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A088617
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Triangle read by rows: T(n,k) = C(n+k,n)*C(n,k)/(k+1), for n >= 0, k = 0..n.
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38
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1, 1, 1, 1, 3, 2, 1, 6, 10, 5, 1, 10, 30, 35, 14, 1, 15, 70, 140, 126, 42, 1, 21, 140, 420, 630, 462, 132, 1, 28, 252, 1050, 2310, 2772, 1716, 429, 1, 36, 420, 2310, 6930, 12012, 12012, 6435, 1430, 1, 45, 660, 4620, 18018, 42042, 60060, 51480, 24310, 4862
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OFFSET
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0,5
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COMMENTS
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Row sums: A006318 (Schroeder numbers). Essentially same as triangle A060693 transposed.
T(n,k) is number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k U's. E.g., T(2,1)=3 because we have UHD, UDH and HUD. - Emeric Deutsch, Dec 06 2003
Conjecture: The expected number of U's in a Schroeder n-path is asymptotically Sqrt[1/2]*n for large n. - David Callan, Jul 25 2008
T(n, k) is also the number of order-preserving and order-decreasing partial transformations (of an n-chain) of width k (width(alpha) = |Dom(alpha)|). - Abdullahi Umar, Oct 02 2008
The antidiagonals of this lower triangular matrix are the rows of A055151. - Tom Copeland, Jun 17 2015
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REFERENCES
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Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
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LINKS
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FORMULA
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Triangle T(n, k) read by rows; given by [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] DELTA [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} T(n, k)*x^k*(1-x)^(n-k) = A000108(n), A001003(n), A007564(n), A059231(n), A078009(n), A078018(n), A081178(n), A082147(n), A082181(n), A082148(n), A082173(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. - Philippe Deléham, Aug 18 2005
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A107841(n), A080243(n), A000007(n), A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Oct 18 2007
O.g.f. (with initial 1 excluded) is the series reversion with respect to x of (1-t*x)*x/(1+x). Cf. A062991 and A089434. - Peter Bala, Jul 31 2012
G.f.: 1 + (1 - x - T(0))/y, where T(k) = 1 - x*(1+y)/( 1 - x*y/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 03 2013
O.g.f. A(x,t) = ( 1 - x - sqrt((1 - x)^2 - 4*x*t) )/(2*x*t) = 1 + (1 + t)*x + (1 + 3*t + 2*t^2)*x^2 + ....
1 + x*(dA(x,t)/dx)/A(x,t) = 1 + (1 + t)*x + (1 + 4*t + 3*t^2)*x^2 + ... is the o.g.f. for A123160.
For n >= 1, the n-th row polynomial equals (1 + t)/(n+1)*Jacobi_P(n-1,1,1,2*t+1). Removing a factor of 1 + t from the row polynomials gives the row polynomials of A033282. (End)
The o.g.f. G(x,t) = {1 - (2t+1) x - sqrt[1 - (2t+1) 2x + x^2]}/2x = (t + t^2) x + (t + 3t^2 + 2t^3) x^2 + (t + 6t^2 + 10t^3 + 5t^3) x^3 + ... generating shifted rows of this entry, excluding the first, was given in my 2008 formulas for A033282 with an o.g.f. f1(x,t) = G(x,t)/(1+t) for A033282. Simple transformations presented there of f1(x,t) are related to A060693 and A001263, the Narayana numbers. See also A086810.
The inverse of G(x,t) is essentially given in A033282 by x1, the inverse of f1(x,t): Ginv(x,t) = x [1/(t+x) - 1/(1+t+x)] = [((1+t) - t) / (t(1+t))] x - [((1+t)^2 - t^2) / (t(1+t))^2] x^2 + [((1+t)^3 - t^3) / (t(1+t))^3] x^3 - ... . The coefficients in t of Ginv(xt,t) are the o.g.f.s of the diagonals of the Pascal triangle A007318 with signed rows and an extra initial column of ones. The numerators give the row o.g.f.s of signed A074909.
(End)
T(n, k) = [x^k] hypergeom([-n, 1 + n], [2], -x). - Peter Luschny, Apr 26 2022
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EXAMPLE
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Triangle begins:
[0] 1;
[1] 1, 1;
[2] 1, 3, 2;
[3] 1, 6, 10, 5;
[4] 1, 10, 30, 35, 14;
[5] 1, 15, 70, 140, 126, 42;
[6] 1, 21, 140, 420, 630, 462, 132;
[7] 1, 28, 252, 1050, 2310, 2772, 1716, 429;
[8] 1, 36, 420, 2310, 6930, 12012, 12012, 6435, 1430;
[9] 1, 45, 660, 4620, 18018, 42042, 60060, 51480, 24310, 4862;
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MAPLE
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R := n -> simplify(hypergeom([-n, n + 1], [2], -x)):
Trow := n -> seq(coeff(R(n, x), x, k), k = 0..n):
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MATHEMATICA
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Table[Binomial[n+k, n] Binomial[n, k]/(k+1), {n, 0, 10}, {k, 0, n}]//Flatten (* Michael De Vlieger, Aug 10 2017 *)
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PROG
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(PARI) {T(n, k)= if(k+1, binomial(n+k, n)*binomial(n, k)/(k+1))}
(Magma) [[Binomial(n+k, n)*Binomial(n, k)/(k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jun 18 2015
(SageMath) flatten([[binomial(n+k, 2*k)*catalan_number(k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 22 2022
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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