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A217730
Expansion of (1+2*x-x^3)/(1-4*x^2+2*x^4).
2
1, 2, 4, 7, 14, 24, 48, 82, 164, 280, 560, 956, 1912, 3264, 6528, 11144, 22288, 38048, 76096, 129904, 259808, 443520, 887040, 1514272, 3028544, 5170048, 10340096, 17651648, 35303296, 60266496, 120532992, 205762688, 411525376, 702517760, 1405035520, 2398545664, 4797091328, 8189147136, 16378294272, 27959497216
OFFSET
0,2
COMMENTS
In general, a(n,j,m) = Sum_{r=1..m} (2^n*(1-(-1)^r)*cos(Pi*r/(m+1))^n*cot(Pi*r/(2*(m+1)))*sin(j*Pi*r/(m+1)))/(m+1) gives the number of paths of length n starting at the j-th node on the path graph P_m. Here we have the case m=7 and j=3. - Herbert Kociemba, Sep 17 2020
LINKS
Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Math. 332 (2014), 45--54. MR3227977. See Fig. 5. - N. J. A. Sloane, Aug 04 2014
FORMULA
G.f.: (1+x)*(1+x-x^2)/(1-4*x^2+2*x^4).
a(n) = Sum_{k=0..n} A216232(n-k,k).
a(n) = 4*a(n-2) - 2*a(n-4) for n>=4, a(0)=1, a(1)=2, a(2)=4, a(3)=7.
a(2*n) = A007070(n), a(2*n-1) = a(2*n)/2 = A007070(n)/2.
a(n)*a(n+1)-a(n-1)*a(n+2) = (1-(-1)^n)*2^floor(n/2-1) for n>0. - Bruno Berselli, Mar 22 2013
a(n) = Sum_{r=1..7} (2^n*(1-(-1)^r)*cos(Pi*r/8)^n*cot(Pi*r/16)*sin(3*Pi*r/8))/8. - Herbert Kociemba, Sep 17 2020
MATHEMATICA
CoefficientList[Series[(1 + 2 x - x^3)/(1 - 4 x^2 + 2 x^4), {x, 0, 40}], x] (* Bruno Berselli, Mar 22 2013 *)
a[n_, j_, m_]:=Sum[(2^(n+1)Cos[Pi r/(m+1)]^n Cot[Pi r/(2(m+1))] Sin[j Pi r/(m+1)])/(m+1), {r, 1, m, 2}]
Table[a[n, 3, 7], {n, 0, 40}]//Round (* Herbert Kociemba, Sep 17 2020 *)
PROG
(Magma) m:=40; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x-x^3)/(1-4*x^2+2*x^4))); // Bruno Berselli, Mar 22 2013
(Maxima) makelist(coeff(taylor((1+2*x-x^3)/(1-4*x^2+2*x^4), x, 0, n), x, n), n, 0, 40); /* Bruno Berselli, Mar 22 2013 */
CROSSREFS
First differences are in A062113.
Sequence in context: A370636 A056180 A000076 * A360055 A347760 A218576
KEYWORD
nonn,easy
AUTHOR
Philippe Deléham, Mar 22 2013
STATUS
approved