Displaying 1-10 of 23 results found.
a(n) = (1/n)*Sum_{i=0..n-1} C(n,i)*C(n,i+1)*4^i*5^(n-i), a(0) = 1.
+0
7
1, 5, 45, 505, 6345, 85405, 1204245, 17558705, 262577745, 4005148405, 62070886845, 974612606505, 15471084667545, 247876665109005, 4003225107031845, 65101209768055905, 1065128963164067745, 17520376884067071205, 289572455530026439245, 4806489064223483202905
COMMENTS
The Hankel transform of this sequence is 20^C(n+1,2). - Philippe Deléham, Oct 28 2007
FORMULA
G.f.: (1-z-sqrt(z^2-18*z+1))/(8*z).
a(n) = Sum_{k=0..n} A088617(n,k)*4^k. a(n) = Sum_{k=0..n} A060693(n,k)*4^(n-k). a(n) = Sum_{k=0..n} C(n+k, 2k)*4^k*C(k), C(n) given by A000108. a(0) = 1, a(n) = a(n-1) + 4*Sum_{k=0..n-1} a(k)*a(n-1-k). - Philippe Deléham, Oct 23 2007 Conjecture: (n+1)*a(n) + 9*(-2*n+1)*a(n-1) + (n-2)*a(n-2) = 0. - R. J. Mathar, May 23 2014 G.f.: 1/(1 - 5*x/(1 - 4*x/(1 - 5*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017 a(n) = hypergeom([-n, n + 1], [2], -4). - Peter Luschny, Jan 08 2018 a(n) ~ 5^(1/4) * phi^(6*n + 3) / (2^(5/2) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 21 2021
MATHEMATICA
a[n_] := Hypergeometric2F1[-n, n + 1, 2, -4];
CoefficientList[Series[(1-x-Sqrt[x^2-18*x+1])/(8*x), {x, 0, 50}], x] (* G. C. Greubel, Feb 10 2018 *)
PROG
(PARI) x='x+O('x^30); Vec((1-x-sqrt(x^2-18*x+1))/(8*x)) \\ G. C. Greubel, Feb 10 2018 (Magma) Q:=Rationals(); R<x>:=PowerSeriesRing(Q, 40); Coefficients(R!((1-x-Sqrt(x^2-18*x+1))/(8*x))) // G. C. Greubel, Feb 10 2018
Series reversion of x*(1-3*x)/(1-x).
+0
23
1, 2, 10, 62, 430, 3194, 24850, 199910, 1649350, 13879538, 118669210, 1027945934, 9002083870, 79568077034, 708911026210, 6359857112438, 57403123415350, 520895417047010, 4749381474135850, 43489017531266654, 399755692955359630, 3687437532852484442, 34121911117572911410
COMMENTS
In general, the series reversion of x(1-r*x)/(1-x) has g.f. (1+x-sqrt(1+2*(1-2*r)*x+x^2))/(2*r) and general term given by a(n)=(1/(n+1))sum{k=0..n, C(n+1,k)C(2n-k,n)(-1)^k*r^(n-k)}; a(n)=(1/(n+1))sum{k=0..n, C(n+1,k+1)C(n+k,k)(-1)^(n-k)*r^k}; a(n)=sum{k=0..n, (1/(k+1))*C(n,k)C(n+k,k)(-1)^(n-k)*r^k}; a(n)=sum{k=0..n, A088617(n,k)*(-1)^(n-k)*r^k}. The Hankel transform of this sequence is 6^C(n+1,2). - Philippe Deléham, Oct 29 2007 Number of Dyck n-paths with three colors of up (U,a,b) and one color of down (D) avoiding UD. - David Scambler, Jun 24 2013 This sequence is implied in the turbulence solutions of the incompressible Navier-Stokes equations in R^3. a(n) = numbers of realizable vorticity eddies in terms of initial conditions. - Fung Lam, Dec 31 2013 Conjugate sequence to this series is defined by series reversion of x(1+3*x)/(1+x), G.f.: ((x-1)-sqrt(1-10*x+ x^2))/(6*x). Conjugate sequence is the negation of this series except a(0). - Fung Lam, Jan 16 2014 Complete Chebyshev transform is G.f. = 3*F((1-x^2)/(1+x^2)), where F(x) is the g.f. of A107841. Real part of G.f. (= (1 - sqrt(3*x^4-2))/((1+x^2))) generates periodic sequence A056594. In general, for reversion of x*(1-r*x)/(1-x), r>=2, Real part of r*F((1-x^2)/(1+x^2)) (= (1 - sqrt(r*x^4 - r + 1))/(1+x^2)) generates A056594. - Fung Lam, Apr 29 2014 a(n) is the number of small Schröder n-paths with 2 types of up steps (i.e., lattice paths from (0,0) to (2n,0) using steps U1=U2=(1,1), F=(2,0), D=(1,-1), with no F steps on the x-axis). - Yu Hin Au, Dec 07 2019
FORMULA
G.f.: (1+x-sqrt(1-10x+x^2))/(6x).
a(n) = (1/(n+1))sum{k=0..n, C(n+1, k)C(2n-k, n)(-1)^k*3^(n-k)}.
a(n) = (1/(n+1))sum{k=0..n, C(n+1, k+1)C(n+k, k)(-1)^(n-k)*3^k}.
a(n) = sum{k=0..n, (1/(k+1))*C(n, k)C(n+k, k)(-1)^(n-k)*3^k}.
a(n) = sum{k=0..n, A088617(n, k)*(-1)^(n-k)*3^k}. G.f.: 1/(1-2x/(1-3x/(1-2x/(1-3x/(1-2x/(1-3x/(1-2x/(1-3x........ (continued fraction). - Paul Barry, Dec 15 2008 G.f.: 1/(1-2x/(1-x-2x/(1-x-2x/(1-x-2x/(1-x-2x/(1-... (continued fraction).
G.f.: 1/(1-2x-6x^2/(1-5x-6x^2/(1-5x-6x^2/(1-5x-6x^2/(1-... (continued fraction). (End)
G.f.: 1/(1+x-3x/(1+x-3x/(1+x-3x/(1+x-3x/(1+x-3x/(1+... (continued fraction). - Paul Barry, Mar 18 2011 D-finite with recurrence: (n+1)*a(n) = 5*(2*n-1)*a(n-1) - (n-2)*a(n-2). - Vaclav Kotesovec, Oct 17 2012 a(n) ~ sqrt(12+5*sqrt(6))*(5+2*sqrt(6))^n/(6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012 a(n+1) is the coefficient of x^(n+1) in 2*sum{j,1,n}((sum{k,1,n}a(k)x^k)^(j+1)), a(1)=1 with offset by 1. - Fung Lam, Dec 31 2013 The series reversion of x*(1 - r*x)/(1 - x) is D-finite with the general recurrence n*a(n) - (2*r-1)*(2*n-3)*a(n-1) + (n-3)*a(n-2) = 0 and with initial values a(1) = 1, a(2) = r-1, a(3) = (2*r-1)*(r-1). This sequence uses r=3, cf. crossrefs. - Georg Fischer, Sep 14 2024
MAPLE
seq(simplify((-1)^n*hypergeom([-n, n + 1], [2], 3)), n=0..10); # Georg Fischer, Sep 14 2024 (from Peter Luschny's formula in A131763, with last parameter r=3)
MATHEMATICA
CoefficientList[Series[(1+x-Sqrt[1-10*x+x^2])/(6*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)
PROG
(PARI) x='x+O('x^66); Vec(serreverse(x*(1-3*x)/(1-x))) \\ Joerg Arndt, May 15 2013
G.f. satisfies A(x) = 1 + x*(A(x)^2 + 2*A(x)^3).
+0
5
1, 3, 24, 255, 3102, 40854, 566934, 8164263, 120864390, 1827982362, 28122626760, 438720097638, 6923868098820, 110346550539780, 1773394661610258, 28707809007278775, 467677404522668742, 7661583171651546786, 126137791939032756960, 2085923447593966281378
FORMULA
Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
Recurrence: 4*n*(2*n+1)*(19*n-26)*a(n) = (2717*n^3 - 6435*n^2 + 4342*n - 840)*a(n-1) + 2*(n-2)*(2*n-3)*(19*n-7)*a(n-2). - Vaclav Kotesovec, Dec 28 2013 a(n) ~ (3/19)^(1/4) * (5+sqrt(57)) * ((143 + 19*sqrt(57))/16)^n / (16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 28 2013 a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * 2^(n-k) * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)
a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - Peter Bala, Sep 08 2024
EXAMPLE
G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...
A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...
MATHEMATICA
CoefficientList[1/x*InverseSeries[Series[4*x^2/(1-x-Sqrt[1-10*x+x^2]), {x, 0, 20}], x], x] (* Vaclav Kotesovec, Dec 28 2013 *)
PROG
(PARI) /* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */
{a(n)=my(A=1); for(i=1, n, A=1+x*(A^2+2*A^3) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
(PARI) /* Formula using Series Reversion: */
{a(n)=my(A=1, G=(1-x-sqrt(1-10*x+x^2+x^3*O(x^n)))/(4*x)); A=(1/x)*serreverse(x/G); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
(PARI) a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 26 2020 (PARI) a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 26 2020
Triangle read by rows: T(n,k) = C(n+k,n)*C(n,k)/(k+1), for n >= 0, k = 0..n.
+0
38
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
COMMENTS
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
REFERENCES
Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
FORMULA
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
EXAMPLE
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;
MAPLE
R := n -> simplify(hypergeom([-n, n + 1], [2], -x)):
Trow := n -> seq(coeff(R(n, x), x, k), k = 0..n):
MATHEMATICA
Table[Binomial[n+k, n] Binomial[n, k]/(k+1), {n, 0, 10}, {k, 0, n}]//Flatten (* Michael De Vlieger, Aug 10 2017 *)
PROG
(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
Triangle related to guillotine partitions of a k-dimensional box by n hyperplanes.
+0
2
1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 15, 22, 1, 1, 5, 28, 93, 90, 1, 1, 6, 45, 244, 645, 394, 1, 1, 7, 66, 505, 2380, 4791, 1806, 1, 1, 8, 91, 906, 6345, 24868, 37275, 8558, 1, 1, 9, 120, 1477, 13926, 85405, 272188, 299865, 41586, 1, 1, 10, 153, 2248, 26845, 229326, 1204245, 3080596, 2474025, 206098, 1
COMMENTS
Row sums are A107703. Transpose of square array A103209, read by antidiagonals.
FORMULA
Number triangle T(n, k)=if(k<=n, sum{j=0..k, C(k+j, 2j)(n-k)^j*C(j)}, 0), C(n) given by A000108.
EXAMPLE
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 3, 6, 1;
1, 4, 15, 22, 1;
1, 5, 28, 93, 90, 1;
1, 6, 45, 244, 645, 394, 1;
1, 7, 66, 505, 2380, 4791, 1806, 1;
1, 8, 91, 906, 6345, 24868, 37275, 8558, 1;
...
PROG
(PARI) T(n, k) = sum(j=0, k, (n-k)^j*binomial(k+j, 2*j)*binomial(2*j, j)/(j+1)); \\ Seiichi Manyama, Oct 02 2023
Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Sum_{j=0..n} 2^j * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
+0
8
1, 1, 3, 1, 3, 6, 1, 3, 15, 12, 1, 3, 24, 93, 24, 1, 3, 33, 255, 645, 48, 1, 3, 42, 498, 3102, 4791, 96, 1, 3, 51, 822, 8691, 40854, 37275, 192, 1, 3, 60, 1227, 18708, 164937, 566934, 299865, 384, 1, 3, 69, 1713, 34449, 464115, 3305868, 8164263, 2474025, 768
FORMULA
G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k * (1 + 2 * A_k(x)).
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} 2^(n-j) * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = (1/n) * Sum_{j=0..n-1} (-1)^j * 3^(n-j) * binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0.
T(n,k) = (1/n) * Sum_{j=1..n} 3^j * 2^(n-j) * binomial(n,j) * binomial(k*n,j-1) for n > 0. (End)
EXAMPLE
Square array begins:
1, 1, 1, 1, 1, 1, ...
3, 3, 3, 3, 3, 3, ...
6, 15, 24, 33, 42, 51, ...
12, 93, 255, 498, 822, 1227, ...
24, 645, 3102, 8691, 18708, 34449, ...
48, 4791, 40854, 164937, 464115, 1055838, ...
MATHEMATICA
T[n_, k_] := Sum[2^j * Binomial[n, j] * Binomial[k*n + j + 1, n]/(k*n + j + 1), {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 27 2020 *)
PROG
(PARI) T(n, k) = sum(j=0, n, 2^j*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));
(PARI) T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k*(1+2*A)); polcoeff(A, n);
(PARI) T(n, k) = sum(j=0, n, 2^(n-j)*binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1);
G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - 2 * x * A(x)^2)).
+0
6
1, 3, 19, 169, 1753, 19795, 236035, 2923857, 37256881, 485202307, 6429346899, 86405569657, 1174917167881, 16134949855251, 223460304878467, 3117521211476641, 43771643214792033, 618045740600046211, 8770377489446217235, 125013010654218317385, 1789104455068153153849
FORMULA
a(n) = 1 + 2 * Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) = Sum_{k=0..n} binomial(n+k,n-k) * 2^k * binomial(3*k,k) / (2*k+1).
a(n) = hypergeom([1/3, 2/3, -n, n + 1], [1/2, 1, 3/2], -(3/2)^3). - Peter Luschny, Nov 12 2021 a(n) ~ sqrt(315 + 31*sqrt(105)) * (31 + 3*sqrt(105))^n / (9 * sqrt(Pi) * 2^(2*n + 5/2) * n^(3/2)). - Vaclav Kotesovec, Nov 13 2021
MATHEMATICA
nmax = 20; A[_] = 0; Do[A[x_] = 1/((1 - x) (1 - 2 x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[n_] := a[n] = 1 + 2 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]
Table[Sum[Binomial[n + k, n - k] 2^k Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 20}]
a[n_] := HypergeometricPFQ[{1/3, 2/3, -n, n + 1}, {1/2, 1, 3/2}, -(3/2)^3];
G.f. satisfies A(x) = 1 + x*A(x)*(1 + 2*A(x)^3).
+0
3
1, 3, 27, 351, 5319, 87885, 1535517, 27898101, 521740197, 9977087439, 194191054263, 3834392341779, 76619557946475, 1546479815079321, 31482877148802873, 645689728734541929, 13328555370318744777, 276704344407952939131, 5773556701375333682355
FORMULA
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(n+3*k+1,n) / (n+3*k+1).
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n) +(-458*n^3 +201*n^2 +401*n -216)*a(n-1) +3*(-1105*n^3 +6549*n^2 -11384*n +5796)*a(n-2) +18*(-262*n^3 +2877*n^2 -10295*n +12006)*a(n-3) +27*(n-4)*(31*n^2 -314*n +735)*a(n-4) -81*(10*n-51) *(n-4)*(n-5)*a(n-5) +243*(n-5)*(n-6) *(n-4)*a(n-6)=0. - R. J. Mathar, Jul 25 2023
MAPLE
add(2^k* binomial(n, k) * binomial(n+3*k+1, n) / (n+3*k+1), k=0..n) ;
end proc:
PROG
(PARI) a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n+3*k+1, n)/(n+3*k+1));
G.f. A(x) satisfies A(x) = (1 + 2 * x * A(x)^3) / (1 - x).
+0
4
1, 3, 21, 201, 2217, 26535, 335001, 4391553, 59203137, 815580507, 11430639165, 162470033625, 2336381642649, 33930648153615, 496935405133617, 7331179445170689, 108846406625097729, 1625145134034548019, 24385673680861258533, 367546405595389076649, 5561980053932228243529
FORMULA
a(0) = 1; a(n) = a(n-1) + 2 * Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) ~ sqrt(-50 + 30*sqrt(3) + (22 - 12*sqrt(3))*(2*(sqrt(3) - 1))^(1/3) + (2*(sqrt(3) - 1))^(2/3)*(-11 + 7*sqrt(3)))/(4*sqrt(3*Pi)*(-1 + sqrt(3))^(3/2) * n^(3/2) * (1 + (3*(-1 + sqrt(3))^(1/3))/2^(2/3) - 3/(2*(-1 + sqrt(3)))^(1/3))^n). - Vaclav Kotesovec, Nov 04 2021 a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(n+2*k+1,n) / (n+2*k+1). - Seiichi Manyama, Jul 24 2023
MATHEMATICA
nmax = 20; A[_] = 0; Do[A[x_] = (1 + 2 x A[x]^3)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = a[n - 1] + 2 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]
PROG
(PARI) a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n+2*k+1, n)/(n+2*k+1)); \\ Seiichi Manyama, Jul 24 2023
Triangle of Narayana ( A001263) with 0 <= k <= n, read by rows.
+0
41
1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 6, 1, 0, 1, 10, 20, 10, 1, 0, 1, 15, 50, 50, 15, 1, 0, 1, 21, 105, 175, 105, 21, 1, 0, 1, 28, 196, 490, 490, 196, 28, 1, 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 0, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 0, 1, 55, 825, 4950, 13860
COMMENTS
Number of Dyck n-paths with exactly k peaks. - Peter Luschny, May 10 2014
FORMULA
Triangle T(n, k), read by rows, given by [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. T(0, 0) = 1, T(n, 0) = 0 for n>0, T(n, k) = C(n-1, k-1)*C(n, k-1)/k for k>0. Coefficient array of the polynomials P(n,x) = x^n*2F1(-n,-n+1;2;1/x).
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*C(2n-j,j)*C(j,k)* A000108(n-j). (End) T(n, k) = C(n,n-k)*C(n-1,n-k)/(n-k+1). - Peter Luschny, May 10 2014 E.g.f.: 1+Integral((sqrt(t)*exp((1+t)*x)*BesselI(1,2*sqrt(t)*x))/x dx). - Peter Luschny, Oct 30 2014 T(n, k) = [x^k] (((2*n - 1)*(1 + x)*p(n-1, x) - (n - 2)*(x - 1)^2*p(n-2, x))/(n + 1)) with p(0, x) = 1 and p(1, x) = x. - Peter Luschny, Apr 26 2022 Recursion based on rows (see the Python program):
T(n, k) = (((B(k) + B(k-1))*(2*n - 1) - (A(k) - 2*A(k-1) + A(k-2))*(n-2))/(n+1)), where A(k) = T(n-2, k) and B(k) = T(n-1, k), for n >= 3. # Peter Luschny, May 02 2022
EXAMPLE
Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 1, 3, 1;
[4] 0, 1, 6, 6, 1;
[5] 0, 1, 10, 20, 10, 1;
[6] 0, 1, 15, 50, 50, 15, 1;
[7] 0, 1, 21, 105, 175, 105, 21, 1;
[8] 0, 1, 28, 196, 490, 490, 196, 28, 1;
[9] 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
MAPLE
A090181 := (n, k) -> binomial(n, n-k)*binomial(n-1, n-k)/(n-k+1): seq(print( seq( A090181(n, k), k=0..n)), n=0..5); # Peter Luschny, May 10 2014
# Alternatively:
egf := 1+int((sqrt(t)*exp((1+t)*x)*BesselI(1, 2*sqrt(t)*x))/x, x);
s := n -> n!*coeff(series(egf, x, n+2), x, n); seq(print(seq(coeff(s(n), t, j), j=0..n)), n=0..9); # Peter Luschny, Oct 30 2014
MATHEMATICA
Flatten[Table[Sum[(-1)^(j-k) * Binomial[2n-j, j] * Binomial[j, k] * CatalanNumber[n-j], {j, 0, n}], {n, 0, 11}, {k, 0, n}]] (* Indranil Ghosh, Mar 05 2017 *) p[0, _] := 1; p[1, x_] := x; p[n_, x_] := ((2 n - 1) (1 + x) p[n - 1, x] - (n - 2) (x - 1)^2 p[n - 2, x]) / (n + 1);
Table[CoefficientList[p[n, x], x], {n, 0, 9}] // TableForm (* Peter Luschny, Apr 26 2022 *)
PROG
(Sage)
U = [0]*(n+1)
for d in DyckWords(n):
U[d.number_of_peaks()] +=1
return U
(Python) from functools import cache
@cache
def Trow(n):
if n == 0: return [1]
if n == 1: return [0, 1]
if n == 2: return [0, 1, 1]
A = Trow(n - 2) + [0, 0]
B = Trow(n - 1) + [1]
for k in range(n - 1, 1, -1):
B[k] = (((B[k] + B[k - 1]) * (2 * n - 1)
- (A[k] - 2 * A[k - 1] + A[k - 2]) * (n - 2)) // (n + 1))
return B
(PARI)
c(n) = binomial(2*n, n)/ (n+1);
tabl(nn) = {for(n=0, nn, for(k=0, n, print1(sum(j=0, n, (-1)^(j-k) * binomial(2*n-j, j) * binomial(j, k) * c(n-j)), ", "); ); print(); ); };
(Magma) [[(&+[(-1)^(j-k)*Binomial(2*n-j, j)*Binomial(j, k)*Binomial(2*n-2*j, n-j)/(n-j+1): j in [0..n]]): k in [0..n]]: n in [0..10]];
CROSSREFS
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n) for x=0,1,2,3,4,5,6,7,8,9. - Philippe Deléham, Aug 10 2006 Sum_{k=0..n} x^(n-k)*T(n,k) = A090192(n+1), A000012(n), 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, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. - Philippe Deléham, Oct 21 2006 Sum_{k=0..n} T(n,k)*x^k*(x-1)^(n-k) = A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, respectively. - Philippe Deléham, Oct 20 2007
Search completed in 0.357 seconds
|