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A007564
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Shifts left when INVERT transform applied thrice.
(Formerly M3556)
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28
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1, 1, 4, 19, 100, 562, 3304, 20071, 124996, 793774, 5120632, 33463102, 221060008, 1473830308, 9904186192, 67015401391, 456192667396, 3122028222934, 21467769499864, 148246598341018, 1027656663676600, 7148588698592956, 49884553176689584
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OFFSET
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0,3
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COMMENTS
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More generally, coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+2)*x+1))/(2*m*x) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k) where N(n,k) = (1/n)*binomial(n,k)*binomial(n,k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
If y = x*A(x) then 3*y^2 - (1+2*x)*y + x = 0 and x = y*(1-3*y)/(1-2*y). - Michael Somos, Sep 28 2003
The sequence 0,1,4,19,... with g.f. (1-4*x-sqrt(1-8*x+4*x^2))/(6*x) and has a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k)*C(k)*4^(n-1-2*k)*3^k. a(n+1) = Sum_{k=0..floor(n/2)} binomial(n,2*k)*C(k)*4^(n-2*k)*3^k counts Motzkin paths of length n in which the level steps have 4 colors and the up steps have 3. It is the binomial transform of A107264 and corresponds to the series reversion of x/(1+4*x+3*x^2). - Paul Barry, May 18 2005
The Hankel transform of this sequence is 3^binomial(n+1,2). - Philippe Deléham, Oct 29 2007
a(n) is the number of Schroder paths of semilength n in which there are no (2,0)-steps at level 0 and at a higher level they come in 2 colors. Example: a(2)=4 because we have UDUD, UUDD, UBD, and URD, where U=(1,1), D=(1,-1), while B (R) is a blue (red) (2,0)-step. - Emeric Deutsch, May 02 2011
a(n) is the number of Schroder paths of semilength n-1 in which the (2,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=19 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 3^2 = 9 paths of shape HH, 3 paths of shape HUD, 3 paths of shape UDH, 2 paths of shape UHD, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) = number of (left) planted binary trees with n edges in which each vertex has a designated favorite neighbor. Planted binary trees are counted by the Catalan numbers A000108.
Example: for n=2, there are 2 planted binary trees: edges LL and LR from the root (L=left, R=right). Each has just one vertex with 2 neighbors, and so a(2)=4.
Proof outline: each vertex has 1,2 or 3 neighbors. Let X (resp. Y) denote the number of vertices with 2 (resp. 3) neighbors. Then X + 2Y = n - 1 (split the non-root edges into pairs with a common parent vertex and singletons). Thus the number of choices for designating favorite neighbors is 2^X * 3^Y = 2^(n-1)(3/4)^Y. The distribution for Y is known because, under the rotation correspondence, a.k.a. the deBruijn-Morselt bijection, vertices with 2 children in an n-edge planted binary tree correspond to DDUs in a Dyck path, and DDUs have the Touchard distribution (A091894) with gf F(x,y) = (1-2x+2xy - sqrt(1-4x+4x^2-4x^2 y))/(2xy). The desired g.f., Sum_{n>=1} a(n)*x^n, is therefore 1/2*(F(2x,3/4)-1). (End)
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
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FORMULA
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a(0)=1; for n>=1, a(n) = Sum_{k=0..n} 3^k*N(n,k) where N(n,k) = (1/n)*binomial(n, k)*binomial(n, k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
With offset 1: a(1) = 1, a(n) = -2*a(n-1) + 3*Sum_{i=1..n-1} a(i)*a(n-i)). - Benoit Cloitre, Mar 16 2004
D-finite with recurrence a(n) = (4*(2n-1)*a(n-1) - 4*(n-2)*a(n-2)) / (n+1) for n>=2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
G.f.: 1/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x........ (continued fraction).
The g.f. of a(n+1) is 1/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2.... (continued fraction). (End)
a(0) = 1, for n>=1, 3a(n) = A047891(n). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 3
1, 1, 1, 1
3, 3, 3, 3, 3
1, 1, 1, 1, 1, 1
...
G.f.: A(x)= (1+2*x-sqrt(1-8*x+4*x^2))/(6*x)= 1/G(0); G(k)= 1 + 2*x - 3*x/G(k+1); (continued fraction, 1-step ). - Sergei N. Gladkovskii, Jan 05 2012
a(n) ~ sqrt(6+4*sqrt(3))*(4+2*sqrt(3))^n/(6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
a(n) = 2^n/sqrt(3)*LegendreP(n,-1,2) for n >= 1, where LegendreP is the associated Legendre function of the first kind, in Maple's notation. - Robert Israel, Mar 24 2015
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EXAMPLE
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G.f. = 1 + x + 4*x^2 + 19*x^3 + 100*x^4 + 562*x^5 + 3304*x^6 + 20071*x^7 + 124996*x^8 + ...
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MAPLE
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A007564_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w] := a[w-1]+3*add(a[j]*a[w-j-1], j=1..w-1) od;
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MATHEMATICA
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a[0]=1; a[1]=1; a[n_]/; n>=2 := a[n] = a[n-1] + 3 Sum[a[k-1]a[n-k], {k, n-1}] ; Table[a[n], {n, 0, 10}] (* David Callan, Aug 25 2009 *)
Table[(2^n (LegendreP[n+1, 2] - LegendreP[n-1, 2]) + 2 KroneckerDelta[n])/(6n+3), {n, 0, 20}] (* Vladimir Reshetnikov, Nov 01 2015 *)
CoefficientList[Series[(1+2x-Sqrt[1-8x+4x^2])/(6x), {x, 0, 30}], x] (* Harvey P. Dale, Feb 07 2016 *)
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PROG
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(PARI) {a(n) = if( n<1, n==0, sum( k=0, n, 3^k * binomial( n, k) * binomial( n, k+1)) / n)} /* Michael Somos, Sep 28 2003 */
(PARI) {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x * (1 - 3*x) / (1 - 2*x) + x * O(x^n)), n))} /* Michael Somos, Sep 28 2003 */
(PARI) a(n) = (2^n*(pollegendre(n+1, 2)-pollegendre(n-1, 2)) + 2*(n==0))/(6*n+3); \\ Michel Marcus, Nov 02 2015
(PARI) x='x+O('x^100); Vec((1+2*x-sqrt(1-8*x+4*x^2))/(6*x)) \\ Altug Alkan, Nov 02 2015
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CROSSREFS
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KEYWORD
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nonn,nice,eigen
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AUTHOR
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STATUS
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approved
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