Displaying 1-10 of 15 results found.
Matrix inverse of A181543 (cubes of entries of Pascal's triangle).
+20
0
1, -1, 1, 7, -8, 1, -163, 189, -27, 1, 8983, -10432, 1512, -64, 1, -966751, 1122875, -163000, 7000, -125, 1, 179781181, -208818216, 30317625, -1304000, 23625, -216, 1, -53090086057, 61664945083, -8953081011, 385146125, -6988625, 64827, -343, 1, 23402291822743, -27182124061184, 3946556485312, -169776943616, 3081169000, -28625408, 153664, -512, 1
COMMENTS
Is the leftmost column derived from A212856?
FORMULA
Sum_{j=k..n} A181543(n,j)*T(j,k) = delta(n,k).
EXAMPLE
1;
-1,1;
7,-8,1;
-163,189,-27,1;
8983,-10432,1512,-64,1;
-966751,1122875,-163000,7000,-125,1;
179781181,-208818216,30317625,-1304000,23625,-216,1;
The Franel number a(n) = Sum_{k = 0..n} binomial(n,k)^3.
(Formerly M1971 N0781)
+10
137
1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, 38165260, 278415920, 2046924400, 15148345760, 112738423360, 843126957056, 6332299624282, 47737325577620, 361077477684436, 2739270870994736, 20836827035351596, 158883473753259752, 1214171997616258240
COMMENTS
Cusick gives a general method of deriving recurrences for the r-th order Franel numbers (this is the sequence of third-order Franel numbers), with floor((r+3)/2) terms.
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004 An identity of V. Strehl states that a(n) = Sum_{k = 0..n} C(n,k)^2 * binomial(2*k,n). Zhi-Wei Sun conjectured that for every n = 2,3,... the polynomial f_n(x) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) * x^(n-k) is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013 Conjecture: a(n) == 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013 a(p) == 2 (mod p^3) for any prime p since p | C(p,k) for all k = 1,...,p-1. - Zhi-Wei Sun, Aug 14 2013 a(n) is the maximal number of totally mixed Nash equilibria in games of 3 players, each with n+1 pure options. - Raimundas Vidunas, Jan 22 2014 This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017 Diagonal of rational functions 1/(1 - x*y - y*z - x*z - 2*x*y*z), 1/(1 - x - y - z + 4*x*y*z), 1/(1 + y + z + x*y + y*z + x*z + 2*x*y*z), 1/(1 + x + y + z + 2*(x*y + y*z + x*z) + 4*x*y*z). - Gheorghe Coserea, Jul 04 2018 a(n) is the constant term in the expansion of ((1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019 Diagonal of rational function 1 / ((1-x)*(1-y)*(1-z) - x*y*z). - Seiichi Manyama, Jul 11 2020 Named after the Swiss mathematician Jérôme Franel (1859-1939). - Amiram Eldar, Jun 15 2021 It appears that a(n) is equal to the coefficient of (x*y*z)^n in the expansion of (1 + x + y - z)^n * (1 + x - y + z)^n * (1 - x + y + z)^n. Cf. A036917. - Peter Bala, Sep 20 2021
REFERENCES
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Jérôme Franel, On a question of Laisant, Intermédiaire des Mathématiciens, vol 1 1894 pp 45-47
H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 56.
Murray Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 193.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
P. Barrucand, A combinatorial identity, Problem 75-4, SIAM Rev., Vol. 17 (1975), p. 168. Solution by D. R. Breach, D. McCarthy, D. Monk and P. E. O'Neil, SIAM Rev., Vol. 18 (1976), p. 303. Juan Pla, Problem H-505, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 33, No. 5 (1995), p. 473; Sum Formulae!, Solution to Problem H-505 by Paul S. Bruckman, ibid., Vol. 35, No. 1 (1997), pp. 93-95. Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683v9 [math.CO] 2013; Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. the 6th China-Japan Sem. (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.
FORMULA
A002893(n) = Sum_{m = 0..n} binomial(n, m)*a(m) [Barrucand].
Sum_{k = 0..n} C(n, k)^3 = (-1)^n*Integral_{x = 0..infinity} L_k(x)^3 exp(-x) dx. - from Askey's book, p. 43
D-finite with recurrence (n + 1)^2*a(n+1) = (7*n^2 + 7*n + 2)*a(n) + 8*n^2*a(n-1) [Franel]. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jan 31 2001
a(n) ~ 2*3^(-1/2)*Pi^-1*n^-1*2^(3*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
O.g.f.: A(x) = Sum_{n >= 0} (3*n)!/n!^3 * x^(2*n)/(1 - 2*x)^(3*n+1). - Paul D. Hanna, Oct 30 2010 G.f.: hypergeom([1/3, 2/3], [1], 27 x^2 / (1 - 2x)^3) / (1 - 2x). - Michael Somos, Dec 17 2010 G.f.: Sum_{n >= 0} a(n)*x^n/n!^3 = [ Sum_{n >= 0} x^n/n!^3 ]^2. - Paul D. Hanna, Jan 19 2011 G.f.: A(x) = 1/(1-2*x)*(1+6*(x^2)/(G(0)-6*x^2)),
with G(k) = 3*(x^2)*(3*k+1)*(3*k+2) + ((1-2*x)^3)*((k+1)^2) - 3*(x^2)*((1-2*x)^3)*((k+1)^2)*(3*k+4)*(3*k+5)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011 In 2011 Zhi-Wei Sun found the formula Sum_{k = 0..n} C(2*k,n)*C(2*k,k)*C(2*(n-k),n-k) = (2^n)*a(n) and proved it via the Zeilberger algorithm. - Zhi-Wei Sun, Mar 20 2013 0 = a(n)*(a(n+1)*(-2048*a(n+2) - 3392*a(n+3) + 768*a(n+4)) + a(n+2)*(-1280*a(n+2) - 2912*a(n+3) + 744*a(n+4)) + a(n+3)*(+288*a(n+3) - 96*a(n+4))) + a(n+1)*(a(n+1)*(-704*a(n+2) - 1232*a(n+3) + 288*a(n+4)) + a(n+2)*(-560*a(n+2) - 1372*a(n+3) + 364*a(n+4)) + a(n+3)*(+154*a(n+3) - 53*a(n+4))) + a(n+2)*(a(n+2)*(+24*a(n+2) + 70*a(n+3) - 20*a(n+4)) + a(n+3)*(-11*a(n+3) + 4*a(n+4))) for all n in Z. - Michael Somos, Jul 16 2014 For r a nonnegative integer, Sum_{k = r..n} C(k,r)^3*C(n,k)^3 = C(n,r)^3*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016 a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^2. - Peter Luschny, May 31 2017 a(n) = Sum_{k=0..floor(n/2)} (n+k)!/(k!^3*(n-2*k)!) * 2^(n-2*k).
G.f. y=A(x) satisfies: 0 = x*(x + 1)*(8*x - 1)*y'' + (24*x^2 + 14*x - 1)*y' + 2*(4*x + 1)*y. (End)
a(n) = [x^n] (1 - x^2)^n*P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 56. - Peter Bala, Mar 24 2022 a(n) = (2^n/(4*Pi^2)) * Integral_{x,y=0..2*Pi} (1+cos(x)+cos(y)+cos(x+y))^n dx dy = (8^n/(Pi^2)) * Integral_{x,y=0..Pi} (cos(x)*cos(y)*cos(x+y))^n dx dy (Pla, 1995). - Amiram Eldar, Jul 16 2022 The g.f. T(x) obeys a period-annihilating ODE:
0=2*(1 + 4*x)*T(x) + (-1 + 14*x + 24*x^2)*T'(x) + x*(1 + x)*(-1 + 8*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (4/243)*(1 - 8*x + 240*x^2 - 464*x^3 + 16*x^4);
g3 = -(8/19683)*(1 - 12*x - 480*x^2 + 3080*x^3 - 12072*x^4 + 4128*x^5 +
64*x^6);
which determine an elliptic surface with four singular fibers. (End)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^2 * binomial(n-1, k). Cf. A361716. For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n + 1], [1, 1], -1). (End)
EXAMPLE
O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 56*x^3 + 346*x^4 + 2252*x^5 + ...
O.g.f.: A(x) = 1/(1-2*x) + 3!*x^2/(1-2*x)^4 + (6!/2!^3)*x^4/(1-2*x)^7 + (9!/3!^3)*x^6/(1-2*x)^10 + (12!/4!^3)*x^8/(1-2*x)^13 + ... - Paul D. Hanna, Oct 30 2010 Let g.f. A(x) = Sum_{n >= 0} a(n)*x^n/n!^3, then
A(x) = 1 + 2*x + 10*x^2/2!^3 + 56*x^3/3!^3 + 346*x^4/4!^3 + ... where
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 + ...]^2. - Paul D. Hanna
MAPLE
add(binomial(n, k)^3, k=0..n) ;
end proc:
A000172_list := proc(len) series(hypergeom([], [1, 1], x)^2, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
MATHEMATICA
Table[Sum[Binomial[n, k]^3, {k, 0, n}], {n, 0, 30}] (* Harvey P. Dale, Aug 24 2011 *) Table[ HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -1], {n, 0, 20}] (* Jean-François Alcover, Jul 16 2012, after symbolic sum *) a[n_] := Sum[ Binomial[2k, n]*Binomial[2k, k]*Binomial[2(n-k), n-k], {k, 0, n}]/2^n; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 20 2013, after Zhi-Wei Sun *) a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/3, 2/3, 1, 27 x^2 / (1 - 2 x)^3] / (1 - 2 x), {x, 0, n}]; (* Michael Somos, Jul 16 2014 *)
PROG
(PARI) {a(n)=polcoeff(sum(m=0, n, (3*m)!/m!^3*x^(2*m)/(1-2*x+x*O(x^n))^(3*m+1)), n)} \\ Paul D. Hanna, Oct 30 2010 (PARI) {a(n)=n!^3*polcoeff(sum(m=0, n, x^m/m!^3+x*O(x^n))^2, n)} \\ Paul D. Hanna, Jan 19 2011 (Haskell)
a000172 = sum . map a000578 . a007318_row
(Sage)
x, y, n = 1, 2, 1
while True:
yield x
n += 1
x, y = y, (8*(n-1)^2*x + (7*n^2-7*n + 2)*y) // n^2
(PARI) A000172(n)={sum(k=0, (n-1)\2, binomial(n, k)^3)*2+if(!bittest(n, 0), binomial(n, n\2)^3)} \\ M. F. Hasler, Sep 21 2015
CROSSREFS
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.) For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275- A291284 and A133370 respectively. Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Square the entries of Pascal's triangle.
+10
71
1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 36, 16, 1, 1, 25, 100, 100, 25, 1, 1, 36, 225, 400, 225, 36, 1, 1, 49, 441, 1225, 1225, 441, 49, 1, 1, 64, 784, 3136, 4900, 3136, 784, 64, 1, 1, 81, 1296, 7056, 15876, 15876, 7056, 1296, 81, 1, 1, 100, 2025, 14400, 44100, 63504, 44100, 14400, 2025, 100, 1
COMMENTS
Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1), having k right turns. - Emeric Deutsch, Nov 23 2003 Narayana numbers of type B. Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for associahedra of type B_n. See A001263 for the h-vectors for associahedra of type A_n. The Hilbert transform of this triangular array is A108625 (see A145905 for the definition of this term). 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 h-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the corresponding array of f-vectors for these type A_n polytopes. See A086645 for the array of h-vectors for type C_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes. (End)
The n-th row consists of the coefficients of the polynomial P_n(t) = Integral_{s = 0..2*Pi} (1 + t^2 - 2*t*cos(s))^n/Pi/2 ds. For example, when n = 3, we get P_3(t) = t^6 + 9*t^4 + 9*t^2 + 1; the coefficients are 1, 9, 9, 1. - Theodore Kolokolnikov, Oct 26 2010 Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013 Let s denote West's stack-sorting map. T(n,k) is the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 231, and 321. T(n,k) is also the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 312, and 321.
T(n,k) is the number of permutations of [n+1] with k descents that avoid the patterns 1342, 3142, 3412, and 3421. (End)
The number of convex polyominoes whose smallest bounding rectangle has size (k+1)*(n+1-k) and which contain the lower left corner of the bounding rectangle (directed convex polyominoes). - Günter Rote, Feb 27 2019 Let P be the poset [n] X [n] ordered by the product order. T(n,k) is the number of antichains in P containing exactly k elements. Cf. A063746. - Geoffrey Critzer, Mar 28 2020
REFERENCES
T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.
J. Riordan, An introduction to combinatorial analysis, Dover Publications, Mineola, NY, 2002, page 191, Problem 15. MR1949650
P. G. Tait, On the Linear Differential Equation of the Second Order, Proceedings of the Royal Society of Edinburgh, 9 (1876), 93-98 (see p. 97) [From Tom Copeland, Sep 09 2010, vol number corrected Sep 10 2010]
LINKS
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
FORMULA
E.g.f.: exp((1+y)*x)*BesselI(0, 2*sqrt(y)*x). - Vladeta Jovovic, Nov 17 2003 G.f.: 1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2); g.f. for row n: (1-t)^n P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been modified with the roles of x and y interchanged so that now x corresponds to n and y to k. - Petros Hadjicostas, Oct 22 2017] G.f. for column k is Sum_{j = 0..k} C(k, j)^2*x^(k+j)/(1 - x)^(2*k+1). - Paul Barry, Nov 15 2005 Column k has g.f. (x^k)*Legendre_P(k, (1+x)/(1-x))/(1 - x)^(k+1) = (x^k)*Sum_{j = 0..k} C(k, j)^2*x^j/(1 - x)^(2*k+1). - Paul Barry, Nov 19 2005 Let E be the operator D*x*D, where D denotes the derivative operator d/dx. Then (1/n!^2) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} binomial(n+k, k)^2*x^k. For example, when n = 3 we have (1/3!)^2*E^3(1/(1 - x)) = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7 = (1/3!)^2 * Sum_{k >= 0} ((k+1)*(k+2)*(k+3))^2*x^k. - Peter Bala, Oct 23 2008 G.f.: A(x, y) = Sum_{n >= 0} (2*n)!/n!^2 * x^(2*n)*y^n/(1 - x - x*y)^(2*n+1). - Paul D. Hanna, Oct 31 2010 Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Generating function: E(y)*E(x*y) = 1 + (1 + x)*y + (1 + 4*x + x^2)*y^2/2!^2 + (1 + 9*x + 9*x^2 + x^3)*y^3/3!^2 + .... Cf. the unsigned version of A021009 with generating function exp(y)*E(x*y). The n-th power of this array has the generating function E(y)^n*E(x*y). In particular, the matrix inverse A055133 has the generating function E(x*y)/E(y). (End) T(n,k) = T(n-1,k)*(n+k)/(n-k) + T(n-1,k-1), T(n,0) = T(n,n) = 1. - Vladimir Kruchinin, Oct 18 2014 Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives Pascal's triangle A007318. - Peter Bala, Dec 21 2014 n-th row polynomial R(n, x) = [z^n] (1 + (1 + x)*z + x*z^2)^n. Note that 1/n*[z^(n-1)] (1 + (1 + x)*z + x*z^2)^n gives the row polynomials of A001263. - Peter Bala, Jun 24 2015 Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2*(1 + t)*x + (1 - t)^2*x^2) denotes the o.g.f. of this array then 1 + x*d/dx log(G(x,t)) = 1 + (1 + t)*x + (1 + 6*t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter Bala, Sep 06 2015 T(n,k) = Sum_{i=0..n} C(n-i,k)*C(n,i)*C(n+i,i)*(-1)^(n-i-k). - Vladimir Kruchinin, Jan 14 2018 G.f. satisfies A(x,y) = x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2). - Vladimir Kruchinin, Oct 23 2020
EXAMPLE
Pascal's triangle begins
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...
so the present triangle begins
1
1 1
1 4 1
1 9 9 1
1 16 36 16 1
1 25 100 100 25 1
1 36 225 400 225 36 1
1 49 441 1225 1225 441 49 1
...
MAPLE
seq(seq(binomial(n, k)^2, k=0..n), n=0..10);
MATHEMATICA
Table[Binomial[n, k]^2, {n, 0, 11}, {k, 0, n}]//Flatten (* Alonso del Arte, Dec 08 2013 *)
PROG
(PARI) {T(n, k) = if( k<0 || k>n, 0, binomial(n, k)^2)}; /* Michael Somos, May 03 2004 */ (PARI) {T(n, k)=polcoeff(polcoeff(sum(m=0, n, (2*m)!/m!^2*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(2*m+1)), n, x), k, y)} \\ Paul D. Hanna, Oct 31 2010 (Maxima) create_list(binomial(n, k)^2, n, 0, 12, k, 0, n); \\ Emanuele Munarini, Mar 11 2011 (Maxima) T(n, k):=if n=k then 1 else if k=0 then 1 else T(n-1, k)*(n+k)/(n-k)+T(n-1, k-1); /* Vladimir Kruchinin, Oct 18 2014 */ (Magma) /* As triangle */ [[Binomial(n, k)^2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 15 2016 (GAP) Flat(List([0..10], n->List([0..n], k->Binomial(n, k)^2))); # Muniru A Asiru, Mar 30 2018 (Maxima)
A(x, y):=1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2);
taylor(x*A(x, y)+x*y*A(x, y)+sqrt(1+4*x^2*y*A(x, y)^2), x, 0, 7, y, 0, 7); /* Vladimir Kruchinin, Oct 23 2020 */
CROSSREFS
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558, A108625 (Hilbert transform), A145903, A181543, A086645 (logarithmic derivative), A105868 (inverse binomial transform), A093118.
Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3*x^k]*(1-x)^(3n+1), read by rows of k=0..2n terms.
+10
10
1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 54, 405, 760, 405, 54, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 490, 35623, 818300, 7917371, 37215794, 91789005, 123519792, 91789005, 37215794, 7917371, 818300, 35623, 490, 1, 1, 704, 73200, 2430400, 34657700, 246781248, 955910032, 2116980800, 2751843600, 2116980800, 955910032, 246781248, 34657700, 2430400, 73200, 704, 1
FORMULA
Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n). Using these coefficients we can obtain formulas for the sums
Sum_{i=1..n} C(e-1+i,e)^3. Let us define b(k,e,3) = sum_{i=0..k-e} (-1)^i*C(3*e+1,i)*C(k-i,e)^3, where k=e+i.
For example:
b(e,e,3) = 1;
b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);
b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);
b(e+3,e,3) = C(e+3,e)^3 - (3*e+1)*C(e+2,e)^3 + C(3*e+1,2)*C(e+1,e)^3 - C(3*e+1,3);
b(e+4,e,3) = C(e+4,e)^3 - (3*e+1)*C(e+3,e)^3 + C(3*e+1,2)*C(e+2,e) - C(3*e+1,3)*C(e+1,e)^3 + C(3*e+1,4).
Then we have the formula: Sum_{i=1..n} C(e-1+i,e)^3 = Sum_{i=0..2*e} b(e+i,e,3)*C(n+e+i,3*e+1).
Example: Sum_{i=1..7} C(2+i,3)^3 = C(10,10) + 54*C(11,10) + 405*C(12,10) + 760*C(13,10) + 405*C(14,10) + 54*C(15,10) + C(16,10) = 820260.
(End)
EXAMPLE
Triangle begins:
1;
1, 4, 1;
1, 20, 48, 20, 1;
1, 54, 405, 760, 405, 54, 1;
1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
Row g.f.s begin:
(1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
(1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
(1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
(1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...
MATHEMATICA
t[n_, k_] := SeriesCoefficient[Sum[Binomial[n+j, j]^3*x^j, {j, 0, n+k}]*(1-x)^(3*n+1), {x, 0, k}]; Table[t[n, k], {n, 0, 9}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Feb 04 2014, after PARI *)
PROG
(PARI) {T(n, k)=polcoeff(sum(j=0, n+k, binomial(n+j, j)^3*x^j)*(1-x)^(3*n+1), k)}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))
Triangle defined by g.f.: Sum_{n>=0} (4*n)!/n!^4 * x^(2*n)*y^n/(1-x-x*y)^(4*n+1), read by rows.
+10
3
1, 1, 1, 1, 26, 1, 1, 123, 123, 1, 1, 364, 3246, 364, 1, 1, 845, 25210, 25210, 845, 1, 1, 1686, 120135, 606500, 120135, 1686, 1, 1, 3031, 430941, 6082475, 6082475, 430941, 3031, 1, 1, 5048, 1277668, 38698856, 137915470, 38698856, 1277668, 5048, 1, 1, 7929
COMMENTS
Compare the g.f. of this triangle with the g.f.s of triangles:
* A008459: Sum_{n>=0} (2n)!/n!^2 * x^(2n)*y^n/(1-x-xy)^(2n+1), * A181543: Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-xy)^(3n+1),
EXAMPLE
G.f.: A(x,y) = 1/(1-x-xy) + 4!*x^2*y/(1-x-xy)^5 + (8!/2!^4)*x^4*y^2/(1-x-xy)^9 + (12!/3!^4)*x^6*y^3/(1-x-xy)^13 +...
Triangle begins:
1;
1, 1;
1, 26, 1;
1, 123, 123, 1;
1, 364, 3246, 364, 1;
1, 845, 25210, 25210, 845, 1;
1, 1686, 120135, 606500, 120135, 1686, 1;
1, 3031, 430941, 6082475, 6082475, 430941, 3031, 1;
1, 5048, 1277668, 38698856, 137915470, 38698856, 1277668, 5048, 1; ...
PROG
(PARI) {T(n, k)=polcoeff(polcoeff(sum(m=0, n, (4*m)!/m!^4*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(4*m+1)), n, x), k, y)}
-log( Sum_{n>=0} (-x)^n/n!^3 ) = Sum_{n>=1} a(n)*x^n/n!^3.
+10
3
1, 3, 46, 1899, 163476, 25333590, 6412369860, 2473269931755, 1379817056827720, 1069150908119474628, 1113779885682143602440, 1518901247410616194635510, 2651993653876241574715172280, 5817640695573490720735010689620
FORMULA
Equals column 0 of the matrix log of triangle T(n,k) = (-1)^(n-k)*C(n,k)^3.
a(n) = -(-1)^n + (1/n) * Sum_{k=1..n-1} (-1)^(n-k-1) * binomial(n,k)^3 * k * a(k). - Ilya Gutkovskiy, Jul 15 2021
EXAMPLE
L(x) = -log(1 - x + x^2/2!^3 - x^3/3!^3 + x^4/4!^3 - x^5/5!^3 +-...)
where
L(x) = x + 3*x^2/2!^3 + 46*x^3/3!^3 + 1899*x^4/4!^3 + 163476*x^5/5!^3 +...
ALTERNATE GENERATING METHOD.
A signed version of A181543(n,k) = C(n,k)^3 begins: 1;
1, 1;
1, 8, 1;
1, 27, 27, 1;
1, 64, 216, 64, 1;
1, 125, 1000, 1000, 125, 1; ...
The matrix log of triangle A181543 begins: 0;
1, 0;
-3, 8, 0;
46, -81, 27, 0;
-1899, 2944, -648, 64, 0;
163476, -237375, 46000, -3000, 125, 0; ...
in which this sequence (signed) is found in column 0.
PROG
(PARI) {a(n)=n!^3*polcoeff(-log(sum(m=0, n, (-x)^m/m!^3)+x*O(x^n)), n)}
(PARI) /* As Column 0 of the Matrix Log of signed Triangle A181543 */ {a(n)=local(L, M=matrix(n+1, n+1, r, c, if(r>=c, (-1)^(r-c)*binomial(r-1, c-1)^3)));
L=sum(n=1, #M, (M^0-M)^n/n); if(n<0, 0, L[n+1, 1])}
G.f. satisfies: A(x) = Sum_{n>=0} x^n*[Sum_{k=0..n} C(n,k)^3 *x^k* A(x)^k].
+10
2
1, 1, 2, 10, 39, 147, 639, 2857, 12725, 58081, 270250, 1268444, 6009439, 28736727, 138401100, 670641714, 3268021317, 16004012529, 78716657052, 388701645264, 1926266491659, 9576792342099, 47753368809171, 238759903786041
FORMULA
G.f. satisfies:
(1) A(x) = Sum_{n>=0} x^(2n)*A(x)^n*[Sum_{k>=0} C(n+k,k)^3*x^k].
(2) A(x) = Sum_{n>=0} (3n)!/n!^3 * x^(3n)*A(x)^n/(1-x-x^2*A(x))^(3n+1).
EXAMPLE
G.f.: A(x) = 1 + x + 2*x^2 + 10*x^3 + 39*x^4 + 147*x^5 + 639*x^6 +...
where g.f. A(x) satisfies:
* A(x) = 1 + x*(1 + x*A(x)) + x^2*(1 + 8*x*A(x) + x^2*A(x)^2) + x^3*(1 + 27*x*A(x) + 27*x^2*A(x)^2 + x^3*A(x)^3) + x^4*(1 + 64*x*A(x) + 216*x^2*A(x)^2 + 64*x^3*A(x)^3 + x^4*A(x)^4) +...;
PROG
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^3*x^k*(A+x*O(x^n))^k))); polcoeff(A, n)}
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n\2, x^(2*m)*(A+x*O(x^n))^m*sum(k=0, n, binomial(m+k, k)^3*x^k))); polcoeff(A, n)}
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n\3, (3*m)!/m!^3*x^(3*m)*A^m/(1-x-x^2*A+x*O(x^n))^(3*m+1))); polcoeff(A, n)}
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^3 * x^(2*k).
+10
2
1, 1, 1, 2, 9, 28, 66, 153, 433, 1345, 3952, 10991, 30954, 90988, 271845, 804153, 2361457, 6979690, 20842285, 62493914, 187274712, 561448399, 1688263179, 5093148285, 15393417178, 46570446829, 141063389488, 427979185898, 1300470246165, 3956367018001, 12048354848013, 36728336040306
COMMENTS
Limit a(n)/a(n+1) = 1 - t = t^3 = 0.3176721961... where t = ((sqrt(93)+9)/18)^(1/3) - ((sqrt(93)-9)/18)^(1/3).
FORMULA
G.f.: A(x) = Sum_{n>=0} (3*n)!/(n!)^3 * x^(4*n) / (1-x-x^3)^(3*n+1).
a(n) = Sum_{k=0..[n/3]} C(n-2*k,k)^3.
G.f.: A(x) = G( x^4/(1-x-x^3)^3 )/(1-x-x^3) where G(x) satisfies:
* G(x^3) = G( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x) and G(x) is the g.f. of A006480.
EXAMPLE
G.f. A(x) = 1 + x + x^2 + 2*x^3 + 9*x^4 + 28*x^5 + 66*x^6 + 153*x^7 +...
which equals the series:
A(x) = 1/(1-x-x^3) + 3!/1!^3*x^4/(1-x-x^3)^4 + 6!/2!^3*x^8/(1-x-x^3)^7 + 9!/3!^3*x^12/(1-x-x^3)^10 + 12!/4!^3*x^16/(1-x-x^3)^13 +...
The g.f. also equals the series:
A(x) = 1 +
x*(1 + x^2) +
x^2*(1 + 2^3*x^2 + x^4) +
x^3*(1 + 3^3*x^2 + 3^3*x^4 + x^6) +
x^4*(1 + 4^3*x^2 + 6^3*x^4 + 4^3*x^6 + x^8) +
x^5*(1 + 5^3*x^2 + 10^3*x^4 + 10^3*x^6 + 5^3*x^8 + x^10) +...
MATHEMATICA
Table[Sum[Binomial[n-2*k, k]^3, {k, 0, Floor[n/3]}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 15 2014 *)
PROG
(PARI) {a(n)=local(A=1); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^3*x^(2*k)) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))
(PARI) {a(n)=polcoeff(sum(m=0, n, x^(4*m)/(1-x-x^3 +x*O(x^n))^(3*m+1)*(3*m)!/(m!)^3), n)}
for(n=0, 40, print1(a(n), ", "))
(PARI) {a(n)=sum(k=0, n\3, binomial(n-2*k, k)^3)}
for(n=0, 40, print1(a(n), ", "))
T(n, k) = (m*n)!/(k!*(n-k)!)^m with m = 3; triangle read by rows, 0 <= k <= n.
+10
2
1, 6, 6, 90, 720, 90, 1680, 45360, 45360, 1680, 34650, 2217600, 7484400, 2217600, 34650, 756756, 94594500, 756756000, 756756000, 94594500, 756756, 17153136, 3705077376, 57891834000, 137225088000, 57891834000, 3705077376, 17153136
FORMULA
T(n, k) = ((3*n)!/(n!)^3) * binomial(n, k)^3 = A006480(n)* A181543(n, k).
EXAMPLE
Triangle starts:
[0] 1;
[1] 6, 6;
[2] 90, 720, 90;
[3] 1680, 45360, 45360, 1680;
[4] 34650, 2217600, 7484400, 2217600, 34650;
[5] 756756, 94594500, 756756000, 756756000, 94594500, 756756;
MAPLE
T := (n, k, m) -> (m*n)!/(k!*(n-k)!)^m:
seq(seq(T(n, k, 3), k=0..n), n=0..7);
MATHEMATICA
Table[((3*n)!/(n!)^3)*Binomial[n, k]^3, {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Oct 27 2018 *)
PROG
(PARI) t(n, k) = (3*n)!/(k!*(n-k)!)^3
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 6 rows of triangle as follows: */
(Magma) [[(Factorial(3*n)/(Factorial(n))^3)*Binomial(n, k)^3: k in [0..n]]: n in [0..15]]; // G. C. Greubel, Oct 27 2018 (GAP) Flat(List([0..6], n->List([0..n], k->Factorial(3*n)/(Factorial(k)*Factorial(n-k))^3))); # Muniru A Asiru, Oct 27 2018
G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3 * x^k*A(x)^(n-k)] * x^n/n ).
+10
1
1, 1, 3, 11, 42, 174, 763, 3457, 16075, 76351, 368767, 1805682, 8943948, 44736096, 225646033, 1146461185, 5862224756, 30144922281, 155791900727, 808773877919, 4215675455503, 22054576750972, 115765182718467, 609508331610920, 3218059655553030, 17034314889643633
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 42*x^4 + 174*x^5 + 763*x^6 +...
where the logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (A + x)*x + (A^2 + 2^3*x*A + x^2)*x^2/2 +
(A^3 + 3^3*x*A^2 + 3^3*x^2*A + x^3)*x^3/3 +
(A^4 + 4^3*x*A^3 + 6^3*x^2*A^2 + 4^3*x^3*A + x^4)*x^4/4 +
(A^5 + 5^3*x*A^4 + 10^3*x^2*A^3 + 10^3*x^3*A^2 + 5^3*x^4*A + x^5)*x^5/5 +
(A^6 + 6^3*x*A^5 + 15^3*x^2*A^4 + 20^3*x^3*A^3 + 15^3*x^4*A^2 + 6^3*x^5*A + x^6)*x^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 25*x^3/3 + 117*x^4/4 + 581*x^5/5 + 2987*x^6/6 + 15499*x^7/7 + 81213*x^8/8 +...
PROG
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^3*x^j/A^j)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
Search completed in 0.031 seconds
| 