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 A055217 a(n) = sum of the first n coefficients of (1+x+x^2)^n. 5
 1, 3, 10, 31, 96, 294, 897, 2727, 8272, 25048, 75747, 228826, 690691, 2083371, 6280650, 18925047, 57002616, 171633840, 516632307, 1554702516, 4677501237, 14069962041, 42314975352, 127240600050, 382555886571, 1150026301089 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..1000 FORMULA Binomial transform of A117186; g.f.: (1+x-sqrt(1-2x-3x^2))/(2x(1-2x-3x^2)); a(n)=(3^(n+1)+A002426(n+1))/2; - Paul Barry, Jan 20 2008 Logarithm g.f. log(1/(1-M(x))=sum(n>0, a(n)/n*x^n), M(x) - o.g.f Motzkin numbers (A001006). a(n)=sum(sum(binomial(n,j)*binomial(j,2*j-n-k),j,ceiling((n+k)/2),n),k,1,n), n>0. - Vladimir Kruchinin, Aug 11 2010 Conjecture: (n+1)*a(n) -(5*n+1)*a(n-1) +3*(n-1)*a(n-2) +9*(n-1)*a(n-3)=0. - R. J. Mathar, Nov 14 2011 a(n) = 3^n * 3/2 + O(3^n/sqrt(n)). - Charles R Greathouse IV, Dec 02 2015 From Peter Luschny, May 12 2016: (Start) a(n) = (3^(n+1) - hypergeom([-(n+1)/2, -n/2], [1], 4))/2. a(n) = (3^(n+1) - GegenbauerC(n+1,-n-1,-1/2))/2. (End) MAPLE a := n -> simplify((3^(n+1) - GegenbauerC(n+1, -n-1, -1/2))/2): seq(a(n), n=0..25); # Peter Luschny, May 12 2016 MATHEMATICA Total/@Table[Take[CoefficientList[Expand[(1+x+x^2)^n], x], n], {n, 30}] (* Harvey P. Dale, Aug 14 2011 *) PROG (Maxima) a(n):=sum(sum(binomial(n, j)*binomial(j, 2*j-n-k), j, ceiling((n+k)/2), n), k, 1, n); \\ Vladimir Kruchinin, Aug 11 2010 (Haskell) a055217 n = sum \$ take (n + 1) \$ a027907_row (n + 1) -- Reinhard Zumkeller, Jan 22 2013 CROSSREFS T(2n+1, n), array T as in A055216. Cf. A027914. Sequence in context: A237930 A192337 A106517 * A097472 A068094 A100058 Adjacent sequences:  A055214 A055215 A055216 * A055218 A055219 A055220 KEYWORD nonn AUTHOR Clark Kimberling, May 07 2000 EXTENSIONS New description from Paul D. Hanna, Oct 09 2003 STATUS approved

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Last modified April 1 04:03 EDT 2020. Contains 333155 sequences. (Running on oeis4.)