G(k)= 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
a(n) ~ 3^(n+3/2)/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 23 2013
a(n) = Sum_{k = 0..floor(n/2)} binomial(n + 1,2*k + 1) *binomial(2*k,k). Cf. A025178. - Peter Bala, Oct 28 2015
a(n) = (n+1)*hypergeom([1/2,(1-n)/2,-n/2],[1,3/2],4). - Peter Luschny, Oct 29 2015
MAPLE
ser:=series(1/(1-z)/sqrt(1-2*z-3*z^2), z=0, 32): 1, seq(coeff(ser, z^n), n=1..31);
a := n -> (n+1)*hypergeom([1/2, (1-n)/2, -n/2], [1, 3/2], 4):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Oct 29 2015
MATHEMATICA
Table[ Sum[ Binomial[n, k]*Binomial[k, n-k], {k, 0, n}], {n, 0, 26}] // Accumulate (* Jean-François Alcover, Jul 10 2013 *)
CoefficientList[Series[1/((1-x)*Sqrt[1-2*x-3*x^2]), {x, 0, 50}], x] (* G. C. Greubel, Dec 21 2017 *)
PROG
(PARI) a(n)=sum(i=0, n, sum(j=0, i, binomial(i, i-j)*binomial(j, i-j)))
(Haskell)
a097893 n = a097893_list !! n
a097893_list = scanl1 (+) a002426_list
-- Reinhard Zumkeller, Jan 22 2013
(PARI) vector(30, n, n--; sum(k=0, n\2, binomial(n+1, 2*k+1)* binomial(2*k, k))) \\ Altug Alkan, Oct 29 2015
(PARI) x='x+O('x^30); Vec(1/((1-x)*sqrt(1-2*x-3*x^2))) \\ G. C. Greubel, Dec 21 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Emeric Deutsch, Sep 03 2004
STATUS
approved