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%I #144 Oct 07 2022 06:10:48
%S 1,2,1,3,4,1,4,10,6,1,5,20,21,8,1,6,35,56,36,10,1,7,56,126,120,55,12,
%T 1,8,84,252,330,220,78,14,1,9,120,462,792,715,364,105,16,1,10,165,792,
%U 1716,2002,1365,560,136,18,1,11,220,1287,3432,5005,4368,2380,816,171,20,1
%N Triangle read by rows: T(n, k) = binomial(n+k-1, 2*k-1).
%C Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.
%C Apart from signs, identical to A053122.
%C Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - _Philippe Deléham_, Feb 16 2004
%C T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - _Emeric Deutsch_, Apr 09 2005
%C T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - _Abdullahi Umar_, Oct 02 2008
%C This sequence is jointly generated with A085478 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = u(n-1,x) + x*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - _Clark Kimberling_, Feb 25 2012
%C Concerning Kimberling's recursion relations, see A102426. - _Tom Copeland_, Jan 19 2016
%C Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 27 2012
%C From _Wolfdieter Lang_, Aug 30 2012: (Start)
%C With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.
%C The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.
%C For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)
%C From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - _Tom Copeland_, Oct 11 2014
%C For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - _Milan Janjic_, Dec 20 2016
%C The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - _Robert B Fowler_, Feb 23 2022
%C The convolution triangle of the nonnegative integers. - _Peter Luschny_, Oct 07 2022
%H T. D. Noe, <a href="/A078812/b078812.txt">Rows n = 0..50 of triangle, flattened</a>
%H J. P. Allouche and M. Mendes France, <a href="http://arxiv.org/abs/1202.0211">Stern-Brocot polynomials and power series</a>, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html">Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays</a>, JIS 12 (2009) 09.8.6.
%H Paul Barry, <a href="http://dx.doi.org/10.1016/j.laa.2015.10.032">Riordan arrays, generalized Narayana triangles, and series reversion</a>, Linear Algebra and its Applications, 491 (2016) 343-385.
%H Paul Barry, <a href="https://arxiv.org/abs/2011.10827">Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers</a>, arXiv:2011.10827 [math.CO], 2020.
%H P. Damianou, <a href="http://arxiv.org/abs/1110.6620">On the characteristic polynomials of Cartan matrices and Chebyshev polynomials</a>, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
%H Nour-Eddine Fahssi, <a href="https://arxiv.org/abs/1808.00045">On the combinatorics of exclusion in Haldane fractional statistics</a>, arXiv:1808.00045 [cond-mat.stat-mech], 2018.
%H Milan Janjić, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Janjic/janjic93.html">Words and Linear Recurrences</a>, J. Int. Seq. 21 (2018), #18.1.4.
%H A. Laradji, and A. Umar, <a href="http://dx.doi.org/10.1007/s00233-005-0553-6">Combinatorial results for semigroups of order-preserving full transformations</a>, Semigroup Forum 72 (2006), 51-62.
%H Yidong Sun, <a href="http://www.fq.math.ca/Papers1/43-4/paper43-4-10b.pdf">Numerical triangles and several classical sequences</a>, Fib. Quart. 43, no. 4, (2005) 359-370.
%F G.f.: x*y / (1 - (2 + y)*x + x^2). To get row n, expand this in powers of x then expand the coefficient of x^n in increasing powers of y.
%F From _Philippe Deléham_, Feb 16 2004: (Start)
%F If indexing begins at 0 we have
%F T(n,k) = (n+k+1)!/((n-k)!*(2k+1))!.
%F T(n,k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k.
%F T(n,k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j) with T(n,k) = 0 if k < 0, T(0, 0)=1 and T(0, k) = 0 for k > 0.
%F G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2).
%F Row sums: Sum_{k>=0} T(n, k) = A001906(n+1). (End)
%F Antidiagonal sums are A000079(n) = Sum_{k=0..floor(n/2)} binomial(n+k+1, n-k). - _Paul Barry_, Jun 21 2004
%F Riordan array (1/(1-x)^2, x/(1-x)^2). - _Paul Barry_, Oct 22 2006
%F T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n, T(n,k) = T(n-1,k-1) + 2*T(n-1,k) - T(n-2,k). - _Philippe Deléham_, Jan 26 2010
%F For another version see A128908. - _Philippe Deléham_, Mar 27 2012
%F T(n,m) = Sum_{k=0..n-m} (binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)* binomial(n+1,m+k+1)). - _Vladimir Kruchinin_, Apr 13 2016
%e Triangle begins, 1 <= k <= n:
%e 1
%e 2 1
%e 3 4 1
%e 4 10 6 1
%e 5 20 21 8 1
%e 6 35 56 36 10 1
%e 7 56 126 120 55 12 1
%e 8 84 252 330 220 78 14 1
%p for n from 1 to 11 do seq(binomial(n+k-1,2*k-1),k=1..n) od; # yields sequence in triangular form; _Emeric Deutsch_, Apr 09 2005
%p # Uses function PMatrix from A357368. Adds a row and column above and to the left.
%p PMatrix(10, n -> n); # _Peter Luschny_, Oct 07 2022
%t (* First program *)
%t u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
%t u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
%t v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
%t Table[Expand[u[n, x]], {n, 1, z/2}]
%t Table[Expand[v[n, x]], {n, 1, z/2}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A085478 *)
%t Table[Expand[v[n, x]], {n, 1, z}]
%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t TableForm[cv]
%t Flatten[%] (* A078812 *) (* _Clark Kimberling_, Feb 25 2012 *)
%t (* Second program *)
%t Table[Binomial[n+k+1, 2*k+1], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Aug 01 2019 *)
%o (PARI) {T(n, k) = if( n<0, 0, binomial(n+k-1, 2*k-1))};
%o (PARI) {T(n, k) = polcoeff( polcoeff( x*y / (1 - (2 + y) * x + x^2) + x * O(x^n), n), k)};
%o (Haskell)
%o a078812 n k = a078812_tabl !! n !! k
%o a078812_row n = a078812_tabl !! n
%o a078812_tabl = [1] : [2, 1] : f [1] [2, 1] where
%o f us vs = ws : f vs ws where
%o ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
%o (us ++ [0, 0])
%o -- _Reinhard Zumkeller_, Dec 16 2013
%o (Sage)
%o @cached_function
%o def T(k,n):
%o if k==n: return 1
%o if k==0: return 0
%o return sum(i*T(k-1,n-i) for i in (1..n-k+1))
%o A078812 = lambda n,k: T(k,n)
%o [[A078812(n,k) for k in (1..n)] for n in (1..8)] # _Peter Luschny_, Mar 12 2016
%o (Sage) [[binomial(n+k+1, 2*k+1) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Aug 01 2019
%o (Maxima)
%o T(n,m):=sum(binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)*binomial(n+1,m+k+1),k,0,n-m); /* _Vladimir Kruchinin_, Apr 13 2016 */
%o (Magma) /* As triangle */ [[Binomial(n+k-1, 2*k-1): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Jun 01 2018
%o (GAP) Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k+1, 2*k+1) ))); # _G. C. Greubel_, Aug 01 2019
%Y This triangle is formed from odd-numbered rows of triangle A011973 read in reverse order.
%Y Row sums give A001906. With signs: A053122.
%Y The column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for k=1..6, resp. For k=7..24 they are A010966..(+2)..A011000 and for k=25..50 they are A017713..(+2)..A017763.
%Y Cf. A128908, A053123, A049310, A119900, A085478, A029653, A106195.
%K easy,nice,nonn,tabl
%O 0,2
%A _Michael Somos_, Dec 05 2002
%E Edited by _N. J. A. Sloane_, Apr 28 2008
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