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 A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). 131

%I

%S 1,1,1,2,3,1,5,9,5,1,14,28,20,7,1,42,90,75,35,9,1,132,297,275,154,54,

%T 11,1,429,1001,1001,637,273,77,13,1,1430,3432,3640,2548,1260,440,104,

%U 15,1,4862,11934,13260,9996,5508,2244,663,135,17,1

%N Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

%C T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example : T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - _Philippe Deléham_, May 23 2005

%C The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - _Philippe Deléham_, May 26 2005

%C Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - _Philippe Deléham_, May 31 2005

%C Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - _Emeric Deutsch_, May 06 2006

%C Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - _Philippe Deléham_, Feb 12 2007

%C The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - _Philippe Deléham_, Feb 26 2007

%C Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - _Philippe Deléham_, Feb 26 2007

%C Number of standard tableaux of shape (n+k,n-k). - _Philippe Deléham_, Mar 22 2007

%C From _Philippe Deléham_, Mar 30 2007: (Start)

%C This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):

%C (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970

%C (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;

%C (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;

%C (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;

%C (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;

%C (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)

%C The table U(n,k)=Sum_{j, 0<=j<=n} T(n,j)*k^j is given in A098474. - _Philippe Deléham_, Mar 29 2007

%C Sequence read mod 2 gives A127872. - _Philippe Deléham_, Apr 12 2007

%C Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - _Philippe Deléham_, Apr 16 2007, Apr 17 2007, Apr 18 2007

%C Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - _Philippe Deléham_, Jun 19 2007

%C Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2)= o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - _Philippe Deléham_, Nov 16 2009

%C The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - _Johannes W. Meijer_, Apr 20 2011

%C 4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - _Gary W. Adamson_, Jun 13 2011

%C The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2,... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - _Gary W. Adamson_, Sep 21 2011

%C Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - _Jayanta Basu_, Apr 30 2013

%C From _Wolfdieter Lang_, Sep 20 2013: (Start)

%C T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):

%C rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.

%C For the odd powers of rho(n) see A039598. (End)

%C Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - _Tom Copeland_, May 26 2016

%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.

%D T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

%H T. D. Noe, <a href="/A039599/b039599.txt">Rows n=0..50 of triangle, flattened</a>

%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

%H Quang T. Bach, Jeffrey B. Remmel, <a href="http://arxiv.org/abs/1510.04319">Generating functions for descents over permutations which avoid sets of consecutive patterns</a>, arXiv:1510.04319 [math.CO], 2015 (see p.25).

%H M. Barnabei, F. Bonetti and M. Silimbani, <a href="http://arxiv.org/abs/1301.1790">Two permutation classes enumerated by the central binomial coefficients</a>, arXiv preprint arXiv:1301.1790 [math.CO], 2013 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Silimbani/silimbani3.html">J. Int. Seq. 16 (2013) #13.3.8</a>

%H Paul Barry, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Barry/barry84.html">A Catalan Transform and Related Transformations on Integer Sequences</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

%H P. Barry, A. Hennessy, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barry2/barry94r.html">The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences</a>, J. Int. Seq. 13 (2010) # 10.8.2, example 15.

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Barry5/barry223.html">On the Hurwitz Transform of Sequences</a>, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.

%H P. Barry, <a href="http://arxiv.org/abs/1311.7161">Comparing two matrices of generalized moments defined by continued fraction expansions</a>, arXiv preprint arXiv:1311.7161 [math.CO], 2013 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Barry3/barry291.html">J. Int. Seq. 17 (2014) # 14.5.1</a>.

%H Jonathan E. Beagley, Paul Drube, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i2p44">Combinatorics of Tableau Inversions</a>, Electron. J. Combin., 22 (2015), #P2.44.

%H S. Chakravarty and Y. Kodama, <a href="http://arxiv.org/abs/0802.0524v2">A generating function for the N-soliton solutions of the Kadomtsev-Petviashvili II equation</a>, arXiv preprint arXiv:0802.0524v2 [nlin.SI], 2008.

%H Wun-Seng Chou, Tian-Xiao He, Peter J.-S. Shiue, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/He/he61.html">On the Primality of the Generalized Fuss-Catalan Numbers</a>, J. Int. Seqs., Vol. 21 (2018), #18.2.1.

%H Johann Cigler, <a href="https://arxiv.org/abs/1611.05252">Some elementary observations on Narayana polynomials and related topics</a>, arXiv:1611.05252 [math.CO], 2016. See p. 11.

%H Paul Drube, <a href="http://arxiv.org/abs/1606.04869">Generating Functions for Inverted Semistandard Young Tableaux and Generalized Ballot Numbers</a>, arXiv:1606.04869 [math.CO], 2016.

%H T.-X. He, L. W. Shapiro, <a href="http://dx.doi.org/10.1016/j.laa.2017.06.025">Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group</a>, Lin. Alg. Applic. 532 (2017) 25-41, example p 32.

%H Aoife Hennessy, <a href="http://repository.wit.ie/1693/1/AoifeThesis.pdf">A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths</a>, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

%H Thomas Koshy, <a href="http://jointmathematicsmeetings.org/meetings/national/jmm/1046-z1-620.pdf">Lobb's generalization of Catalan's parenthesization problem</a>, The College Mathematics Journal 40 (2), March 2009, 99-107.

%H Huyile Liang, Jeffrey Remmel, Sainan Zheng, <a href="https://arxiv.org/abs/1710.05795">Stieltjes moment sequences of polynomials</a>, arXiv:1710.05795 [math.CO], 2017, see page 11.

%H Andrew Lobb, <a href="http://www.jstor.org/stable/3618696">Deriving the n-th Catalan number</a>, Mathematical Gazette 83 (8), March 1999, 109-110.

%H Donatella Merlini and Renzo Sprugnoli, <a href="https://doi.org/10.1016/j.disc.2016.08.017">Arithmetic into geometric progressions through Riordan arrays</a>, Discrete Mathematics 340.2 (2017): 160-174. See page 161.

%H Pedro J. Miana, Hideyuki Ohtsuka, Natalia Romero, <a href="http://arxiv.org/abs/1602.04347">Sums of powers of Catalan triangle numbers</a>, arXiv:1602.04347 [math.NT], 2016 (see 2.8).

%H A. Papoulis, <a href="/A000108/a000108_8.pdf">A new method of inversion of the Laplace transform</a>, Quart. Appl. Math 14 (1957), 405-414. [Annotated scan of selected pages]

%H Athanasios Papoulis, <a href="http://www.jstor.org/stable/43636019">A new method of inversion of the Laplace transform</a>, Quart. Appl. Math 14.405-414 (1957): 124. [NB: there is a typo]

%H Yidong Sun and Fei Ma, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i1p33">Some new binomial sums related to the Catalan triangle</a>, Electronic Journal of Combinatorics 21(1) (2014), #P1.33

%H Yidong Sun and Fei Ma, <a href="http://arxiv.org/abs/1305.2017">Four transformations on the Catalan triangle</a>, arXiv preprint arXiv:1305.2017 [math.CO], 2013.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lobb_number">Lobb number</a>.

%H W.-J. Woan, L. Shapiro and D. G. Rogers, <a href="http://www.jstor.org/stable/2974473">The Catalan numbers, the Lebesgue integral and 4^{n-2}</a>, Amer. Math. Monthly, 104 (1997), 926-931.

%H Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, <a href="http://dx.doi.org/10.1016/j.disc.2017.07.006">Some matrix identities on colored Motzkin paths</a>, Discrete Mathematics 340.12 (2017): 3081-3091.

%F T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.

%F From _Emeric Deutsch_, May 06 2006: (Start)

%F T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).

%F G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)

%F The following formulas were added by _Philippe Deléham_ during 2003 to 2009: (Start)

%F Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.

%F T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.

%F T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n) T(n-j, k-1)*A000108(j).

%F T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).

%F G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.

%F T(0, 0) = 1, T(n, k) = 0 if n<0 or n<k; T(n, 0) = T(n-1, 0) + T(n-1, 1); for k>=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).

%F a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.

%F T(n, k) = A050165(n, n-k).

%F Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).

%F Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).

%F Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.

%F Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n .

%F T(n, k)*(-2)^(n-k) = A114193(n, k).

%F Sum_{k>=h} T(n,k) = binomial(2n,n-h).

%F Sum_{k=0..n} T(n,k)*5^k = A127628(n) .

%F Sum_{k=0..n} T(n,k)*7^k = A115970(n) .

%F T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).

%F Sum_{k=0..n} T(n,k)*6^k = A126694(n).

%F Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).

%F Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).

%F Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).

%F Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).

%F T(2*n,n) = A126596(n).

%F Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.

%F Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).

%F T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).

%F T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k}.

%F Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).

%F T(n,k) = A053121(2*n,2*k).

%F Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).

%F T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).

%F Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).

%F Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).

%F Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.

%F Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).

%F Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).

%F Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).

%F Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).

%F Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.

%F Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.

%F Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).

%F Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively .

%F Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).

%F Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).

%F T(n,k) + T(n,k+1) = A039598(n,k).

%F T(n,k) = A128899(n,k)+A128899(n,k+1).

%F Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.

%F Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.

%F Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.

%F Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).

%F Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).

%F Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).

%F Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).

%F Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).

%F Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.

%F T(n,2*k)+T(n,2*k+1) = A118919(n,k).

%F Sum_{k=0..j} T(n,k) = A050157(n,j).

%F Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).

%F Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).

%F Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).

%F Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).

%F Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).

%F Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).

%F Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).

%F Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).

%F (End)

%F T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - _Paul Barry_, Feb 17 2011

%F Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - _Philippe Deléham_, Feb 01 2014

%F Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - _Werner Schulte_, Jul 22 2015

%F Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - _Werner Schulte_, Jul 22 2015

%F Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - _Werner Schulte_, Dec 03 2015

%F T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - _Peter Luschny_, May 13 2016

%e Triangle T(n, k) begins:

%e n\k 0 1 2 3 4 5 6 7 8 9

%e 0: 1

%e 1: 1 1

%e 2: 2 3 1

%e 3: 5 9 5 1

%e 4: 14 28 20 7 1

%e 5: 42 90 75 35 9 1

%e 6: 132 297 275 154 54 11 1

%e 7: 429 1001 1001 637 273 77 13 1

%e 8: 1430 3432 3640 2548 1260 440 104 15 1

%e 9: 4862 11934 13260 9996 5508 2244 663 135 17 1

%e ... Reformatted by _Wolfdieter Lang_, Dec 21 2015

%e From _Paul Barry_, Feb 17 2011: (Start)

%e Production matrix begins

%e 1, 1,

%e 1, 2, 1,

%e 0, 1, 2, 1,

%e 0, 0, 1, 2, 1,

%e 0, 0, 0, 1, 2, 1,

%e 0, 0, 0, 0, 1, 2, 1,

%e 0, 0, 0, 0, 0, 1, 2, 1 (End)

%e From _Wolfdieter Lang_, Sep 20 2013: (Start)

%e Example for rho(N) = 2*cos(Pi/N) powers:

%e n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =

%e 2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)

%p T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # _Emeric Deutsch_, May 06 2006

%t Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* _Geoffrey Critzer_, Dec 18 2011 *)

%t Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* _Harvey P. Dale_, Dec 18 2011 *)

%t Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* _Jayanta Basu_, Apr 30 2013 *)

%o (Sage) # Algorithm of L. Seidel (1877)

%o # Prints the first n rows of the triangle

%o def A039599_triangle(n) :

%o D = [0]*(n+2); D[1] = 1

%o b = True ; h = 1

%o for i in range(2*n-1) :

%o if b :

%o for k in range(h,0,-1) : D[k] += D[k-1]

%o h += 1

%o else :

%o for k in range(1,h, 1) : D[k] += D[k+1]

%o if b : print [D[z] for z in (1..h-1)]

%o b = not b

%o A039599_triangle(10) # _Peter Luschny_, May 01 2012

%o (MAGMA) /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Oct 16 2015

%o (PARI) a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)

%o trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))

%o trianglerows(10) \\ _Felix Fröhlich_, Jun 24 2016

%Y Diagonals give: A000108 A000245 A000344 A000588 A001392 A000589 A000590, A000012 A005408 A014107(n>1) Row sums: A000984

%Y Cf. A008313 A039598 A084938 A000007

%Y Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

%Y Cf. A067311.

%K nonn,tabl,easy,nice

%O 0,4

%A _N. J. A. Sloane_

%E Corrected by _Philippe Deléham_, Nov 26 2009, Dec 14 2009

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Last modified August 21 22:52 EDT 2018. Contains 313958 sequences. (Running on oeis4.)