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A059398
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Row sums of triangle in A059397.
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4
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1, 2, 6, 17, 51, 154, 473, 1464, 4568, 14332, 45187, 143024, 454217, 1446604, 4618576, 14777451, 47371177, 152110326, 489165277, 1575211177, 5078690936, 16392526502, 52963765321, 171282782902, 554393341371, 1795821017014
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OFFSET
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0,2
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COMMENTS
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Number of paths in the first quadrant from (0,0) to the line x=n, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0) (in other words, left factors of the paths in A128720). Example: a(2)=6 because we have hh, H, UD, hU, Uh and UU. Row sums of triangle in A132276. - Emeric Deutsch, Sep 03 2007
Row sums of the Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2) (A132276). - Emanuele Munarini, May 05 2011
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LINKS
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W. Klostermeyer et al., A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
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FORMULA
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a(n) = (1/2)*sum(binomial(2*k,k)*(-1)^(n-k+1)*sum(binomial(i+k-1,i)*binomial(i,n-k-i+1),i=0..n-k+1),k=0..n+1). - Emanuele Munarini, May 05 2011
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MAPLE
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g:=(1/2)*(sqrt((1+x-x^2)/(1-3*x-x^2))-1)/x: gser:=series(g, x=0, 30): seq(coeff(gser, x, n), n=0..25); # Emeric Deutsch, Sep 03 2007
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MATHEMATICA
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Table[Sum[Binomial[2k, k](-1)^(n-k+1)Sum[Binomial[i+k-1, i]Binomial[i, n-k-i+1], {i, 0, n-k+1}], {k, 0, n+1}]/2, {n, 0, 28}] (* Emanuele Munarini, May 05 2011 *)
With[{nn = 50}, CoefficientList[Series[(Sqrt[(1 + x - x^2)/(1 - 3*x - x^2)] - 1)/x/2, {x, 0, nn}], x]] (* G. C. Greubel, Jan 29 2018 *)
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PROG
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(Maxima) makelist(sum(binomial(2*k, k)*(-1)^(n-k+1)*sum(binomial(i+k-1, i)*binomial(i, n-k-i+1), i, 0, n-k+1), k, 0, n+1)/2, n, 0, 28); /* Emanuele Munarini, May 05 2011 */
(PARI) x='x+O('x^30); Vec((sqrt((1+x-x^2)/(1-3*x-x^2))-1)/x/2) \\ G. C. Greubel, Jan 29 2018
(Magma) Q:=Rationals(); R<x>:=PowerSeriesRing(Q, 40); Coefficients(R!(Sqrt((1+x-x^2)/(1-3*x-x^2))-1)/(2*x)) // G. C. Greubel, Jan 29 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001
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STATUS
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approved
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