Displaying 1-10 of 11 results found.
Antidiagonal sums of triangle A104684.
+20
4
1, 2, 7, 26, 101, 404, 1645, 6784, 28243, 118442, 499601, 2117366, 9008969, 38458644, 164643197, 706574780, 3038800419, 13093784762, 56513880913, 244283771986, 1057348164103, 4582148496448, 19879232544027, 86331108851932, 375262802895691, 1632570339730086, 7108008200622949
FORMULA
a(n) = Sum_{k=0..floor(n/2)} A104684(n-k, k).
MATHEMATICA
nterms=30; Table[Sum[Binomial[r=n-k, k]Binomial[2r-k, r], {k, 0, Floor[n/2]}], {n, 0, nterms-1}] (* Paolo Xausa, Nov 26 2021 *)
PROG
(PARI) T(n, k) = binomial(n, k)*binomial(2*n-k, n); \\ A104684a(n) = sum(k=0, n\2, T(n-k, k));
The simplest sequence of positive numbers: the all 1's sequence.
(Formerly M0003)
+10
2531
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
COMMENTS
Number of ways of writing n as a product of primes.
Number of ways of writing n as a sum of distinct powers of 2.
Continued fraction for golden ratio A001622. An example of an infinite sequence of positive integers whose distinct pairwise concatenations are all primes! - Don Reble, Apr 17 2005 For n >= 0, let M(n) be the matrix with first row = (n n+1) and 2nd row = (n+1 n+2). Then a(n) = absolute value of det(M(n)). - K.V.Iyer, Apr 11 2009 a(n) is also tau_1(n) where tau_2(n) is A000005. a(n) is a completely multiplicative arithmetical function.
a(n) is both squarefree and a perfect square. See A005117 and A000290. (End) a(n) is also the number of complete graphs on n nodes. - Pablo Chavez (pchavez(AT)cmu.edu), Sep 15 2009
Totally multiplicative sequence with a(p) = 1 for prime p. Totally multiplicative sequence with a(p) = a(p-1) for prime p. - Jaroslav Krizek, Oct 18 2009 n-th prime minus phi(prime(n)); number of divisors of n-th prime minus number of perfect partitions of n-th prime; the number of perfect partitions of n-th prime number; the number of perfect partitions of n-th noncomposite number. - Juri-Stepan Gerasimov, Oct 26 2009 For all n>0, the sequence of limit values for a(n) = n!*Sum_{k>=n} k/(k+1)!. Also, a(n) = n^0. - Harlan J. Brothers, Nov 01 2009 a(n) is also the number of 0-regular graphs on n vertices. - Jason Kimberley, Nov 07 2009 1) When sequence is read as a regular triangular array, T(n,k) is the coefficient of the k-th power in the expansion of (x^(n+1)-1)/(x-1).
2) Sequence can also be read as a uninomial array with rows of length 1, analogous to arrays of binomial, trinomial, etc., coefficients. In a q-nomial array, T(n,k) is the coefficient of the k-th power in the expansion of ((x^q -1)/(x-1))^n, and row n has a sum of q^n and a length of (q-1)*n + 1. (End)
The number of maximal self-avoiding walks from the NW to SW corners of a 2 X n grid.
a(n) = A007310(n+1) (Modd 3) := A193680( A007310(n+1)), n>=0. For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the three residue classes Modd 3, called [0], [1], and [2], are shown in the array A088520, if there the third row is taken as class [0] after inclusion of 0. - Wolfdieter Lang, Feb 09 2012 Let M = Pascal's triangle without 1's ( A014410) and V = a variant of the Bernoulli numbers A027641 but starting [1/2, 1/6, 0, -1/30, ...]. Then M*V = [1, 1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012 As a lower triangular array, T is an example of the fundamental generalized factorial matrices of A133314. Multiplying each n-th diagonal by t^n gives M(t) = I/(I-t*S) = I + t*S + (t*S)^2 + ... where S is the shift operator A129184, and T = M(1). The inverse of M(t) is obtained by multiplying the first subdiagonal of T by -t and the other subdiagonals by zero, so A167374 is the inverse of T. Multiplying by t^n/n! gives exp(t*S) with inverse exp(-t*S). - Tom Copeland, Nov 10 2012 The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole. According to that historical definition, the length of one degree of latitude, that is, 60 nautical miles, would be exactly 111111.111... meters. - Jean-François Alcover, Jun 02 2013 Consider n >= 1 nonintersecting spheres each with surface area S. Define point p on sphere S_i to be a "public point" if and only if there exists a point q on sphere S_j, j != i, such that line segment pq INTERSECT S_i = {p} and pq INTERSECT S_j = {q}; otherwise, p is a "private point". The total surface area composed of exactly all private points on all n spheres is a(n)*S = S. ("The Private Planets Problem" in Zeitz.) - Rick L. Shepherd, May 29 2014 A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016 a(n) is also the determinant of the (n+1) X (n+1) matrix M defined by M(i,j) = binomial(i,j) for 0 <= i,j <= n, since M is a lower triangular matrix with main diagonal all 1's. - Jianing Song, Jul 17 2018 a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j) for 1 <= i,j <= n (see Xavier Merlin reference). - Bernard Schott, Dec 05 2018 a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = tau(gcd(i,j)) for 1 <= i,j <= n (see De Koninck & Mercier reference). - Bernard Schott, Dec 08 2020
REFERENCES
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 692 pp. 90 and 297, Ellipses, Paris, 2004.
Xavier Merlin, Méthodix Algèbre, Exercice 1-a), page 153, Ellipses, Paris, 1995.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
Paul Zeitz, The Art and Craft of Mathematical Problem Solving, The Great Courses, The Teaching Company, 2010 (DVDs and Course Guidebook, Lecture 6: "Pictures, Recasting, and Points of View", pp. 32-34).
FORMULA
a(n) = 1.
G.f.: 1/(1-x).
E.g.f.: exp(x).
G.f.: Product_{k>=0} (1 + x^(2^k)). - Zak Seidov, Apr 06 2007 Completely multiplicative with a(p^e) = 1.
Regarded as a square array by antidiagonals, g.f. 1/((1-x)(1-y)), e.g.f. Sum T(n,m) x^n/n! y^m/m! = e^{x+y}, e.g.f. Sum T(n,m) x^n y^m/m! = e^y/(1-x). Regarded as a triangular array, g.f. 1/((1-x)(1-xy)), e.g.f. Sum T(n,m) x^n y^m/m! = e^{xy}/(1-x). - Franklin T. Adams-Watters, Feb 06 2006 a(n) = Sum_{l=1..n} (-1)^(l+1)*2*cos(Pi*l/(2*n+1)) = 1 identically in n >= 1 (for n=0 one has 0 from the undefined sum). From the Jolley reference, (429) p. 80. Interpretation: consider the n segments between x=0 and the n positive zeros of the Chebyshev polynomials S(2*n, x) (see A049310). Then the sum of the lengths of every other segment starting with the one ending in the largest zero (going from the right to the left) is 1. - Wolfdieter Lang, Sep 01 2016
EXAMPLE
1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...)))) = A001622. 1/9 = 0.11111111111111...
Modd 7 for nonnegative odd numbers not divisible by 3:
A007310: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
Modd 3: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
(End)
MATHEMATICA
Array[1 &, 50] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
PROG
(Magma) [1 : n in [0..100]];
(PARI) {a(n) = 1};
(Haskell)
a000012 = const 1
(Maxima) makelist(1, n, 1, 30); /* Martin Ettl, Nov 07 2012 */
CROSSREFS
Cf. A000004, A007395, A010701, A000027, A027641, A014410, A211216, A212393, A060544, A051801, A104684.
De Bruijn's S(3,n): (3n)!/(n!)^3.
(Formerly M4284)
+10
101
1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
COMMENTS
Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011 a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013 a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013 a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016 Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016 At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0<x<4/27), generate this sequence via the period-energy function T(x) = 2*Pi*2F1(1/3,2/3; 1; (27/4)*x). - Bradley Klee, Feb 25 2018 The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018
REFERENCES
L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
FORMULA
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001 O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005 G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010 D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012 0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(+36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2 for all n in Z. - Michael Somos, Oct 22 2014 0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016 a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894. This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End) F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018 a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459. a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023 a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
The following hold for n >= 1:
a(n) = Sum_{k = 0.. 2*n} (-1)^(n+k) * k/n * binomial(2*n, k)^3 = 3/2 * Sum_{k = 0.. 2*n} (-1)^(n+k) * (k/n)^2 * binomial(2*n, k)^3.
a(n) = 3/2 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)^2*binomial(2*n-1, k).
a(n) = 3 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)*binomial(2*n-1, k)^2. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(2*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024
EXAMPLE
G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
MATHEMATICA
Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *) CoefficientList[Series[Hypergeometric2F1[1/3, 2/3, 1, 27*x], {x, 0, 5}], x] (* Bradley Klee, Feb 28 2018 *)
PROG
(PARI) {a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */ (PARI) {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */ (Magma) [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011 (Maxima) makelist(multinomial_coeff(n, n, n), n, 0, 24); /* Emanuele Munarini, Oct 25 2016 */ (GAP) List([0..20], n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018 (Python)
from math import factorial
EXTENSIONS
Terms a(17) and beyond from T. D. Noe, Jun 29 2008
T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.
+10
64
1, 1, 2, 1, 6, 6, 1, 12, 30, 20, 1, 20, 90, 140, 70, 1, 30, 210, 560, 630, 252, 1, 42, 420, 1680, 3150, 2772, 924, 1, 56, 756, 4200, 11550, 16632, 12012, 3432, 1, 72, 1260, 9240, 34650, 72072, 84084, 51480, 12870, 1, 90, 1980, 18480, 90090, 252252, 420420, 411840, 218790, 48620
COMMENTS
T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003 T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004 Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005 Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.
Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n. An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes. The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End) T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021 The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022 Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023
REFERENCES
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 366.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, Table I, p. 92.
D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.
FORMULA
T(n, k) = (n+k)!/(k!^2*(n-k)!) = T(n-1, k)*(n+k)/(n-k) = T(n, k-1)*(n+k)*(n-k+1)/k^2 = T(n-1, k-1)*(n+k)*(n+k-1)/k^2.
binomial(x, n)^2 = Sum_{k>=0} T(n,k) * binomial(x, n+k). - Michael Somos, May 11 2012 G.f.: G(t, z) = 1/sqrt(1-2*z-4*t*z+z^2). Row generating polynomials = P_n(1+2z), i.e., T(n, k) = [z^k] P_n(1+2*z), where P_n are the Legendre polynomials. - Emeric Deutsch, Apr 20 2004 1 + z*d/dz(log(G(t,z)) = 1 + (1 + 2*t)*z + (1 + 8*t + 8*t^2)*z^2 + ... is the o.g.f. for a signed version of A127674. - Peter Bala, Sep 02 2015 If R(n,t) denotes the n-th row polynomial then x^3 * exp( Sum_{n >= 1} R(n,t)*x^n/n ) = x^3 + (1 + 2*t)*x^4 + (1 + 5*t + 5*t^2)*x^5 + (1 + 9*t + 21*t^2 + 14*t^3)*x^6 + ... is an o.g.f for A033282. - Peter Bala, Oct 19 2015 P(n,x) := 1/(1 + x)*Integral_{t = 0..x} R(n,t) dt are (modulo differences of offset) the row polynomials of A033282. - Peter Bala, Jun 23 2016 R(n,x) = Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)*x^k.
R(n,x) = Sum_{k = 0..n} binomial(n,k)^2*x^k*(1 + x)^(n-k).
n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x).
R(n,x) = (-1)^n*R(n,-1 - x).
R(n,x) = 1/n! * (d/dx)^n ((x^2 + x)^n). (End)
The row polynomials are R(n,x) = hypergeom([-n, n + 1], [1], -x). - Peter Luschny, Mar 09 2018 If we let A(n,k) = (-1)^(n+k)*(2*k+1)*(n*(n-1)*...*(n-(k-1)))/((n+1)*...*(n+(k+1))) for n >= 0 and k = 0..n, and we consider both T(n,k) and A(n,k) as infinite lower triangular arrays, then they are inverses of one another. (Empty products are by definition 1.) See the example below. The rational numbers |A(n,k)| appear in Table II on p. 92 in Ser's (1933) book. - Petros Hadjicostas, Jul 11 2020 Row polynomial R(n,x) = Sum_{k >= n} binomial(k,n)^2 * x^(k-n)/(1+x)^(k+1) for x > -1/2.
R(n,x) = 1/(1 + x)^(n+1) * hypergeom([n+1, n+1], [1], x/(1 + x)).
R(n,x) = (1 + x)^n * hypergeom([-n, -n], [1], x/(1 + x)).
R(n,x) = hypergeom([(n+1)/2, -n/2], [1], -4*x*(1 + x)).
If we set R(-1,x) = 1, we can run the recurrence n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x) backwards to give R(-n,x) = R(n-1,x).
R(n,x) = [t^n] ( (1 + t)*(1 + x*(1 + t)) )^n. (End)
n*T(n,k) = (2*n-1)*T(n-1,k) + (4*n-2)*T(n-1,k-1) - (n-1)*T(n-2,k). - Fabián Pereyra, Jun 30 2022 n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n, k) * x^k o (1 + x)^(n-k), where o denotes the black diamond product of power series as defined by Dukes and White (see Bala, Section 4.4, exercise 3).
Denote this triangle by T. Then T * transpose(T) = A143007, the square array of crystal ball sequences for the A_n X A_n lattices. Let S denote the triangle ((-1)^(n+k)*T(n, k))n,k >= 0, a signed version of this triangle. Then S^(-1) * T = A007318, Pascal's triangle; it appears that T * S^(-1) = A110098.
EXAMPLE
The triangle T(n, k) starts:
n\k 0 1 2 3 4 5 6 7
0: 1
1: 1 2
2: 1 6 6
3: 1 12 30 20
4: 1 20 90 140 70
5: 1 30 210 560 630 252
6: 1 42 420 1680 3150 2772 924
7: 1 56 756 4200 11550 16632 12012 3432
row n = 8: 1 72 1260 9240 34650 72072 84084 51480 12870,
row n = 9: 1 90 1980 18480 90090 252252 420420 411840 218790 48620,
row n = 10: 1 110 2970 34320 210210 756756 1681680 2333760 1969110 923780 184756.
Its inverse (from Table II, p. 92, in Ser's book) is
1;
-1/2, 1/2;
1/3, -1/2, 1/6;
-1/4, 9/20, -1/4, 1/20;
1/5, -2/5, 2/7, -1/10, 1/70;
-1/6, 5/14, -25/84, 5/36, -1/28, 1/252;
1/7, -9/28, 25/84, -1/6, 9/154, -1/84, 1/924;
... (End)
MAPLE
p := (n, x) -> orthopoly[P](n, 1+2*x): seq(seq(coeff(p(n, x), x, k), k=0..n), n=0..9);
MATHEMATICA
Flatten[Table[Binomial[n, k]Binomial[n + k, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Dec 24 2011 *) Table[CoefficientList[Hypergeometric2F1[-n, n + 1, 1, -x], x], {n, 0, 9}] // Flatten
PROG
(PARI) {T(n, k) = local(t); if( n<0, 0, t = (x + x^2)^n; for( k=1, n, t=t'); polcoeff(t, k) / n!)} /* Michael Somos, Dec 19 2002 */ (PARI) {T(n, k) = binomial(n, k) * binomial(n+k, k)} /* Michael Somos, Sep 22 2013 */ (PARI) {T(n, k) = if( k<0 || k>n, 0, (n+k)! / (k!^2 * (n-k)!))} /* Michael Somos, Sep 22 2013 */ (Haskell)
a063007 n k = a063007_tabl !! n !! k
a063007_row n = a063007_tabl !! n
a063007_tabl = zipWith (zipWith (*)) a007318_tabl a046899_tabl
(Magma) /* As triangle: */ [[Binomial(n, k)*Binomial(n+k, k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 03 2015
CROSSREFS
See A331430 for an essentially identical triangle, except with signed entries. The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692, A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Triangle (0 <= k <= n) read by rows: T(n, k) is the number of Schröder paths from (0,0) to (2n,0) having k peaks.
+10
25
1, 1, 1, 2, 3, 1, 5, 10, 6, 1, 14, 35, 30, 10, 1, 42, 126, 140, 70, 15, 1, 132, 462, 630, 420, 140, 21, 1, 429, 1716, 2772, 2310, 1050, 252, 28, 1, 1430, 6435, 12012, 12012, 6930, 2310, 420, 36, 1, 4862, 24310, 51480, 60060, 42042, 18018, 4620, 660, 45, 1, 16796
COMMENTS
The rows sum to A006318 (Schroeder numbers), the left column is A000108 (Catalan numbers); the next-to-left column is A001700, the alternating sum in each row but the first is 0. T(n,k) is the 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 peaks. Example: T(2,1)=3 because we have UU*DD, U*DH and HU*D, the peaks being shown by *. E.g., T(n,k) = binomial(n,k)*binomial(2n-k,n-1)/n for n>0. - Emeric Deutsch, Dec 06 2003 T(n,k) is also the number of rooted plane trees with maximal degree 3 and k vertices of degree 2 (a node may have at most 2 children, and there are exactly k nodes with 1 child). Equivalently, T(n,k) is the number of syntactically different expressions that can be formed that use a unary operation k times, a binary operation n-k times, and nothing else (sequence of operands is fixed). - Lars Hellstrom (Lars.Hellstrom(AT)residenset.net), Dec 08 2009
FORMULA
Triangle T(n, k) (0 <= k <= n) read by rows; given by [1, 1, 1, 1, 1, ...] DELTA [1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 12 2003 If C_n(x) is the g.f. of row n of the Narayana numbers ( A001263), C_n(x) = Sum_{k=1..n} binomial(n,k-1)*(binomial(n-1,k-1)/k) * x^k and T_n(x) is the g.f. of row n of T(n,k), then T_n(x) = C_n(x+1), or T(n,k) = [x^n]Sum_{k=1..n}( A001263(n,k)*(x+1)^k). - Mitch Harris, Jan 16 2007, Jan 31 2007 G.f.: (1 - t*y - sqrt((1-y*t)^2 - 4*y)) / 2.
T(n, k) = binomial(2n-k, n)*binomial(n, k)/(n-k+1). - Philippe Deléham, Dec 07 2003 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 = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Apr 01 2007 Sum_{k=0..n} T(n,k)*x^(n-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 G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-xy-x/(1-.... (continued fraction);
G.f.: 1/(1-(x+xy)/(1-x/(1-(x+xy)/(1-x/(1-(x+xy)/(1-.... (continued fraction). (End)
T(n,k) = [k<=n]*(Sum_{j=0..n} binomial(n,j)^2*binomial(j,k))/(n-k+1). - Paul Barry, May 28 2009 With F(x,t) = (1-(2+t)*x-sqrt(1-2*(2+t)*x+(t*x)^2))/(2*x) an o.g.f. (nulling the n=0 term) in x for the A060693 polynomials in t, G(x,t) = x/(1+t+(2+t)*x+x^2) is the compositional inverse in x.
Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+t+(2+t)*x+x^2)^2 / (1+t-x^2), the n-th A060693 polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n) x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/d) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). (End)
Rows of this entry are non-vanishing antidiagonals of A097610. See p. 14 of Agapito et al. for a bivariate generating function and its inverse. - Tom Copeland, Feb 03 2016 Sum_{k=0..n} (-1)^k*(1+x*(n-k))*T(n,k) = x + (1-x)* A000007(n). (End)
Conjecture: Sum_{k=0..n} (-1)^k*T(n,k)*(n+1-k)^2 = 1+n+n^2. - Werner Schulte, Jan 11 2017
EXAMPLE
Triangle begins:
00: [ 1]
01: [ 1, 1]
02: [ 2, 3, 1]
03: [ 5, 10, 6, 1]
04: [ 14, 35, 30, 10, 1]
05: [ 42, 126, 140, 70, 15, 1]
06: [ 132, 462, 630, 420, 140, 21, 1]
07: [ 429, 1716, 2772, 2310, 1050, 252, 28, 1]
08: [ 1430, 6435, 12012, 12012, 6930, 2310, 420, 36, 1]
09: [ 4862, 24310, 51480, 60060, 42042, 18018, 4620, 660, 45, 1]
10: [16796, 92378, 218790, 291720, 240240, 126126, 42042, 8580, 990, 55, 1]
...
MATHEMATICA
t[n_, k_] := Binomial[n, k]*Binomial[2 n - k, n]/(n - k + 1); Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]] (* Robert G. Wilson v, May 30 2011 *)
PROG
(PARI) T(n, k) = binomial(n, k)*binomial(2*n - k, n)/(n - k + 1);
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print); \\ Indranil Ghosh, Jul 28 2017 (Python)
from sympy import binomial
def T(n, k): return binomial(n, k) * binomial(2 * n - k, n) / (n - k + 1)
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Jul 28 2017
Number of (1,1)-steps in all Delannoy paths of length n.
+10
10
0, 1, 8, 57, 384, 2505, 16008, 100849, 628736, 3888657, 23900040, 146146473, 889928064, 5399971161, 32668236552, 197123362785, 1186790473728, 7131032334369, 42773183020296, 256161548120857, 1531966218561920, 9150330147133161, 54591847064667528, 325361790187810257
COMMENTS
A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E =(1,0), N = (0,1) and D = (1,1).
FORMULA
a(n) = Sum_{k=1..n} k*binomial(n, k)*binomial(2*n-k, n).
G.f.: x*(1-x)/(1-6*x+x^2)^(3/2).
D-finite with recurrence (n-1)*(2*n-3)*a(n) = 4*(3*n^2-6*n+2)*a(n-1) - (n-1)*(2*n-1)*a(n-2). - Vaclav Kotesovec, Oct 18 2012 a(n) ~ (3+2*sqrt(2))^n*sqrt(n)/(2^(7/4)*sqrt(Pi)). - Vaclav Kotesovec, Oct 18 2012 a(n) = n^2*hypergeom([-n+1, -n+1], [2], 2). - Peter Luschny, Jan 20 2020
EXAMPLE
a(2)=8 because in the 13 (= A001850(2)) Delannoy paths of length 2, namely, DD, DNE,DEN,NED,END,NDE,EDN,NENE,NEEN,ENNE,ENEN,NNEE and EENN, we have a total of eight D steps.
MAPLE
a := n -> add(k*binomial(n, k)*binomial(2*n-k, n), k=1..n): seq(a(n), n=0..24);
# Alternative:
a := n -> n^2*hypergeom([-n+1, -n+1], [2], 2):
MATHEMATICA
CoefficientList[Series[x*(1-x)/(1-6*x+x^2)^(3/2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 18 2012 *)
PROG
(PARI) for(n=0, 25, print1(sum(k=0, n, k*binomial(n, k)*binomial(2*n-k, n)), ", ")) \\ G. C. Greubel, Jan 31 2017 (Python)
from math import comb
def A108666(n): return sum(comb(n, k)**2*k<<k-1 for k in range(1, n+1)) if n else 0 # Chai Wah Wu, Mar 22 2023
Triangle read by rows: T(n, k) (0<=k<=2n) is the number of Delannoy paths of length n, having k steps.
+10
6
1, 0, 1, 2, 0, 0, 1, 6, 6, 0, 0, 0, 1, 12, 30, 20, 0, 0, 0, 0, 1, 20, 90, 140, 70, 0, 0, 0, 0, 0, 1, 30, 210, 560, 630, 252, 0, 0, 0, 0, 0, 0, 1, 42, 420, 1680, 3150, 2772, 924, 0, 0, 0, 0, 0, 0, 0, 1, 56, 756, 4200, 11550, 16632, 12012, 3432
COMMENTS
A Delannoy path of length n is a path from (0, 0) to (n, n), consisting of steps E = (1,0), N = (0,1) and D = (1,1).
Row n has 2*n+1 terms, the first n of which are 0.
Row sums are the central Delannoy numbers ( A001850). Column sums are the central trinomial coefficients ( A002426).
FORMULA
T(n, k) = binomial(n, 2*n-k) binomial(k, n).
G.f.: 1/sqrt((1 - t*z)^2 - 4*z*t^2).
T(n, 2*n) = binomial(2*n, n) ( A000984). Sum_{k=0..n} k*T(n, k) = A109984(n).
EXAMPLE
T(2, 3) = 6 because we have DNE, DEN, NED, END, NDE and EDN.
Triangle begins
1;
0,1,2;
0,0,1,6,6;
0,0,0,1,12,30,20;
...
MAPLE
T := (n, k)->binomial(n, 2*n-k)*binomial(k, n):
for n from 0 to 8 do seq(T(n, k), k=0..2*n) od; # yields sequence in triangular form
# Alternative:
gf := ((1 - x*y)^2 - 4*x^2*y)^(-1/2):
yser := series(gf, y, 12): ycoeff := n -> coeff(yser, y, n):
row := n -> seq(coeff(expand(ycoeff(n)), x, k), k=0..2*n):
PROG
(PARI) {T(n, k) = binomial(n, k-n) * binomial(k, n)} /* Michael Somos, Sep 22 2013 */ (Haskell)
a109983 n k = a109983_tabf !! n !! k
a109983_row n = a109983_tabf !! n
a109983_tabf = zipWith (++) (map (flip take (repeat 0)) [0..]) a063007_tabl
Expansion of (1+6*x)/(1-4*x)^(7/2).
+10
3
1, 20, 210, 1680, 11550, 72072, 420420, 2333760, 12471030, 64664600, 327202876, 1622493600, 7909656300, 38003792400, 180324117000, 846321189120, 3934071152550, 18132120329400, 82937661506700
FORMULA
a(n) = binomial(2n+3, n) * binomial(n+3, 3). - Paul Barry, May 02 2005 G.f.: G(0) where G(k) = 1 + 4*x*(k+1)*(4*k+5)/((2*k+1)^2 - x*(2*k+1)^2*(2*k+3)*(4*k+7)/(x*(2*k+3)*(4*k+7) + 2*(k+1)^2/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 12 2012
D-finite with recurrence: n*a(n) + 2*(n-11)*a(n-1) + 12*(-2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012
MATHEMATICA
Array[Binomial[2 # + 3, #]*Binomial[# + 3, 3] &, 19, 0] (* Michael De Vlieger, Aug 18 2021 *)
PROG
(Magma) [Binomial(2*n+3, n)*Binomial(n+3, 3): n in [0..20]]; // Vincenzo Librandi, Aug 20 2011
a(n) = binomial(2n+4,n)*binomial(n+4,4).
+10
1
1, 30, 420, 4200, 34650, 252252, 1681680, 10501920, 62355150, 355655300, 1963217256, 10546208400, 55367594100, 285028443000, 1442592936000, 7193730107520, 35406640372950, 172255143129300, 829376615067000
FORMULA
G.f.: (1+12x+6x^2)/(1-4x)^(9/2).
D-finite with recurrence n^2*a(n) -2*(n+2)*(2*n+3)*a(n-1)=0. - R. J. Mathar, Feb 20 2015
MATHEMATICA
Table[Binomial[2n+4, n]Binomial[n+4, 4], {n, 0, 20}] (* Harvey P. Dale, May 03 2019 *)
Beukers integral int(int( -log(x*y) / (1-x*y) * P_n(2*x-1) * P_n(2*y-1) ,x=0..1,y=0..1)) = (A(n) + B(n)*zeta(3)) / A003418(n)^3. This sequence gives negated values of A(n).
+10
1
0, 12, 1404, 750372, 137096340, 425299945236, 11144361386340, 104074481089949004, 23323094579273069340, 18031967628526215059268, 525443267415363230379732, 20671296686851400981142679500
COMMENTS
Values of B(n) are given in A171485. P_n(x) are the Legendre Polynomials (see A008316) defined by n!*P_n(x) = (d/dx)^n (x^n*(1-x)^n).
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